hcju/proofwikips huggingface dataset:
Tau Function of 245
== Example of Use of $\tau$ Function ==  :$\map \tau {245} = 6$  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}}
Tau Function of 44
== Example of Use of $\tau$ Function ==  :$\map \tau {44} = 6$  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}}
Tau Function of 45
== Example of Use of $\tau$ Function ==  :$\map \tau {45} = 6$  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}}
Tau Function of 75
== Example of Use of $\tau$ Function ==  :$\map \tau {75} = 6$  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}}
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 ==  :$\tau \left({76}\right) = 6$  where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({76}\right) = 6$  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}}
Tau Function of 98
== Example of Use of $\tau$ Function ==  :$\map \tau {98} = 6$  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}}
Tau Function of 99
== Example of Use of $\tau$ Function ==  :$\map \tau {99} = 6$  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}}
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 ==  :$\tau \left({116}\right) = 6$  where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({116}\right) = 6$  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}}
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 ==  :$\tau \left({117}\right) = 6$  where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({117}\right) = 6$  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}}
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 ==  :$\tau \left({147}\right) = 6$  where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({147}\right) = 6$  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}}
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 ==  :$\tau \left({148}\right) = 6$  where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({148}\right) = 6$  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}}
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 ==  :$\tau \left({171}\right) = 6$  where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({171}\right) = 6$  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}}
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 ==  :$\tau \left({172}\right) = 6$  where $\tau$ denotes the Definition:Tau Funct..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({172}\right) = 6$  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}}
Tau Function of 332
== Example of Use of $\tau$ Function ==  :$\map \tau {332} = 6$  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}}
Tau Function of 333
== Example of Use of $\tau$ Function ==  :$\map \tau {333} = 6$  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}}
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 ==  :$\phi \left({255}\right) = 128$  where..." wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer ==  :$\phi \left({255}\right) = 128$  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}}
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 ==  The value of the Euler $\phi$ function for the..." wikitext text/x-wiki == Example of Use of Euler $\phi$ Function ==  The value of the Euler $\phi$ function for the integer $256$ is $128$.  == 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}}
Tau Function of 20
== Example of Use of $\tau$ Function ==  :$\tau \left({20}\right) = 6$  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}}
Tau Function of 836
== Example of Use of $\tau$ Function ==  :$\map \tau {836} = 12$  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}}
Sigma Function of 4030
== Example of Sigma Function of Square-Free Integer ==  :$\map \sigma {4030} = 8064$  where $\sigma$ denotes the $\sigma$ function. == Proof ==  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}}
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 ==  :$\tau \left({4030}\right) = 16$  where $\tau$ denotes the Definition:Tau Fun..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({4030}\right) = 16$  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: :$1, 2, 5, 10, 13, 26, 31, 62, 65, 130, 155, 310, 403, 806, 2015, 4030$
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 ==  :$\tau \left({490}\right) = 12$  where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({490}\right) = 12$  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: :$490 = 2 \times 5 \times 7^2$ 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}}
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 ==  :$\tau \left({550}\right) = 12$  where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({550}\right) = 12$  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: :$550 = 2 \times 5^2 \times 11$ 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}}
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 ==  :$\tau \left({572}\right) = 12$  where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({572}\right) = 12$  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: :$572 = 2^2 \times 11 \times 13$ 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}}
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 ==  :$\tau \left({650}\right) = 12$  where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({650}\right) = 12$  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: :$650 = 2 \times 5^2 \times 13$ 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}}
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 ==  :$\tau \left({88}\right) = 8$  where $\tau$ denotes the Definition:Tau Functi..." wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({88}\right) = 8$  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}}