hcju/proofwikips huggingface dataset:
Sigma Function of 217
== Example of Sigma Function of Non-Square Semiprime ==  :$\sigma \left({217}\right) = 256$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$217 = 7 \times 31$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \sigma \left({217}\right)       | r = \left({7 + 1}\right) \left({31 + 1}\right)       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 8 \times 32       | c =  }} {{eqn | r = 2^3 \times 2^5       | c =  }} {{eqn | r = 2^8       | c =  }} {{eqn | r = \left({2^4}\right)^2       | c =  }} {{eqn | r = 16^2       | c =  }} {{eqn | r = 256       | c =  }} {{end-eqn}}
Sigma Function of 214
== Example of Sigma Function of Non-Square Semiprime ==  :$\map \sigma {214} = 324$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$214 = 2 \times 107$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \map \sigma {214}       | r = \paren {2 + 1} \paren {107 + 1}       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 3 \times 108       | c =  }} {{eqn | r = 3 \times \paren {2^2 \times 3^3}       | c =  }} {{eqn | r = 2^3 \times 3^4       | c =  }} {{eqn | r = \paren {2 \times 3^2}^2       | c =  }} {{eqn | r = 18^2       | c =  }} {{eqn | r = 324       | c =  }} {{end-eqn}}
Sigma Function of 210
== Example of Sigma Function of Square-Free Integer ==  :$\map \sigma {210} = 576$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$210 = 2 \times 3 \times 5 \times 7$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {210}       | r = \paren {2 + 1} \paren {3 + 1} \paren {5 + 1} \paren {7 + 1}       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 4 \times 6 \times 8       | c =  }} {{eqn | r = 3 \times 2^2 \times \paren {2 \times 3} \times 2^3       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \paren {2^3 \times 3}^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}}
Sigma Function of 265
== Example of Sigma Function of Non-Square Semiprime ==  :$\sigma \left({265}\right) = 324$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$265 = 5 \times 53$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({265}\right)       | r = \left({5 + 1}\right) \left({53 + 1}\right)       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 6 \times 54       | c =  }} {{eqn | r = \left({2 \times 3}\right) \times \left({2 \times 3^3}\right)       | c =  }} {{eqn | r = 2^2 \times 3^4       | c =  }} {{eqn | r = \left({2 \times 3^2}\right)^2       | c =  }} {{eqn | r = 18^2       | c =  }} {{eqn | r = 324       | c =  }} {{end-eqn}}
Sigma Function of 282
== Example of Sigma Function of Square-Free Integer ==  :$\map \sigma {282} = 576$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$282 = 2 \times 3 \times 47$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {282}       | r = \paren {2 + 1} \paren {3 + 1} \paren {47 + 1}       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 4 \times 48       | c =  }} {{eqn | r = 3 \times 2^2 \times \paren {2^4 \times 3}       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \paren {2^3 \times 3}^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}}
Sigma Function of 310
== Example of Sigma Function of Square-Free Integer ==  :$\sigma \left({310}\right) = 576$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$310 = 2 \times 5 \times 31$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({310}\right)       | r = \left({2 + 1}\right) \left({5 + 1}\right) \left({31 + 1}\right)       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 6 \times 32       | c =  }} {{eqn | r = 3 \times \left({2 \times 3}\right) \times 2^5       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}}
Sigma Function of 322
== Example of Sigma Function of Square-Free Integer ==  :$\sigma \left({322}\right) = 576$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$322 = 2 \times 7 \times 23$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({322}\right)       | r = \left({2 + 1}\right) \left({7 + 1}\right) \left({23 + 1}\right)       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 3 \times 8 \times 24       | c =  }} {{eqn | r = 3 \times 2^3 \times \left({2^3 \times 3}\right)       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}}
Sigma Function of 343
== Example of Sigma Function of Power of Prime ==  :$\map \sigma {343} = 400$  where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Power of Prime: :$\map \sigma {p^k} = \dfrac {p^{k + 1} - 1} {p_i - 1}$ We have that: :$343 = 7^3$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {343}       | r = \frac {7^4 - 1} {7 - 1}       | c =  }} {{eqn | r = \frac {2400} 6       | c =  }} {{eqn | r = 400       | c =  }} {{eqn | r = 20^2       | c =  }} {{end-eqn}} Thus we have that: :$7^0 + 7^2 + 7^2 + 7^3 = 20^2$  == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 400|next = Numbers whose Sigma is Square/Examples/400}}: $400$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 400|next = Numbers whose Sigma is Square/Examples/400}}: $400$
Sigma Function of 345
== Example of Sigma Function of Square-Free Integer ==  :$\sigma \left({345}\right) = 576$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$345 = 3 \times 5 \times 23$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({345}\right)       | r = \left({3 + 1}\right) \left({5 + 1}\right) \left({23 + 1}\right)       | c = Sigma Function of Square-Free Integer }} {{eqn | r = 4 \times 6 \times 24       | c =  }} {{eqn | r = 2^2 \times \left({2 \times 3}\right) \times \left({2^3 \times 3}\right)       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}}
Sigma Function of 382
== Example of Sigma Function of Non-Square Semiprime ==  :$\sigma \left({382}\right) = 576$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$382 = 2 \times 191$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \sigma \left({382}\right)       | r = \left({2 + 1}\right) \left({191 + 1}\right)       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 3 \times 192       | c =  }} {{eqn | r = 3 \times \left({2^6 \times 3}\right)       | c =  }} {{eqn | r = 2^6 \times 3^2       | c =  }} {{eqn | r = \left({2^3 \times 3}\right)^2       | c =  }} {{eqn | r = 24^2       | c =  }} {{eqn | r = 576       | c =  }} {{end-eqn}}
Legendre's Condition/Lemma 1/Dependent on N Functions
 == 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: :$F \in C^2 \closedint a b$   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 ==  == 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
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 ==  The value of the $\tau$ function for the integer $1$ is $1$.  == 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.
Sigma Function of 1
== Example of Sigma Function of Integer ==  :$\map \sigma 1 = 1$  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.
Parenthesization/Examples/4
== Example of Parenthesization ==  A word of $4$ elements can be parenthesized in $5$ distinct ways: :$\quad a_1 \paren {a_2 \paren {a_3 a_4} }$ :$\quad a_1 \paren {\paren {a_2 a_3} a_4}$ :$\quad \paren {a_1 a_2} \paren {a_3 a_4}$ :$\quad \paren {a_1 \paren {a_2 a_3} } a_4$ :$\quad \paren {\paren {a_1 a_2} a_3} a_4$  == 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}}  == 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}}: $42$
Complex Exponential is Uniformly Continuous on Half-Planes/Corollary
== Corollary to Complex Exponential is Uniformly Continuous on Half-Planes ==  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 $\exp g$.  == 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  == 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 $5.7$
Quasiamicable Numbers/Examples/48,75
== Examples of Quasiamicable Pair ==  $48$ and $75$ form a quasiamicable pair.  == 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.
Sigma Function of 140
== Example of Sigma Function of Integer ==  :$\map \sigma {140} = 336$  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}}
Partial Products of Uniformly Convergent Product Converge Uniformly
== 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.  == Also see == * Uniform Product of Continuous Functions is Continuous * Infinite Product of Analytic Functions is Analytic
Logarithm of Infinite Product of Complex Functions/Corollary
 == Corollary to Logarithm of Infinite Product of Complex Functions ==  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 $g = \displaystyle \prod_{n \mathop = n_0}^\infty f_n$.  == 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 $U$.
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 ==  :$\tau \left({60}\right) = 12$  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}}
Sigma Function of 104
== Example of Sigma Function of Integer ==  :$\map \sigma {104} = 210$  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}}
Sigma Function of 105
== Example of Sigma Function of Square-Free Integer ==  :$\map \sigma {105} = 192$  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}}
Continued Fraction Expansion of Irrational Square Root/Examples/61
== Examples of Continued Fraction Expansion of Irrational Square Root ==  The continued fraction expansion of the square root of $61$ is given by: :$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$  {{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: :$\sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}$
Magic Square/Examples/Order 5
== Examples of Order $5$ Magic Squares ==  :$\begin{array}{|c|c|c|c|c|} \hline 23 &  6 & 19 &  2 & 15 \\ \hline 10 & 18 &  1 & 14 & 22 \\ \hline 17 &  5 & 13 & 21 &  9 \\ \hline  4 & 12 & 25 &  8 & 16 \\ \hline 11 & 23 &  7 & 20 &  3 \\ \hline \end{array}$  == Also see == * Magic Constant of Order 5 Magic Square
Numbers with Euler Phi Value of 72
== Example of Use of Euler $\phi$ Function ==  There are $17$  positive integers for which the value of the Euler $\phi$ function is $72$: :$73, 91, 95, 111, 117, 135, 146, 148, 152, 182, 190, 216, 222, 228, 234, 252, 270$  == 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}}  == 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}}: $72$
Square of 1 Less than Number Base/Examples/5
== Example of Square of 1 Less than Number Base ==  The square of $5$ is expressed in base $6$ as: :$5^2 = \left[{41}\right]_6$  == 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.
Sigma Function of 836
== Example of Sigma Function of Integer ==  :$\map \sigma {836} = 1680$  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}}
Euler Phi Function of 104
== Example of Use of Euler $\phi$ Function ==  :$\map \phi {104} = 48$  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}}