Legendre's Condition/Lemma 1/Dependent on N Functions 0 55435 496503 470503 2020-10-25T13:44:39Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks}} == Lemma == Let $\mathbf y = \paren {\sequence {\map {y_i} x}_{1 \mathop \le i \mathop \le N} }$ be a vector real function, such that: :$\map {\mathbf y} a = A$ :$\map {\mathbf y} b = B$ Let $J \sqbrk {\mathbf y}$ be a functional, such that: :$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$ where: :$F \in C^2 \closedint a b$  {{WRT}} all its variables. Then: :$\ds \delta^2 J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$ where: :$\mathbf P = \dfrac 1 2 F_{y_i'y_j'}$ :$\mathbf Q = \dfrac 1 2 \paren {F_{ y_i y_j} - \dfrac \d {\d x} F_{y_i y_j'} }$ == Proof == {{ProofWanted}} == Sources == {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin|prev = Definition:Twice Differentiable/Functional/Dependent on N functions|next = Legendre's Condition/Lemma 2/Dependent on N Functions}}: $\S 5.29$: Generalization to n Unknown Functions Category:Calculus of Variations nom8ugjj6nbhrzoyzl65vluwk9memcs


Divisor Counting Function/Examples/1

Divisor Counting Function/Examples/1 0 55440 451771 292209 2020-03-02T09:43:28Z Prime.mover 59 Prime.mover moved page Tau Function/Examples/1 to Divisor Counting Function/Examples/1 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> The value of the $\tau$ function for the integer $1$ is $1$. </onlyinclude> == Proof == By definition, the $\tau$ function of an integer $n$ is the number of positive integer divisors of $n$. There is only one positive integer which is a divisor of $1$, and that is $1$ itself. Hence the result. {{qed}} Category:Tau Function p6llhtqkzlgph5ch80n8f8yooghmbhd


Sigma Function of 1

Sigma Function of 1 0 55443 379705 317546 2018-12-02T09:42:58Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma 1 = 1$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == By definition, the $\sigma$ function of an integer $n$ is the sum of the positive integer divisors of $n$. There is only one positive integer which is a divisor of $1$, and that is $1$ itself. Hence the result. {{qed}} Category:Sigma Function of Integer lqa21r4lnidrvam1iqgugfvslb3sys4


Parenthesization/Examples/4

Parenthesization/Examples/4 0 55471 392225 382884 2019-02-12T21:45:51Z Prime.mover 59 wikitext text/x-wiki == Example of Parenthesization == <onlyinclude> A word of $4$ elements can be parenthesized in $5$ distinct ways: :$\quad a_1 \paren {a_2 \paren {a_3 a_4} }$ :$\quad a_1 \paren {\paren {a_2 a_3} a_4}$ :$\quad \paren {a_1 a_2} \paren {a_3 a_4}$ :$\quad \paren {a_1 \paren {a_2 a_3} } a_4$ :$\quad \paren {\paren {a_1 a_2} a_3} a_4$ </onlyinclude> == Proof == From Number of Distinct Parenthesizations on Word, the number of distinct parenthesizations of a word $w$ of $n$ elements is the Catalan number $C_{n - 1}$: :$C_{n - 1} = \dfrac 1 n \dbinom {2 \paren {n - 1} } {n - 1}$ For $n = 4$ we have: {{begin-eqn}} {{eqn | l = C_4       | r = \dfrac 1 4 \dbinom {2 \times 3} 3       | c =  }} {{eqn | r = \dfrac 1 4 \times \dfrac {6!} {3! \times 3!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \dfrac 1 4 \times \dfrac {6 \times 5 \times 4} {3 \times 2 \times 1}       | c = {{Defof|Factorial}} }} {{eqn | r = 5       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Power Series Expansion for Exponential Function|next = Number of Distinct Parenthesizations on Word}}: $\text {3-4}$ Generating Functions * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Parenthesization/Examples/3|next = Number of Paths on Graph along X-axis using Diagonal Steps}}: $42$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Parenthesization/Examples/3|next = Number of Paths on Graph along X-axis using Diagonal Steps}}: $42$ Category:Parenthesization Category:Catalan Numbers Category:5 0zj4kbos3u8p9pg4ng0n991004dq51p


Complex Exponential is Uniformly Continuous on Half-Planes/Corollary 0 55581 454378 437840 2020-03-14T09:10:09Z Prime.mover 59 wikitext text/x-wiki == Corollary to Complex Exponential is Uniformly Continuous on Half-Planes == <onlyinclude> Let $X$ be a set. Let $\family {g_n}$ be a family of mappings $g_n : X \to \C$. Let $g_n$ converge uniformly to $g: X \to \C$. Let there be a constant $a \in \R$ such that $\map \Re {\map g x} \le a$ for all $x \in X$. Then $\exp g_n$ converges uniformly to $\exp g$. </onlyinclude> == Proof == By uniform convergence, there exists $N > 0$ such that $\cmod {\map {g_n} x - \map g x} \le 1$ for all $n > N$. Then $\map \Re {\map {g_n} x} \le a + 1$. The result now follows from: :Complex Exponential is Uniformly Continuous on Half-Planes, applied to the half-plane $\set {z \in \C : \map \Re z \le a + 1}$ :Uniformly Continuous Function Preserves Uniform Convergence {{qed}} == Sources == * {{BookReference|Functions of One Complex Variable|1973|John B. Conway|next = Equivalence of Definitions of Uniform Absolute Convergence of Product of Complex Functions}} $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Lemma $5.7$ Category:Exponential Function Category:Uniform Continuity 3aina9ra6ungqocjtrdi605lkz8q7ml


Quasiamicable Numbers/Examples/48,75

Quasiamicable Numbers/Examples/48,75 0 55614 461023 318187 2020-04-10T11:22:48Z Prime.mover 59 wikitext text/x-wiki == Examples of Quasiamicable Pair == <onlyinclude> $48$ and $75$ form a quasiamicable pair. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map \sigma {48}       | r = 124       | c = {{SigmaLink|48}} }} {{eqn | l = \map \sigma {75}       | r = 124       | c = {{SigmaLink|75}} }} {{eqn | l = 48 + 75 + 1       | r = 124       | c =  }} {{end-eqn}} Hence the result by definition of quasiamicable pair. {{qed}} Category:Quasiamicable Numbers Category:48 Category:75 m2x4gsxvwcsq82aaucf03d9ukeig1tr


Sigma Function of 140

Sigma Function of 140 0 55646 415146 317552 2019-07-29T12:31:00Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {140} = 336$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \map \sigma n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$140 = 2^2 \times 5 \times 7$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {140}       | r = \frac {2^3 - 1} {2 - 1} \times \frac {5^2 - 1} {5 - 1} \times \frac {7^2 - 1} {7 - 1}       | c =  }} {{eqn | r = \frac 7 1 \times \frac {6 \times 4} 4 \times \frac {8 \times 6} 6       | c =  }} {{eqn | r = 7 \times 6 \times 8       | c =  }} {{eqn | r = 7 \times \paren {2 \times 3} \times 2^3       | c =  }} {{eqn | r = 2^4 \times 3 \times 7       | c =  }} {{eqn | r = 336       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:140 hi2p6sdll9aixcvg1wj6hxm44t6lo32


Partial Products of Uniformly Convergent Product Converge Uniformly 0 55686 390030 375099 2019-01-26T23:41:40Z Leigh.Samphier 3031 wikitext text/x-wiki == Definition == Let $X$ be a set. Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field. Let $\left\langle{f_n}\right\rangle$ be a sequence of bounded mappings $f_n:X\to \mathbb K$. Let the product $\displaystyle \prod_{n \mathop = 1}^\infty f_n$ converge uniformly. Then the sequence of partial products converges uniformly. == Proof == Let $n_0\in\N$ be such that the sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty f_n$ converges uniformly. By Product of Bounded Functions is Bounded, $\displaystyle \prod_{n \mathop = 1}^{n_0-1}f_n$ is bounded. By Uniformly Convergent Sequence Multiplied with Function, the sequence of partial products converges uniformly. {{qed}} == Also see == * Uniform Product of Continuous Functions is Continuous * Infinite Product of Analytic Functions is Analytic Category:Uniform Convergence Category:Infinite Products hb2okugmrd5y4few6v4pdi5lygaqlxu


Logarithm of Infinite Product of Complex Functions/Corollary 0 55695 458325 340546 2020-03-29T14:19:39Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks}} == Corollary to Logarithm of Infinite Product of Complex Functions == <onlyinclude> Let $X$ be a locally compact and locally connected metric space. Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to \C$. Let the product $\displaystyle \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$. Let $x_0 \in X$. Then there exist $n_0 \in \N$, $k \in \Z$ and a neighborhood $U$ of $x_0$ such that: :$(1): \quad \map {f_n} x \ne 0$ for $n \ge n_0$ and $x \in U$ :$(2): \quad$ The series $\displaystyle \sum_{n \mathop = n_0}^\infty \ln f_n$ converges uniformly on $U$ to $\ln g + 2 k \pi i$, where $g = \displaystyle \prod_{n \mathop = n_0}^\infty f_n$. </onlyinclude> == Outline of Proof == We construct a neighborhood of $x_0$ on which $\displaystyle \sum_{n \mathop = n_0}^\infty \ln f_n$ and $\ln g$ are continuous, so that $k$ is continuous and thus constant. == Proof == Let $K$ be a compact neighborhood of $x_0$. By Tail of Uniformly Convergent Product Converges Uniformly to One, there exists $N \in \N$ such that $\displaystyle \prod_{n \mathop = N}^\infty \map {f_n} x \notin \R^-$ for all $x \in K$. By Factors in Uniformly Convergent Product Converge Uniformly to One, there exists $M \in \N$ such that $\size {\map {f_n} x - 1} \le \dfrac 1 2$ for $n \ge M$ and $x \in K$. Let $n_0 = \map \max {N, M}$. Let $g = \displaystyle \prod_{n \mathop = n_0}^\infty f_n$. By Logarithm of Infinite Product of Complex Functions, there exists $k: K \to \Z$ such that $\displaystyle \sum_{n \mathop = n_0}^\infty \ln f_n = \ln g + 2 k \pi i$ uniformly on $K$. We show that $k$ is constant on some neighborhood $U \subset K$. By Uniform Product of Continuous Functions is Continuous and Complex Logarithm is Continuous Outside Branch, $\ln g$ is continuous on $K$. By the Heine-Cantor Theorem, $\ln$ is uniformly continuous on $\map {\overline B} {1, \dfrac 1 2}$. By Uniformly Continuous Function Preserves Uniform Convergence, $\displaystyle \sum_{n \mathop = n_0}^\infty \ln f_n$ converges uniformly on $K$. {{questionable|we're using here that partial products belong to $\map {\overline B} {1, \dfrac 1 2}$. Need to use the Cauchy criterion for uniform products here}} By the Uniform Limit Theorem, $\displaystyle \sum_{n \mathop = n_0}^\infty \ln f_n$ is continuous. Because $\ln g$ and $\displaystyle \sum_{n \mathop = n_0}^\infty \ln f_n$ are continuous, so is $k$. Let $U\subset K$ be a connected neighborhood of $x_0$. By Continuous Mapping from Connected to Discrete Space is Constant $k$ is constant on $U$. {{qed}} Category:Infinite Products aua5s84b252zwf90pyleti5pwtb2l3v


Divisor Counting Function/Examples/60

Divisor Counting Function/Examples/60 0 56068 451785 319151 2020-03-02T09:43:29Z Prime.mover 59 Prime.mover moved page Tau Function/Examples/60 to Divisor Counting Function/Examples/60 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({60}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$60 = 2^2 \times 3 \times 5$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({60}\right)       | r = \tau \left({2^2 \times 3^1 \times 5^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $60$ can be enumerated as: :$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$ {{OEIS|A018266}}{{qed}} Category:Tau Function Category:60 sb03tbvnrfiqeamd4dl20bo0izexchv


Sigma Function of 104

Sigma Function of 104 0 56216 479684 317755 2020-07-24T19:45:17Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {104} = 210$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$104 = 2^3 \times 13$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {104}       | r = \frac {2^4 - 1} {2 - 1} \times \frac {13^2 - 1} {13 - 1}       | c = Sigma Function of Integer }} {{eqn | r = \frac {16 - 1} 1 \times \frac {169 - 1} {12}       | c =  }} {{eqn | r = 15 \times 14       | c =  }} {{eqn | r = 210       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:104 6cyvd2sqdk5b0s78odnr4n8pvjo8lkb


Sigma Function of 105

Sigma Function of 105 0 56217 442172 317715 2020-01-03T23:38:18Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\map \sigma {105} = 192$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$105 = 3 \times 5 \times 7$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {105}       | r = \paren {3 + 1} \paren {5 + 1} \paren {7 + 1}       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 4 \times 6 \times 8       | c =  }} {{eqn | r = 2^2 \times \paren {2 \times 3} \times 2^3       | c =  }} {{eqn | r = 2^6 \times 3       | c =  }} {{eqn | r = 192       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Square-Free Integer Category:105 bfc5pmjgfhijkmj2camptgunxtax9kh


Continued Fraction Expansion of Irrational Square Root/Examples/61 0 56247 470651 470637 2020-05-26T06:42:40Z Prime.mover 59 wikitext text/x-wiki == Examples of Continued Fraction Expansion of Irrational Square Root == <onlyinclude> The continued fraction expansion of the square root of $61$ is given by: :$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$ </onlyinclude> {{OEIS|A010145}} === Convergents === {{:Continued Fraction Expansion of Irrational Square Root/Examples/61/Convergents}} == Proof == Let $\sqrt {61} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$ From Partial Quotients of Continued Fraction Expansion of Irrational Square Root, the partial quotients of this continued fraction can be calculated as: :$a_r = \floor {\dfrac {\floor {\sqrt {61} } + P_r} {Q_r} }$ where: :$P_r = \begin {cases} 0 & : r = 0 \\ a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\ \end {cases}$ :$Q_r = \begin {cases} 1 & : r = 0 \\ \dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\ \end {cases}$ {{PartialQuotientCalculator-Start | n = 61}} {{PartialQuotientCalculator | n = 61 | r =  1 | ar-1 =  7 | Qr-1 =  1 | Pr-1 = 0}} {{PartialQuotientCalculator | n = 61 | r =  2 | ar-1 =  1 | Qr-1 = 12 | Pr-1 = 7}} {{PartialQuotientCalculator | n = 61 | r =  3 | ar-1 =  4 | Qr-1 =  3 | Pr-1 = 5}} {{PartialQuotientCalculator | n = 61 | r =  4 | ar-1 =  3 | Qr-1 =  4 | Pr-1 = 7}} {{PartialQuotientCalculator | n = 61 | r =  5 | ar-1 =  1 | Qr-1 =  9 | Pr-1 = 5}} {{PartialQuotientCalculator | n = 61 | r =  6 | ar-1 =  2 | Qr-1 =  5 | Pr-1 = 4}} {{PartialQuotientCalculator | n = 61 | r =  7 | ar-1 =  2 | Qr-1 =  5 | Pr-1 = 6}} {{PartialQuotientCalculator | n = 61 | r =  8 | ar-1 =  1 | Qr-1 =  9 | Pr-1 = 4}} {{PartialQuotientCalculator | n = 61 | r =  9 | ar-1 =  3 | Qr-1 =  4 | Pr-1 = 5}} {{PartialQuotientCalculator | n = 61 | r = 10 | ar-1 =  4 | Qr-1 =  3 | Pr-1 = 7}} {{PartialQuotientCalculator | n = 61 | r = 11 | ar-1 =  1 | Qr-1 = 12 | Pr-1 = 5}} |} and the cycle is complete: :$\sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}$ {{qed}} Category:Continued Fractions Category:61 7tgz1nya9cmwb8mbta2ho1dtiwyxxbm


Continued Fraction Expansion of Irrational Square Root/Examples/61/Convergents 0 56249 470652 470639 2020-05-26T06:43:41Z Prime.mover 59 wikitext text/x-wiki == Convergents to Continued Fraction Expansion of $\sqrt {61}$ == <onlyinclude> The sequence of convergents to the continued fraction expansion of the square root of $61$ begins: :$\dfrac 7 1, \dfrac {8} 1, \dfrac {39} 5, \dfrac {125} {16}, \dfrac {164} {21}, \dfrac {453} {58}, \dfrac {1070} {137}, \dfrac {1523} {195}, \dfrac {5639} {722}, \dfrac {24079} {3083}, \ldots$ </onlyinclude> {{OEIS-Numerators|A041106}} {{OEIS-Denominators|A041107}} == Proof == Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion. Let $\sequence {p_n}_{n \ge \mathop 0}$ and $\sequence {q_n}_{n \ge \mathop 0}$ be its numerators and denominators. Then the $n$th convergent is $p_n / q_n$. By definition: :$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$ :$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$ From Continued Fraction Expansion of $\sqrt {61}$: :$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$ Thus the convergents are assembled: {{ConvergentCalculator-Start | a0 = 7 | a1 = 1}} {{ConvergentCalculator | k =  2 | ak =  4 | pk-1 =          8 | pk-2 =          7 | qk-1 =         1 | qk-2 =         1}} {{ConvergentCalculator | k =  3 | ak =  3 | pk-1 =         39 | pk-2 =          8 | qk-1 =         5 | qk-2 =         1}} {{ConvergentCalculator | k =  4 | ak =  1 | pk-1 =        125 | pk-2 =         39 | qk-1 =        16 | qk-2 =         5}} {{ConvergentCalculator | k =  5 | ak =  2 | pk-1 =        164 | pk-2 =        125 | qk-1 =        21 | qk-2 =        16}} {{ConvergentCalculator | k =  6 | ak =  2 | pk-1 =        453 | pk-2 =        164 | qk-1 =        58 | qk-2 =        21}} {{ConvergentCalculator | k =  7 | ak =  1 | pk-1 =       1070 | pk-2 =        453 | qk-1 =       137 | qk-2 =        58}} {{ConvergentCalculator | k =  8 | ak =  3 | pk-1 =       1523 | pk-2 =       1070 | qk-1 =       195 | qk-2 =       137}} {{ConvergentCalculator | k =  9 | ak =  4 | pk-1 =       5639 | pk-2 =       1523 | qk-1 =       722 | qk-2 =       195}} {{ConvergentCalculator | k = 10 | ak =  1 | pk-1 =      24079 | pk-2 =       5639 | qk-1 =      3083 | qk-2 =       722}} {{ConvergentCalculator | k = 11 | ak = 14 | pk-1 =      29718 | pk-2 =      24079 | qk-1 =      3805 | qk-2 =      3083}} {{ConvergentCalculator | k = 12 | ak =  1 | pk-1 =     440131 | pk-2 =      29718 | qk-1 =     56353 | qk-2 =      3805}} {{ConvergentCalculator | k = 13 | ak =  4 | pk-1 =     469849 | pk-2 =     440131 | qk-1 =     60158 | qk-2 =     56353}} {{ConvergentCalculator | k = 14 | ak =  3 | pk-1 =    2319527 | pk-2 =     469849 | qk-1 =    296985 | qk-2 =     60158}} {{ConvergentCalculator | k = 15 | ak =  1 | pk-1 =    7428430 | pk-2 =    2319527 | qk-1 =    951113 | qk-2 =    296985}} {{ConvergentCalculator | k = 16 | ak =  2 | pk-1 =    9747957 | pk-2 =    7428430 | qk-1 =   1248098 | qk-2 =    951113}} {{ConvergentCalculator | k = 17 | ak =  2 | pk-1 =   26924344 | pk-2 =    9747957 | qk-1 =   3447309 | qk-2 =   1248098}} {{ConvergentCalculator | k = 18 | ak =  1 | pk-1 =   63596645 | pk-2 =   26924344 | qk-1 =   8142716 | qk-2 =   3447309}} {{ConvergentCalculator | k = 19 | ak =  3 | pk-1 =   90520989 | pk-2 =   63596645 | qk-1 =  11590025 | qk-2 =   8142716}} {{ConvergentCalculator | k = 20 | ak =  4 | pk-1 =  335159612 | pk-2 =   90520989 | qk-1 =  42912791 | qk-2 =  11590025}} {{ConvergentCalculator | k = 21 | ak =  1 | pk-1 = 1431159437 | pk-2 =  335159612 | qk-1 = 183241189 | qk-2 =  42912791}} |} {{qed}} Category:Continued Fractions Category:61 5f0pmtfjlgu37o58x5s1l6eti82bo01


Magic Square/Examples/Order 5

Magic Square/Examples/Order 5 0 56292 300654 300567 2017-06-11T11:02:42Z Prime.mover 59 Prime.mover moved page Magic Square/Example/Order 5 to Magic Square/Examples/Order 5 wikitext text/x-wiki == Examples of Order $5$ Magic Squares == <onlyinclude> :$\begin{array}{|c|c|c|c|c|} \hline 23 &  6 & 19 &  2 & 15 \\ \hline 10 & 18 &  1 & 14 & 22 \\ \hline 17 &  5 & 13 & 21 &  9 \\ \hline  4 & 12 & 25 &  8 & 16 \\ \hline 11 & 23 &  7 & 20 &  3 \\ \hline \end{array}$ </onlyinclude> == Also see == * Magic Constant of Order 5 Magic Square Category:Magic Squares ahx3qr6080nia7iuamy6fgpvaedp5ti


Numbers with Euler Phi Value of 72

Numbers with Euler Phi Value of 72 0 56365 475849 455827 2020-06-24T16:21:09Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler $\phi$ Function == <onlyinclude> There are $17$  positive integers for which the value of the Euler $\phi$ function is $72$: :$73, 91, 95, 111, 117, 135, 146, 148, 152, 182, 190, 216, 222, 228, 234, 252, 270$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = 72       | r = \map \phi {73}       | c = Euler Phi Function of Prime }} {{eqn | r = \map \phi {91}       | c = {{EulerPhiLink|91}} }} {{eqn | r = \map \phi {95}       | c = {{EulerPhiLink|95}} }} {{eqn | r = \map \phi {111}       | c = {{EulerPhiLink|111}} }} {{eqn | r = \map \phi {117}       | c = {{EulerPhiLink|117}} }} {{eqn | r = \map \phi {135}       | c = {{EulerPhiLink|135}} }} {{eqn | r = \map \phi {146}       | c = {{EulerPhiLink|146}} }} {{eqn | r = \map \phi {148}       | c = {{EulerPhiLink|148}} }} {{eqn | r = \map \phi {152}       | c = {{EulerPhiLink|152}} }} {{eqn | r = \map \phi {182}       | c = {{EulerPhiLink|182}} }} {{eqn | r = \map \phi {190}       | c = {{EulerPhiLink|190}} }} {{eqn | r = \map \phi {216}       | c = {{EulerPhiLink|216}} }} {{eqn | r = \map \phi {222}       | c = {{EulerPhiLink|222}} }} {{eqn | r = \map \phi {228}       | c = {{EulerPhiLink|228}} }} {{eqn | r = \map \phi {234}       | c = {{EulerPhiLink|234}} }} {{eqn | r = \map \phi {252}       | c = {{EulerPhiLink|252}} }} {{eqn | r = \map \phi {270}       | c = {{EulerPhiLink|270}} }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 4 Positive Integers in Arithmetic Sequence which have Same Euler Phi Value/Mistake|next = Positive Integers which are Euler Phi Value for 17 Integers}}: $72$ Category:Euler Phi Function Category:72 36cpgnjrma7zyvwd49i6mkrq7mpnlfy


Square of 1 Less than Number Base/Examples/5 0 56424 296925 296921 2017-05-10T13:33:26Z Prime.mover 59 wikitext text/x-wiki == Example of Square of 1 Less than Number Base == <onlyinclude> The square of $5$ is expressed in base $6$ as: :$5^2 = \left[{41}\right]_6$ </onlyinclude> == Proof == In base $10$: {{begin-eqn}} {{eqn | l = 5^2       | r = \left({6 - 1}\right)^2       | c =  }} {{eqn | r = 6^2 - 2 \times 6 + 1       | c =  }} {{eqn | r = 4 \times 6 + 1       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Square of 1 Less than Number Base etjz2uat44omvcyi8tz7rc2lkk253mj


Sigma Function of 836

Sigma Function of 836 0 56474 488970 318343 2020-09-18T22:38:27Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {836} = 1680$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$836 = 2^2 \times 11 \times 19$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {836}       | r = \frac {2^3 - 1} {2 - 1} \times \paren {11 + 1} \times \paren {19 + 1}       | c = Sigma Function of Integer }} {{eqn | r = \frac 7 1 \times 12 \times 20       | c =  }} {{eqn | r = 7 \times \paren {2^2 \times 3} \times \paren {2^2 \times 5}       | c =  }} {{eqn | r = 2^4 \times 3 \times 5 \times 7       | c =  }} {{eqn | r = 1680       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:836 r1t5ilql58pwkirxakqji1rijx8wigf


Euler Phi Function of 104

Euler Phi Function of 104 0 56574 415511 297622 2019-07-31T15:42:32Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler $\phi$ Function == <onlyinclude> :$\map \phi {104} = 48$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Integer: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ where $p \divides n$ denotes the primes which divide $n$. We have that: :$104 = 2^3 \times 13$ Thus: {{begin-eqn}} {{eqn | l = \map \phi {104}       | r = 104 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 {13} }       | c =  }} {{eqn | r = 104 \times \frac 1 2 \times \frac {12} {13}       | c =  }} {{eqn | r = 4 \times 1 \times 12       | c =  }} {{eqn | r = 48       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function Category:104 c5bntb3ll1mwk80jojr9g504forcbok


Euler Phi Function of 105

Euler Phi Function of 105 0 56575 394760 297623 2019-03-05T16:45:05Z Prime.mover 59 wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\map \phi {105} = 48$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Square-Free Integer: :$\displaystyle \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$ where $p \divides n$ denotes the primes which divide $n$. We have that: :$105 = 3 \times 5 \times 7$ and so is square-free. Thus: {{begin-eqn}} {{eqn | l = \map \phi {105}       | r = \paren {3 - 1} \paren {5 - 1} \paren {7 - 1}       | c =  }} {{eqn | r = 2 \times 4 \times 6       | c =  }} {{eqn | r = 48       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of Square-Free Integer Category:105 p9zwsy94v795fymhtmghu8rg3t14cai


Euler Phi Function of 15

Euler Phi Function of 15 0 56579 297640 2017-05-16T20:46:52Z Prime.mover 59 Created page with "== Example of Euler $\phi$ Function of Non-Square Semiprime == <onlyinclude> :$\phi \left({15}\right) = 8$ </onlyinclude> where..." wikitext text/x-wiki == Example of Euler $\phi$ Function of Non-Square Semiprime == <onlyinclude> :$\phi \left({15}\right) = 8$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == We have that: :$15 = 3 \times 5$ Thus: {{begin-eqn}} {{eqn | l = \phi \left({15}\right)       | r = \left({3 - 1}\right) \left({5 - 1}\right)       | c = Euler $\phi$ Function of Non-Square Semiprime }} {{eqn | r = 2 \times 4       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of Non-Square Semiprime Category:15 38qrkw3pc8084fyd3flsrr96rpcrmzl


Divisor Counting Function/Examples/105

Divisor Counting Function/Examples/105 0 56595 451773 438890 2020-03-02T09:43:28Z Prime.mover 59 Prime.mover moved page Tau Function/Examples/105 to Divisor Counting Function/Examples/105 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {105} = 8$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$105 = 3 \times 5 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {105}       | r = \map \tau {3^1 \times 5^1 \times 7^1}       | c =  }} {{eqn | r = \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} The divisors of $105$ can be enumerated as: :$1, 3, 5, 7, 15, 21, 35, 105$ {{OEIS|A018286}}{{qed}} Category:Tau Function Category:105 2htkwufti6wb7pp0xrnn69im8wdkrev


Euler Phi Function of 165

Euler Phi Function of 165 0 56604 297726 2017-05-17T20:19:14Z Prime.mover 59 Created page with "== Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\phi \left({165}\right) = 80$ </onlyinclude> where..." wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\phi \left({165}\right) = 80$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Square-Free Integer: :$\displaystyle \phi \left({n}\right) = \prod_{\substack {p \mathop \backslash n \\ p \mathop > 2} } \left({p - 1}\right)$ where $p \mathop \backslash n$ denotes the primes which divide $n$. We have that: :$165 = 3 \times 5 \times 11$ and so is square-free. Thus: {{begin-eqn}} {{eqn | l = \phi \left({165}\right)       | r = \left({3 - 1}\right) \left({5 - 1}\right) \left({11 - 1}\right)       | c =  }} {{eqn | r = 2 \times 4 \times 10       | c =  }} {{eqn | r = 80       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of Square-Free Integer Category:165 jiw6cuy5f1dkk50qucprga9b4gcmgmf


Euler Phi Function of 35

Euler Phi Function of 35 0 56612 440076 297741 2019-12-18T14:17:40Z Prime.mover 59 wikitext text/x-wiki == Example of Euler $\phi$ Function of Non-Square Semiprime == <onlyinclude> :$\map \phi {35} = 24$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == We have that: :$35 = 5 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \map \phi {35}       | r = \paren {5 - 1} \paren {7 - 1}       | c = Euler $\phi$ Function of Non-Square Semiprime }} {{eqn | r = 4 \times 6       | c =  }} {{eqn | r = 24       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of Non-Square Semiprime Category:35 mjjv5cjj5ypy7kqkvzap4kg15ce1q85


Tau Function of 35

Tau Function of 35 0 56613 297742 2017-05-17T20:54:37Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({35}\right) = 4$ </onlyinclude> where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({35}\right) = 4$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$35 = 5 \times 7$ Thus: :$\tau \left({35}\right) = \tau \left({5^1 \times 7^1}\right) = \left({1 + 1}\right) \left({1 + 1}\right) = 4$ The divisors of $35$ can be enumerated as: :$1, 5, 7, 35$ {{qed}} Category:Tau Function Category:35 148wl4amvfuqejw7csdysz46kyd7184


Prime Magic Square/Examples/Order 3/Smallest 0 56666 481551 299484 2020-08-13T10:19:40Z RandomUndergrad 3904 wikitext text/x-wiki == Example of Order $3$ Prime Magic Square == This order $3$ prime magic square has the smallest elements: <onlyinclude> :$\begin{array}{|c|c|c|} \hline 67 & 1 & 43 \\ \hline 13 & 37 & 61 \\ \hline 31 & 73 & 7 \\ \hline \end{array}$ </onlyinclude> == Proof == For the purpose of this magic square only, we consider $1$ as a prime. A simple parity argument can show that $2$ cannot be included in a prime magic square: If it is, the row containing $2$ sum to an even number, while a row not containing $2$ will sum to an odd number. {{Improve|I'm drawing a blank on how to present the following result clearly}} We aim to show that all elements of an order $3$ prime magic square has the same remainder when divided by $3$. There are two parts to this: === $3$ cannot be used === For simplicity, we denote the numbers in the cells by their remainders when divided by $3$. Note that $3$ is the only prime divisible by $3$. We define the off-diagonals as: :$\begin{array}{|c|c|c|} \hline * &  &  \\ \hline  & * &  \\ \hline  &  & * \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  & * &  \\ \hline  &  & * \\ \hline * &  &  \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  &  & * \\ \hline * &  &  \\ \hline  & * &  \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  &  & * \\ \hline  & * &  \\ \hline * &  &  \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  & * &  \\ \hline * &  &  \\ \hline  &  & * \\ \hline \end{array} \begin{array}{|c|c|c|} \hline * &  &  \\ \hline  &  & * \\ \hline  & * &  \\ \hline \end{array}$ We also observe that, by switching rows and columns, the numbers in each row and column remains unchanged, while the two diagonals become two off-diagonals sharing one cell. Therefore the position of the numbers do not matter in the most part. Suppose $3$ is used in the square. {{WLOG}} there are only two cases: ==== Case $1$: The row containing $3$ has numbers with all remainders ==== We have: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 2 \\ \hline  &  &  \\ \hline  &  &  \\ \hline \end{array}$ Hence the row sum is divisible by $3$. Since: :$1 + 1 \equiv 2 \pmod 3$ :$2 + 2 \equiv 1 \pmod 3$ :$1 + 2 \equiv 0 \pmod 3$ there is a unique way to fill in the columns: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 2 \\ \hline 1 & 1 & 2 \\ \hline 2 & 1 & 2 \\ \hline \end{array}$ Note that the order of $1$ and $2$ in the leftmost column do not matter due to symmetry. The sums of rows $2$ and $3$ are not divisible by $3$. Hence this case cannot occur. {{qed|lemma}} ==== Case $2$: The row containing $3$ leave out numbers with some remainder ==== {{WLOG}} suppose $2$ is not used. Then: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 1 \\ \hline  &  &  \\ \hline  &  &  \\ \hline \end{array}$ Filling in the columns: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 1 \\ \hline 1 & 2 & 2 \\ \hline 1 & 2 & 2 \\ \hline \end{array}$ All off-diagonals sum to $1$, which is not $1 + 1 = 2$. Hence this case cannot occur. {{qed|lemma}} === Primes of remainder $1, 2$ cannot be mixed === {{WLOG}} suppose there are $2$ $1$'s and $1$ $2$. Then the row sum is not divisible by $3$. We have: :$\begin{array}{|c|c|c|} \hline 1 & 1 & 2 \\ \hline  &  &  \\ \hline  &  &  \\ \hline \end{array}$ Filling in the first and third columns: :$\begin{array}{|c|c|c|} \hline 1 & 1 & 2 \\ \hline 1 &  & 1 \\ \hline 2 &  & 1 \\ \hline \end{array}$ Finally, filling up the rows: :$\begin{array}{|c|c|c|} \hline 1 & 1 & 2 \\ \hline 1 & 2 & 1 \\ \hline 2 & 1 & 1 \\ \hline \end{array}$ There must be an off-diagonal with sum divisible by $3$. Hence this case cannot occur. {{qed|lemma}} Using this result, we divide the primes $\le 73$ into two groups: :Remainder of $1$: $\set {1, 7, 13, 19, 31, 37, 43, 61, 67, 73}$ :Remainder of $2$: $\set {5, 11, 17, 23, 29, 41, 47, 53, 59, 71}$ We only need to show these primes cannot form a smaller magic square. Consider: :$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f \\ \hline g & h & i \\ \hline \end{array}$ Let $C$ be the magic constant. Then: {{begin-eqn}} {{eqn | l = 4C       | r = \paren {a + e + i} + \paren {b + e + h} + \paren {c + e + g} + \paren {d + e + f}       | c = These are all lines passing through the center }} {{eqn | r = \paren {a + b + c + d + e + f + g + h + i} + 3 e       | c = Center counted $4$ times }} {{eqn | r = 3 C + 3 e }} {{end-eqn}} Hence $e = \dfrac C 3$, which is $\dfrac 1 9$ of the sum of all numbers in the square. We have: :$1 + 7 + 13 + 19 + 31 + 37 + 43 + 61 + 67 + 73 = 352$ :$5 + 11 + 17 + 23 + 29 + 41 + 47 + 53 + 59 + 71 = 356$ $352$ and $356$ have remainders $1$ and $5$ when divided by $9$. In the lists: :$1, 19, 37, 73$ have a remainder of $1$ when divided by $9$. :$5, 23, 41, 59$ have a remainder of $5$ when divided by $9$. Omitting each number gives the corresponding center square values: :$39, 37, 35, 31$ for the first list :$39, 37, 35, 33$ for the second list Only $31, 37$ of the first list are possible candidates. However: :$73 + 31 > 93$ Hence $31$ fail to produce a magic square. This leaves $37$, which possibility is demonstrated above. {{qed}} == Also see == * Magic Constant of Smallest Prime Magic Square == Sources == * {{MathWorld|Prime Magic Square|PrimeMagicSquare}} Category:Prime Magic Squares rm8quqhjni9pd5p3v8lxkgilju1wocj


Tau Function of 120

Tau Function of 120 0 56751 311826 298498 2017-08-19T11:36:27Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({120}\right) = 16$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$120 = 2^3 \times 3 \times 5$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({120}\right)       | r = \tau \left({2^3 \times 3^1 \times 5^1}\right)       | c =  }} {{eqn | r = \left({3 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 16       | c =  }} {{end-eqn}} The divisors of $120$ can be enumerated as: :$1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120$ {{OEIS|A018293}}{{qed}} Category:Tau Function Category:120 b9jhzy1sjyd0lx0pivvzu4lw82nskgs


Tau Function of 6

Tau Function of 6 0 56966 304284 299213 2017-07-10T11:16:00Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({6}\right) = 4$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$6 = 2 \times 3$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({6}\right)       | r = \tau \left({2^1 \times 3^1}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} The divisors of $6$ can be enumerated as: :$1, 2, 3, 6$ {{qed}} Category:Tau Function Category:6 m4moxy6pt99zs65yhbvjapb73oboovj


Tau Function of 28

Tau Function of 28 0 56967 304291 299215 2017-07-10T13:15:30Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({28}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$28 = 2^2 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({28}\right)       | r = \tau \left({2^2 \times 7^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $28$ can be enumerated as: :$1, 2, 4, 7, 14, 28$ {{OEIS|A018254}}{{qed}} Category:Tau Function Category:28 kozulcl7ntshl4lk0ccw3yp6kdnjcfa


Tau Function of 140

Tau Function of 140 0 56970 300366 299221 2017-06-07T20:22:04Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({140}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$140 = 2^2 \times 5 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({140}\right)       | r = \tau \left({2^2 \times 5^1 \times 7^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $140$ can be enumerated as: :$1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140$ {{OEIS|A018301}}{{qed}} Category:Tau Function Category:140 03lobprrp437yfmyxrzusvctn5mqow1


Tau Function of 270

Tau Function of 270 0 56972 299223 2017-05-26T21:24:58Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({270}\right) = 16$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({270}\right) = 16$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$270 = 2 \times 3^3 \times 5$ Thus: :$\tau \left({270}\right) = \tau \left({2^1 \times 3^3 \times 5^1}\right) = \left({1 + 1}\right) \left({3 + 1}\right) \left({1 + 1}\right) = 16$ The divisors of $270$ can be enumerated as: :$1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270$ {{qed}} Category:Tau Function Category:270 53q4gcguc1el4g5mz3s3spxvtkbucc3


Sigma Function of 270

Sigma Function of 270 0 56973 318217 318215 2017-09-16T18:25:41Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({270}\right) = 720$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$270 = 2 \times 3^3 \times 5$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({270}\right)       | r = \left({2 + 1}\right) \times \frac {3^4 - 1} {3 - 1} \times \left({5 + 1}\right)       | c =  }} {{eqn | r = 3 \times \frac {81 - 1} 2 \times 6       | c =  }} {{eqn | r = 3 \times 40 \times 6       | c =  }} {{eqn | r = 3 \times \left({2^3 \times 5}\right) \times \left({2 \times 3}\right)       | c =  }} {{eqn | r = 2^4 \times 3^2 \times 5       | c =  }} {{eqn | r = 720       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:270 8f9ogza4vhxcmt0zwjuesjjyatmwm20


Tau Function of 496

Tau Function of 496 0 56975 304296 301321 2017-07-10T21:04:44Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({496}\right) = 10$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$496 = 2^4 \times 31$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({496}\right)       | r = \tau \left({2^4 \times 31^1}\right)       | c =  }} {{eqn | r = \left({4 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 10       | c =  }} {{end-eqn}} The divisors of $496$ can be enumerated as: :$1, 2, 4, 8, 16, 31, 62, 124, 248, 496$ {{OEIS|A018487}}{{qed}} Category:Tau Function Category:496 21pb8k2a3ndqsh3wik0lkwnekrrraa4


Sigma Function of 496

Sigma Function of 496 0 56976 317812 317810 2017-09-16T13:05:07Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({496}\right) = 992$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$496 = 2^4 \times 31$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({496}\right)       | r = \left({2^5 - 1}\right) \times \left({31 + 1}\right)       | c =  }} {{eqn | r = 31 \times 32       | c =  }} {{eqn | r = 31 \times 2^5       | c =  }} {{eqn | r = 992       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:496 nchj6b8510ltl0p3vdpo9zxxjul9xx3


Prime Magic Square/Examples/Order 12/Smallest with Consecutive Primes from 3 0 57043 462309 462295 2020-04-16T08:09:19Z RandomUndergrad 3904 wikitext text/x-wiki == Example of Order $12$ Prime Magic Square == This order $12$ prime magic square is the smallest whose elements are consecutive odd primes starting from $3$ (including $1$). The primes themselves are the $143$ consecutive odd primes from $3$ up to $827$. This magic square has magic constant $4514$. <onlyinclude> :$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline   1 & 823 & 821 & 809 & 811 & 797 &  19 &  29 & 313 &  31 &  23 &  37 \\ \hline  89 &  83 & 211 &  79 & 641 & 631 & 619 & 709 & 617 &  53 &  43 & 739 \\ \hline  97 & 227 & 103 & 107 & 193 & 557 & 719 & 727 & 607 & 139 & 757 & 281 \\ \hline 223 & 653 & 499 & 197 & 109 & 113 & 563 & 479 & 173 & 761 & 587 & 157 \\ \hline 367 & 379 & 521 & 383 & 241 & 467 & 257 & 263 & 269 & 167 & 601 & 599 \\ \hline 349 & 359 & 353 & 647 & 389 & 331 & 317 & 311 & 409 & 307 & 293 & 449 \\ \hline 503 & 523 & 233 & 337 & 547 & 397 & 421 &  17 & 401 & 271 & 431 & 433 \\ \hline 229 & 491 & 373 & 487 & 461 & 251 & 443 & 463 & 137 & 439 & 457 & 283 \\ \hline 509 & 199 &  73 & 541 & 347 & 191 & 181 & 569 & 577 & 571 & 163 & 593 \\ \hline 661 & 101 & 643 & 239 & 691 & 701 & 127 & 131 & 179 & 613 & 277 & 151 \\ \hline 659 & 673 & 677 & 683 &  71 &  67 &  61 &  47 &  59 & 743 & 733 &  41 \\ \hline 827 &   3 &   7 &   5 &  13 &  11 & 787 & 769 & 773 & 419 & 149 & 751 \\ \hline \end{array}$ </onlyinclude> == Proof == It is sufficient to show that for $n \le 11$, there is no order $n$ prime magic square. We will show this fact regardless of whether $1$ is included in the magic square. === Order $2$ === First the order $2$ magic square is eliminated. Consider: :$\begin{array}{|c|c|} \hline  a & b \\ \hline  c & d \\ \hline \end{array}$ Then we must have $a + b = a + c$. So $b = c$, so they are not distinct, so this array cannot be a magic square. {{qed|lemma}} Next, by definition of magic square, each row adds up to the magic constant. Hence the sum of all entries of the magic square of order $n$ must be divisible by $n$. Here is a list of: :$1 + $ the sums of the first $n^2 - 1$ odd primes :sums of the first $n^2$ odd primes :their divisibility by $n$: $\begin{array}{|c|c|c|} \hline    & \text{Including } 1 & \text{Divisible by } n? & \text{Not including } 1 & \text{Divisible by } n? \\ \hline  3 &    99 & \text{Yes} &   127 & \text{No} \\ \hline  4 &   380 & \text{Yes} &   438 & \text{No} \\ \hline  5 &  1059 & \text{No}  &  1159 & \text{No} \\ \hline  6 &  2426 & \text{No}  &  2582 & \text{No} \\ \hline  7 &  4887 & \text{No}  &  5115 & \text{No} \\ \hline  8 &  8892 & \text{No}  &  9204 & \text{No} \\ \hline  9 & 15115 & \text{No}  & 15535 & \text{No} \\ \hline 10 & 24132 & \text{No}  & 24678 & \text{No} \\ \hline 11 & 36887 & \text{No}  & 37559 & \text{No} \\ \hline \end{array}$ So the only potential magic squares are of order $3$ or $4$. These magic squares, if they exist, must have magic constants $33$ and $95$. === Order $3$ === The first $8$ primes are $3, 5, 7, 11, 13, 17, 19, 23$. Because every prime and $1$ appears exactly once in a magic square, each number contributes to at least $2$ sums: the row and column sums. However, there is only one way to express $32$ as a sum of $2$ primes less than $23$: :$32 = 19 + 13$ and so $33$ cannot be made from a sum that includes $1$ in $2$ distinct ways. Thus an order $3$ prime magic square cannot be made. === Order $4$ === Every row of an order $4$ magic square contains $4$ odd numbers. These sum to an even number. But the magic constant of an order $4$ prime magic square, as shown above, is $95$. Hence it is not possible to create an order $4$ prime magic square. {{qed|lemma}} Hence there can be no prime magic square whose order is less than $12$. Thus the order $12$ prime magic square is the smallest whose elements are consecutive odd primes starting from $3$ (including or not including $1$). {{qed}} == Also see == * Magic Constant of Smallest Prime Magic Square with Consecutive Primes from 3 == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square of Reversal of Small-Digit Number/Examples/12|next = Prime Magic Square/Examples/Order 12/Smallest with Consecutive Primes/Mistake}}: $144$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Square of Reversal of Small-Digit Number/Examples/12|next = Prime Magic Square/Examples/Order 12/Smallest with Consecutive Primes/Mistake}}: $144$ * {{MathWorld|Prime Magic Square|PrimeMagicSquare}} Category:Prime Magic Squares aritkhflesvmf45cqcy250otnkv5tim


Euler Phi Function of 164

Euler Phi Function of 164 0 57145 437843 299795 2019-12-05T15:22:47Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler $\phi$ Function == <onlyinclude> :$\map \phi {164} = 80$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Integer: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ where $p \divides n$ denotes the primes which divide $n$. We have that: :$164 = 2^2 \times 41$ Thus: {{begin-eqn}} {{eqn | l = \map \phi {164}       | r = 164 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 {41} }       | c =  }} {{eqn | r = 164 \times \frac 1 2 \times \frac {40} {41}       | c =  }} {{eqn | r = 2 \times 1 \times 40       | c =  }} {{eqn | r = 80       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function Category:164 p4uxlt8z13je88kx51h1ub82f2kob5w


Tau Function of 18

Tau Function of 18 0 57163 437769 299847 2019-12-04T16:52:45Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {18} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$18 = 2 \times 3^2$ Thus: :$\map \tau {18} = \map \tau {2^1 \times 3^2} = \paren {1 + 1} \paren {2 + 1} = 6$ The divisors of $18$ can be enumerated as: :$1, 2, 3, 6, 9, 18$ {{OEIS|A018251}}{{qed}} Category:Tau Function Category:18 6mn7ak5k9wqy7vn876hfo5jym5mrp6f


Tau Function of 27

Tau Function of 27 0 57164 299848 2017-06-01T17:38:30Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({27}\right) = 4$ </onlyinclude> where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({27}\right) = 4$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == {{begin-eqn}} {{eqn | l = \tau \left({27}\right)       | r = \tau \left({3^3}\right)       | c =  }} {{eqn | r = 3 + 1       | c = Tau of Power of Prime }} {{eqn | r = 4       | c =  }} {{end-eqn}} The divisors of $27$ can be enumerated as: :$1, 3, 9, 27$ {{qed}} Category:Tau Function Category:27 2wzd93gzsz3x4pam1hs2hjn0merpq37


Tau Function of 24

Tau Function of 24 0 57165 439339 299851 2019-12-12T16:51:03Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {24} = 8$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$24 = 2^3 \times 3$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {24}       | r = \map \tau {2^3 \times 3^1}       | c =  }} {{eqn | r = \paren {3 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} The divisors of $24$ can be enumerated as: :$1, 2, 3, 4, 6, 8, 12, 24$ {{OEIS|A018253}}{{qed}} Category:Tau Function Category:24 12jn4nnaall53zx6orjfk1t5ittt0va


Tau Function of 32

Tau Function of 32 0 57166 299850 2017-06-01T17:45:57Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({32}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({32}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == {{begin-eqn}} {{eqn | l = \tau \left({32}\right)       | r = \tau \left({2^5}\right)       | c =  }} {{eqn | r = 5 + 1       | c = Tau of Power of Prime }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $32$ can be enumerated as: :$1, 2, 4, 8, 16, 32$ {{qed}} Category:Tau Function Category:32 4pgdl9biib1b9f07pkgmtu3idyvrlv2


Tau Function of 56

Tau Function of 56 0 57167 478128 300358 2020-07-13T20:19:32Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {56} = 8$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$56 = 2^3 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {56}       | r = \map \tau {2^3 \times 7^1}       | c =  }} {{eqn | r = \paren {3 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} The divisors of $56$ can be enumerated as: :$1, 2, 4, 7, 8, 14, 28, 56$ {{OEIS|A018265}}{{qed}} Category:Tau Function Category:56 1ohrcsenhcfs5cro7d76r5byksnjw0r


Tau Function of 64

Tau Function of 64 0 57168 299853 2017-06-01T17:51:50Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({64}\right) = 7$ </onlyinclude> where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({64}\right) = 7$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == {{begin-eqn}} {{eqn | l = \tau \left({64}\right)       | r = \tau \left({2^6}\right)       | c =  }} {{eqn | r = 6 + 1       | c = Tau of Power of Prime }} {{eqn | r = 7       | c =  }} {{end-eqn}} The divisors of $64$ can be enumerated as: :$1, 2, 4, 8, 16, 32, 64$ {{qed}} Category:Tau Function Category:64 929fehhxkun7m814g7j21flfet4nklw


Euler Phi Function of 194

Euler Phi Function of 194 0 57256 300154 300152 2017-06-05T05:53:24Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler $\phi$ Function == <onlyinclude> :$\phi \left({194}\right) = 96$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == We have that: :$194 = 2 \times 97$ Thus: {{begin-eqn}} {{eqn | l = \phi \left({194}\right)       | r = 97 - 1       | c = Euler Phi Function of 2 times Odd Prime }} {{eqn | r = 96       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of 2 times Odd Prime Category:194 p556xb2odwsuha7fesumz8jcf821x1g


Euler Phi Function of 195

Euler Phi Function of 195 0 57257 442119 300153 2020-01-03T17:19:26Z Prime.mover 59 wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\map \phi {195} = 96$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Square-Free Integer: :$\displaystyle \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$ where $p \divides n$ denotes the primes which divide $n$. We have that: :$195 = 3 \times 5 \times 13$ and so is square-free. Thus: {{begin-eqn}} {{eqn | l = \map \phi {195}       | r = \paren {3 - 1} \paren {5 - 1} \paren {13 - 1}       | c =  }} {{eqn | r = 2 \times 4 \times 12       | c =  }} {{eqn | r = 96       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of Square-Free Integer Category:195 2rfjcc03p4hfc5wxk9g35vnqcr4w4qe


Tau Function of 70

Tau Function of 70 0 57331 419360 300361 2019-08-20T15:16:23Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {70} = 8$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$70 = 2 \times 5 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {70}       | r = \map \tau {2^1 \times 5^1 \times 7^1}       | c =  }} {{eqn | r = \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} The divisors of $70$ can be enumerated as: :$1, 2, 5, 7, 10, 14, 35, 70$ {{OEIS|A018270}}{{qed}} Category:Tau Function Category:70 srddfybjijycf4kl1td3oi05t3fmzmt


Amicable Pair/Examples/220-284

Amicable Pair/Examples/220-284 0 57457 495109 445192 2020-10-16T22:16:34Z Prime.mover 59 wikitext text/x-wiki == Example of Amicable Pair == <onlyinclude> $220$ and $284$ are the smallest amicable pair: :$\map \sigma {220} = \map \sigma {284} = 504 = 220 + 284$ </onlyinclude> == Proof == Let $\map s n$ denote the aliquot sum of $n$. By definition: :$\map s n = \map \sigma n - n$ where $\map \sigma n$ denotes the $\sigma$ function. Thus: {{begin-eqn}} {{eqn | l = \map s {220}       | r = \map \sigma {220} - 220       | c =  }} {{eqn | r = 504 - 220       | c = $\sigma$ of $220$ }} {{eqn | r = 284       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map s {284}       | r = \map \sigma {284} - 284       | c =  }} {{eqn | r = 504 - 284       | c = $\sigma$ of $284$ }} {{eqn | r = 220       | c =  }} {{end-eqn}} It can be determined by inspection of the aliquot sums of all smaller integers that there is no smaller amicable pair. {{qed}} == Historical Note == {{:Amicable Pair/Examples/220-284/Historical Note}} == Sources == * {{BookReference|History of the Theory of Numbers|1919|Leonard Eugene Dickson|volume = I|prev = Definition:Aliquot Part|next = Amicable Pair/Examples/220-284/Historical Note}}: Preface * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 230|next = 220}}: $220$ * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Definition:Amicable Pair/Definition 1|next = Symbols:Abbreviations/A/amp|entry = amicable numbers}} * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 230|next = 220}}: $220$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Amicable Pair/Historical Note|next = Mathematician:André-Marie Ampère|entry = amicable numbers}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Amicable Pair/Historical Note|next = Mathematician:André-Marie Ampère|entry = amicable numbers}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Amicable Pair/Definition 1|next = Amicable Pair/Examples/220-284/Historical Note|entry = amicable numbers}} * {{MathWorld|Amicable Pair|AmicablePair}} Category:Amicable Pairs 156m1wuwhpnymqw0qxn9gftolka1wxn


Sigma Function of 220

Sigma Function of 220 0 57461 454512 318160 2020-03-15T06:22:18Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {220} = 504$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \map \sigma n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$220 = 2^2 \times 5 \times 11$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {220}       | r = \frac {2^3 - 1} {2 - 1} \times \frac {5^2 - 1} {5 - 1} \times \frac {11^2 - 1} {11 - 1}       | c =  }} {{eqn | r = \frac 7 1 \times \frac {6 \times 4} 4 \times \frac {12 \times 10} {10}       | c =  }} {{eqn | r = 7 \times 6 \times 12       | c =  }} {{eqn | r = 7 \times \paren {2 \times 3} \times \paren {2^2 \times 3}       | c =  }} {{eqn | r = 2^3 \times 3^2 \times 7       | c =  }} {{eqn | r = 504       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:220 1vcippmcl9c5totyrn7swo390so3i5n


Sigma Function of 284

Sigma Function of 284 0 57462 428849 318226 2019-09-30T15:44:51Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {284} = 504$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \map \sigma n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$284 = 2^2 \times 71$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {284}       | r = \frac {2^3 - 1} {2 - 1} \times \paren {71 + 1}       | c =  }} {{eqn | r = 7 \times 72       | c =  }} {{eqn | r = 7 \times \paren {2^3 \times 3^2}       | c =  }} {{eqn | r = 2^3 \times 3^2 \times 7       | c =  }} {{eqn | r = 504       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:284 6rdlg5a1yonk3gmbtsgxoqqmutizo8w


Tau Function of 242

Tau Function of 242 0 57624 301348 301347 2017-06-16T06:38:54Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({242}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$242 = 2 \times 11^2$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({242}\right)       | r = \tau \left({2^1 \times 11^2}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({2 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $242$ can be enumerated as: :$1, 2, 11, 22, 121, 242$ {{OEIS|A018351}}{{qed}} Category:Tau Function Category:242 cw3x17ukjsvscw56toml4uol58puatc


Tau Function of 243

Tau Function of 243 0 57625 301350 301349 2017-06-16T06:40:14Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({243}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$243 = 3^5$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({243}\right)       | r = \tau \left({3^5}\right)       | c =  }} {{eqn | r = 5 + 1       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $243$ can be enumerated as: :$1, 3, 9, 27, 81, 243$ {{qed}} Category:Tau Function Category:243 gbe4y7w8ust9kft8qcmx5c47zsdlomu


Tau Function of 244

Tau Function of 244 0 57627 442166 301352 2020-01-03T23:24:31Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {244} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$244 = 2^2 \times 61$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {244}       | r = \map \tau {2^2 \times 61^1}       | c =  }} {{eqn | r = \paren {2 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $244$ can be enumerated as: :$1, 2, 4, 61, 122, 244$ {{OEIS|A018352}}{{qed}} Category:Tau Function Category:244 tt5qjt0kh52cpbdoy10801body7eao2


Tau Function of 245

Tau Function of 245 0 57628 437662 301353 2019-12-03T22:27:03Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {245} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$245 = 5 \times 7^2$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {245}       | r = \map \tau {5^1 \times 7^2}       | c =  }} {{eqn | r = \paren {1 + 1} \paren {2 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $245$ can be enumerated as: :$1, 5, 7, 35, 49, 245$ {{OEIS|A018353}}{{qed}} Category:Tau Function Category:245 ax6xkkjmionh27wol5xl7cb1o3aikwv


Tau Function of 44

Tau Function of 44 0 57631 392274 301359 2019-02-13T13:43:41Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {44} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$44 = 2^2 \times 11$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {44}       | r = \map \tau {2^2 \times 11^1}       | c =  }} {{eqn | r = \paren {2 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $44$ can be enumerated as: :$1, 2, 4, 11, 22, 44$ {{OEIS|A018259}}{{qed}} Category:Tau Function Category:44 obchaa6rttld5l92qyw5jn6low7xme2


Tau Function of 45

Tau Function of 45 0 57632 498640 301360 2020-11-11T15:10:51Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {45} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\ds \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$45 = 3^2 \times 5$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {45}       | r = \map \tau {3^2 \times 5^1}       | c =  }} {{eqn | r = \paren {2 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $45$ can be enumerated as: :$1, 3, 5, 9, 15, 45$ {{OEIS|A018260}}{{qed}} Category:Tau Function Category:45 pi8h7w7wkxq59n6ftpw5q77k2hovj9c


Tau Function of 75

Tau Function of 75 0 57633 462846 301361 2020-04-18T19:11:38Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {75} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$75 = 3 \times 5^2$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {75}       | r = \map \tau {3^1 \times 5^2}       | c =  }} {{eqn | r = \paren {1 + 1} \paren {2 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $75$ can be enumerated as: :$1, 3, 5, 15, 25, 75$ {{OEIS|A018272}}{{qed}} Category:Tau Function Category:75 4n8i6x28vhggiyuf2krmi2wt5vf3pbe


Tau Function of 76

Tau Function of 76 0 57634 301362 2017-06-16T07:04:24Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({76}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({76}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$76 = 2^2 \times 19$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({76}\right)       | r = \tau \left({2^2 \times 19^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $76$ can be enumerated as: :$1, 2, 4, 19, 38, 76$ {{OEIS|A018273}}{{qed}} Category:Tau Function Category:76 0v0qr9pw59m0qr4ynosy2tmi4b9xbfs


Tau Function of 98

Tau Function of 98 0 57635 493092 301363 2020-10-07T21:00:26Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {98} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\ds \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$98 = 2 \times 7^2$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {98}       | r = \map \tau {2^1 \times 7^2}       | c =  }} {{eqn | r = \paren {1 + 1} \paren {2 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $98$ can be enumerated as: :$1, 2, 7, 14, 49, 98$ {{OEIS|A018281}}{{qed}} Category:Tau Function Category:98 emweidxaurus9g2kcqsmh9li2ba1aip


Tau Function of 99

Tau Function of 99 0 57636 440180 301364 2019-12-19T14:26:33Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {99} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$99 = 3^2 \times 11$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {99}       | r = \map \tau {3^2 \times 11^1}       | c =  }} {{eqn | r = \paren {2 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $99$ can be enumerated as: :$1, 3, 9, 11, 33, 99$ {{OEIS|A018282}}{{qed}} Category:Tau Function Category:99 jmexhprzj9jku6bnx4x88o7tbhrbct4


Tau Function of 116

Tau Function of 116 0 57637 301365 2017-06-16T07:12:51Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({116}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({116}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$116 = 2^2 \times 29$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({116}\right)       | r = \tau \left({2^2 \times 29^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $116$ can be enumerated as: :$1, 2, 4, 29, 58, 116$ {{OEIS|A018291}}{{qed}} Category:Tau Function Category:116 lb1er71ruxb8hxim1qr1w8j9alfsg7c


Tau Function of 117

Tau Function of 117 0 57638 301366 2017-06-16T07:14:55Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({117}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({117}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$117 = 3^2 \times 13$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({117}\right)       | r = \tau \left({3^2 \times 13^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $117$ can be enumerated as: :$1, 3, 9, 13, 39, 117$ {{OEIS|A018292}}{{qed}} Category:Tau Function Category:117 sa61trstzw6njn8hwvtrkn1aovqug3j


Tau Function of 147

Tau Function of 147 0 57639 301367 2017-06-16T07:17:02Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({147}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({147}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$147 = 3 \times 7^2$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({147}\right)       | r = \tau \left({3^1 \times 7^2}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({2 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $147$ can be enumerated as: :$1, 3, 7, 21, 49, 147$ {{OEIS|A018303}}{{qed}} Category:Tau Function Category:147 ltd5c69ddqp6j2mj4cioynjxd2z1e5o


Tau Function of 148

Tau Function of 148 0 57640 301368 2017-06-16T07:18:53Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({148}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({148}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$148 = 2^2 \times 37$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({148}\right)       | r = \tau \left({2^2 \times 37^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $148$ can be enumerated as: :$1, 2, 4, 37, 74, 148$ {{OEIS|A018304}}{{qed}} Category:Tau Function Category:148 bd1lmvoloq9hgkal0gdmfacusjavqru


Tau Function of 171

Tau Function of 171 0 57641 301369 2017-06-16T07:20:53Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({171}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({171}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$171 = 3^2 \times 19$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({171}\right)       | r = \tau \left({3^2 \times 19^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $171$ can be enumerated as: :$1, 3, 9, 19, 57, 171$ {{OEIS|A018316}}{{qed}} Category:Tau Function Category:171 mxk5jkfzab71vxjdtrxj3uanp3verfz


Tau Function of 172

Tau Function of 172 0 57642 301370 2017-06-16T07:22:22Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({172}\right) = 6$ </onlyinclude> where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({172}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$172 = 2^2 \times 43$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({172}\right)       | r = \tau \left({2^2 \times 43^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $172$ can be enumerated as: :$1, 2, 4, 43, 86, 172$ {{OEIS|A018317}}{{qed}} Category:Tau Function Category:172 f4fyyhnchmz2lbtz7jghqb1nermgh2d


Tau Function of 332

Tau Function of 332 0 57643 453229 301371 2020-03-07T22:54:29Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {332} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$332 = 2^2 \times 83$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {332}       | r = \map \tau {2^2 \times 83^1}       | c =  }} {{eqn | r = \paren {2 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $332$ can be enumerated as: :$1, 2, 4, 83, 166, 332$ {{OEIS|A018397}}{{qed}} Category:Tau Function Category:332 t66rqz4j7izzms654yr849sg0ltcmsr


Tau Function of 333

Tau Function of 333 0 57645 479809 301373 2020-07-26T19:08:37Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {333} = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$333 = 3^2 \times 37$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {333}       | r = \map \tau {3^2 \times 37^1}       | c =  }} {{eqn | r = \paren {2 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $333$ can be enumerated as: :$1, 3, 9, 37, 111, 333$ {{OEIS|A018398}}{{qed}} Category:Tau Function Category:333 1024mwyu3a23yig2nixfbw8izpf3zfr


Euler Phi Function of 255

Euler Phi Function of 255 0 57677 301456 2017-06-17T05:05:03Z Prime.mover 59 Created page with "== Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\phi \left({255}\right) = 128$ </onlyinclude> where..." wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\phi \left({255}\right) = 128$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Square-Free Integer: :$\displaystyle \phi \left({n}\right) = \prod_{\substack {p \mathop \backslash n \\ p \mathop > 2} } \left({p - 1}\right)$ where $p \mathop \backslash n$ denotes the primes which divide $n$. We have that: :$255 = 3 \times 5 \times 17$ and so is square-free. Thus: {{begin-eqn}} {{eqn | l = \phi \left({255}\right)       | r = \left({3 - 1}\right) \left({5 - 1}\right) \left({17 - 1}\right)       | c =  }} {{eqn | r = 2 \times 4 \times 16       | c =  }} {{eqn | r = 2 \times 2^2 \times 2^4       | c =  }} {{eqn | r = 2^7       | c =  }} {{eqn | r = 128       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of Square-Free Integer Category:255 15jcgmcyt9gmekph9k99y8vohk7gi3y


Euler Phi Function of 256

Euler Phi Function of 256 0 57678 301457 2017-06-17T05:08:31Z Prime.mover 59 Created page with "== Example of Use of Euler $\phi$ Function == <onlyinclude> The value of the Euler $\phi$ function for the..." wikitext text/x-wiki == Example of Use of Euler $\phi$ Function == <onlyinclude> The value of the Euler $\phi$ function for the integer $256$ is $128$. </onlyinclude> == Proof == From the corollary to Euler Phi Function of Prime Power: :$\phi \left({2^k}\right) = 2^{k-1}$ Thus: {{begin-eqn}} {{eqn | l = \phi \left({256}\right)       | r = \phi \left({2^8}\right)       | c =  }} {{eqn | r = 2^7       | c =  }} {{eqn | r = 128       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function Category:256 9v9f6py2viisafgp53w6dr79zy9yg8k


Tau Function of 20

Tau Function of 20 0 58327 304238 303115 2017-07-10T09:17:26Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({20}\right) = 6$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$20 = 2^2 \times 5$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({20}\right)       | r = \tau \left({2^2 \times 5^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 6       | c =  }} {{end-eqn}} The divisors of $20$ can be enumerated as: :$1, 2, 4, 5, 10, 20$ {{OEIS|A005018}}{{qed}} Category:Tau Function Category:20 aiul2j4zaqmzkql5jarvm0pqntnpuo0


Non-Square Positive Integers not Sum of Square and Prime 0 58473 326104 303396 2017-11-11T09:49:55Z Barto 3079 "work to do" template wikitext text/x-wiki == Conjecture == The sequence of (strictly) positive integers which are not square and not the sum of a square and a prime is believed to be complete: :$10, 34, 58, 85, 91, 130, 214, 226, 370, 526, 706, 730, 771, 1255, 1351, 1414, 1906, 2986, 3676, 9634, 21679$ {{OEIS|A020495}} == Progress == From Square of n such that 2n-1 is Composite is not Sum of Square and Prime, $n^2$ is the sum of a square and a prime {{iff}} $2 n - 1$ composite. Hence the question is specifically about non-squares. No prime number is in this sequence, as trivially: :$p = p + 0^2$ and so is the sum of a prime (itself), and $0^2$, which is square. Each non-square composite $n \in \Z$ can be tested by subtracting successive squares less than $n$ and investigating whether a prime can result. In the following, the smallest $m$ such that $n - m^2 = p$ is shown where such a $p$ exists. Otherwise the nonexistence of such a $p$ is demonstrated explicitly. As follows: {{begin-eqn}} {{eqn | l = 4 - 1^2       | r = 3       | c = which is prime }} {{eqn | l = 6 - 1^2       | r = 5       | c = which is prime }} {{eqn | l = 8 - 1^2       | r = 7       | c = which is prime }} {{end-eqn}} $10$ cannot be expressed as $10 = m^2 + p$. Thus $10$ is seen to be in this sequence. {{begin-eqn}} {{eqn | l = 12 - 1^2       | r = 11       | c = which is prime }} {{eqn | l = 14 - 1^2       | r = 13       | c = which is prime }} {{eqn | l = 15 - 2^2       | r = 11       | c = which is prime }} {{eqn | l = 16 - 3^2       | r = 7       | c = which is prime }} {{eqn | l = 18 - 1^2       | r = 17       | c = which is prime }} {{eqn | l = 20 - 1^2       | r = 19       | c = which is prime }} {{eqn | l = 21 - 2^2       | r = 17       | c = which is prime }} {{eqn | l = 22 - 3^2       | r = 13       | c = which is prime }} {{eqn | l = 24 - 1^2       | r = 23       | c = which is prime }} {{eqn | l = 26 - 3^2       | r = 17       | c = which is prime }} {{eqn | l = 27 - 2^2       | r = 23       | c = which is prime }} {{eqn | l = 28 - 3^2       | r = 19       | c = which is prime }} {{eqn | l = 30 - 1^2       | r = 29       | c = which is prime }} {{eqn | l = 32 - 1^2       | r = 31       | c = which is prime }} {{eqn | l = 33 - 2^2       | r = 29       | c = which is prime }} {{end-eqn}} $34$ cannot be expressed as $34 = m^2 + p$. Thus $34$ is seen to be in this sequence. Similarly: $58$ cannot be expressed as $58 = m^2 + p$. Thus $58$ is seen to be in this sequence. This establishes the pattern. The algorithm for determining whether a particular $n$ belongs to this sequence can be defined in pseudocode as follows: <pre> For n := 1, loop indefinitely, incrementing by 1:   Is n prime? If so, continue to the next n   Is n square? If so, continue to the next n   For m := 1, incrementing by 1 until m^2 > n:     Is n - n^2 prime? If so, continue to the next n   Next m   Add n to the sequence Next n </pre> {{wtd|Design (or adoption) of a rigorous pseudocode needs to be done. Alternatively we may seek to stick with our existing technique for implement algorithms, and/or use a flow chart.}} {{finish|Tidy up the algorithm above.  Do we demonstrate an instance of an computer program here?}} == Examples == {{:Non-Square Positive Integers not Sum of Square and Prime/Examples}} == Historical Note == {{:Non-Square Positive Integers not Sum of Square and Prime/Historical Note}} Category:Numbers not Sum of Square and Prime Category:Unproven Hypotheses 200j0yl2xxwh429avnf174txlp4nyt8


Tau Function of 836

Tau Function of 836 0 58989 488971 304141 2020-09-18T22:40:48Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {836} = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$836 = 2^2 \times 11 \times 19$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {836}       | r = \map \tau {2^2 \times 11^1 \times 19^1}       | c =  }} {{eqn | r = \paren {2 + 1} \paren {1 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $836$ can be enumerated as: :$1, 2, 4, 11, 19, 22, 38, 44, 76, 209, 418, 836$ {{OEIS|A018674}}{{qed}} Category:Tau Function Category:836 3p7ep19hp9lks5cd2e5mlmuyd81v6bc


Sigma Function of 4030

Sigma Function of 4030 0 58996 393901 317889 2019-02-25T15:01:57Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Square-Free Integer == <onlyinclude> :$\map \sigma {4030} = 8064$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == <onlyinclude> We have that: :$4030 = 2 \times 5 \times 13 \times 31$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {4030}       | r = \paren {2 + 1} \paren {5 + 1} \paren {13 + 1} \paren {31 + 1}       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 6 \times 14 \times 32       | c =  }} {{eqn | r = 3 \times \paren {2 \times 3} \times \paren {2 \times 7} \times 2^5       | c =  }} {{eqn | r = 2^7 \times 3^2 \times 7       | c =  }} {{eqn | r = 8064       | c =  }} {{end-eqn}} {{qed}} </onlyinclude> Category:Sigma Function of Square-Free Integer Category:4030 94fs2bv2gjbu850omhvswl2iscnyp4p


Tau Function of 4030

Tau Function of 4030 0 58997 304156 2017-07-09T19:08:53Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({4030}\right) = 16$ </onlyinclude> where $\tau$ denotes the Definition:Tau Fun..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({4030}\right) = 16$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$4030 = 2 \times 5 \times 13 \times 31$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({4030}\right)       | r = \tau \left({2^1 \times 5^1 \times 13^1 \times 31^1}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 16       | c =  }} {{end-eqn}} The divisors of $4030$ can be enumerated as: :$1, 2, 5, 10, 13, 26, 31, 62, 65, 130, 155, 310, 403, 806, 2015, 4030$ {{qed}} Category:Tau Function Category:4030 h53efoeraud09v5hpgi203y0gm5a3ve


Tau Function of 490

Tau Function of 490 0 59007 304197 2017-07-09T22:41:58Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({490}\right) = 12$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({490}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$490 = 2 \times 5 \times 7^2$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({490}\right)       | r = \tau \left({2^1 \times 5^1 \times 7^2}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({1 + 1}\right) \left({2 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $490$ can be enumerated as: :$1, 2, 5, 7, 10, 14, 35, 49, 70, 98, 245, 490$ {{OEIS|A018483}}{{qed}} Category:Tau Function Category:490 jchifntti6yp12ggx2eiwo8aqnefwvb


Tau Function of 550

Tau Function of 550 0 59008 304202 2017-07-09T22:56:05Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({550}\right) = 12$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({550}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$550 = 2 \times 5^2 \times 11$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({550}\right)       | r = \tau \left({2^1 \times 5^2 \times 11^1}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $550$ can be enumerated as: :$1, 2, 5, 10, 11, 22, 25, 50, 55, 110, 275, 550$ {{OEIS|A018514}}{{qed}} Category:Tau Function Category:550 npbvgbi3gczo3qlsits8hp9d7wm6zvo


Tau Function of 572

Tau Function of 572 0 59010 304205 2017-07-09T23:06:44Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({572}\right) = 12$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({572}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$572 = 2^2 \times 11 \times 13$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({572}\right)       | r = \tau \left({2^2 \times 11^1 \times 13^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $572$ can be enumerated as: :$1, 2, 4, 11, 13, 22, 26, 44, 52, 143, 286, 572$ {{OEIS|A018525}}{{qed}} Category:Tau Function Category:572 rhv4hqbtrzva88kvoefc2n6ngmd4zed


Tau Function of 650

Tau Function of 650 0 59015 304223 2017-07-09T23:30:22Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({650}\right) = 12$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({650}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$650 = 2 \times 5^2 \times 13$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({650}\right)       | r = \tau \left({2^1 \times 5^2 \times 13^1}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $650$ can be enumerated as: :$1, 2, 5, 10, 13, 25, 26, 50, 65, 130, 325, 650$ {{OEIS|A018571}}{{qed}} Category:Tau Function Category:650 5gefu5m1ce8n0dmit0o9j7j003lfdhc


Tau Function of 88

Tau Function of 88 0 59022 304235 2017-07-10T09:09:04Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({88}\right) = 8$ </onlyinclude> where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({88}\right) = 8$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$88 = 2^3 \times 11$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({88}\right)       | r = \tau \left({2^3 \times 11^1}\right)       | c =  }} {{eqn | r = \left({3 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} The divisors of $88$ can be enumerated as: :$1, 2, 4, 8, 11, 22, 44, 88$ {{OEIS|A018277}}{{qed}} Category:Tau Function Category:88 qds6jo0jge6b258ki5stwt2s5aodcq0


Tau Function of 104

Tau Function of 104 0 59024 334107 304239 2018-01-02T16:23:08Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({104}\right) = 8$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$104 = 2^3 \times 13$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({104}\right)       | r = \tau \left({2^3 \times 13^1}\right)       | c =  }} {{eqn | r = \left({3 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} The divisors of $104$ can be enumerated as: :$1, 2, 4, 8, 13, 26, 52, 104$ {{OEIS|A018285}}{{qed}} Category:Tau Function Category:104 euuy536l81o3uo5z62ubdnadh6md7m3


Sigma Function of 272

Sigma Function of 272 0 59026 318220 318218 2017-09-16T18:26:03Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({272}\right) = 558$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$272 = 2^4 \times 17$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({272}\right)       | r = \frac {2^5 - 1} {2 - 1} \times \frac {17^2 - 1} {17 - 1}       | c = Sigma Function of Integer }} {{eqn | r = \frac {31 - 1} 1 \times \frac {289 - 1} {16}       | c =  }} {{eqn | r = 31 \times 18       | c =  }} {{eqn | r = \left({2 \times 3^2}\right) \times 31       | c =  }} {{eqn | r = 558       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:272 rx3scmsglraze6gjw7jksv67w3pcnw9


Tau Function of 272

Tau Function of 272 0 59027 304242 2017-07-10T09:26:58Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({272}\right) = 10$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({272}\right) = 10$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$272 = 2^4 \times 17$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({272}\right)       | r = \tau \left({2^4 \times 17^1}\right)       | c =  }} {{eqn | r = \left({4 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 10       | c =  }} {{end-eqn}} The divisors of $272$ can be enumerated as: :$1, 2, 4, 8, 16, 17, 34, 68, 136, 272$ {{OEIS|A018366}}{{qed}} Category:Tau Function Category:272 s7nn4xxhrkckargd8ql0zio5nglgyc7