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# 15

Please do not rely on any information it contains.

15 is a composite number. In fact, it is the smallest composite number ${\displaystyle n}$ with the property that there is only one group of order ${\displaystyle n}$ (see A000001 and A050384).

## Membership in core sequences

 Odd numbers ..., 9, 11, 13, 15, 17, 19, 21, ... A005408 Composite numbers ..., 10, 12, 14, 15, 16, 18, ... A002808 Semiprimes 4, 6, 9, 10, 14, 15, 21, 22, ... A001358 Partition numbers 1, 1, 2, 3, 5, 7, 11, 15, 22, ... A000041 Triangular numbers 1, 3, 6, 10, 15, 21, 28, 36, ... A000217

In Pascal's triangle, 15 occurs four times, the first two times surrounding the first occurrence of 20, with 6 on either side.

## Sequences pertaining to 15

 Multiples of 15 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, ... A008597 15-gonal numbers 1, 15, 42, 82, 135, 201, 280, 372, 477, 595, 726, 870, ... A051867 ${\displaystyle 3x+1}$ sequence starting at 15 15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, ... A033480

## Partitions of 15

There are 176 partitions of 15.

The Goldbach representations of 15 using distinct primes are: 2 + 13 = 3 + 5 + 7.

## Roots and powers of 15

In the table below, irrational numbers are given truncated to eight decimal places.

 ${\displaystyle {\sqrt {15}}}$ 3.87298334 A010472 15 2 225 ${\displaystyle {\sqrt[{3}]{15}}}$ 2.46621207 A010587 15 3 3375 ${\displaystyle {\sqrt[{4}]{15}}}$ 1.96798967 A011012 15 4 50625 ${\displaystyle {\sqrt[{5}]{15}}}$ 1.71877192 A011100 15 5 759375 ${\displaystyle {\sqrt[{6}]{15}}}$ 1.57041780 A011350 15 6 11390625 ${\displaystyle {\sqrt[{7}]{15}}}$ 1.47235670 A011351 15 7 170859375 ${\displaystyle {\sqrt[{8}]{15}}}$ 1.40285055 A011352 15 8 2562890625 ${\displaystyle {\sqrt[{9}]{15}}}$ 1.35106675 A011353 15 9 38443359375 ${\displaystyle {\sqrt[{10}]{15}}}$ 1.31101942 A011354 15 10 576650390625 ${\displaystyle {\sqrt[{11}]{15}}}$ 1.27913795 A011355 15 11 8649755859375 ${\displaystyle {\sqrt[{12}]{15}}}$ 1.25316311 A011356 15 12 129746337890625 A001024

## Logarithms and fifteenth powers

In the OEIS specifically and mathematics in general, ${\displaystyle \log x}$ refers to the natural logarithm of ${\displaystyle x}$, whereas all other bases are specified with a subscript.

As above, irrational numbers in the following table are truncated to eight decimal places.

TABLE

(See A000584 for the fifth powers of integers).

## Values for number theoretic functions with 15 as an argument

 ${\displaystyle \mu (15)}$ 1 ${\displaystyle M(15)}$ –1 ${\displaystyle \pi (15)}$ 6 ${\displaystyle \sigma _{1}(15)}$ 24 ${\displaystyle \sigma _{0}(15)}$ 4 ${\displaystyle \phi (15)}$ 8 ${\displaystyle \Omega (15)}$ 2 ${\displaystyle \omega (15)}$ 2 ${\displaystyle \lambda (15)}$ 4 This is the Carmichael lambda function. ${\displaystyle \lambda (15)}$ –1 This is the Liouville lambda function. ${\displaystyle \zeta (15)}$ 1.00003058823630702... (see A013673). 15! 1307674368000 ${\displaystyle \Gamma (15)}$ 87178291200

## Factorization of some small integers in a quadratic integer ring adjoining the square roots of −15, 15

Neither ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-15}})}}$ nor ${\displaystyle \mathbb {Z} [{\sqrt {15}}]}$ is a unique factorization domain. ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-15}})}}$ only has two units, which makes things simpler and gives us more confidence in identifying instances of multiple distinct factorizations. But there are still traps for the unwary, such as the presence of so-called "half-integers."

For example, we could say that 16 has three distinct factorizations: 2 4, ${\displaystyle (1-{\sqrt {-15}})(1+{\sqrt {-15}})}$ and ${\displaystyle \left({\frac {7}{2}}-{\frac {\sqrt {-15}}{2}}\right)\left({\frac {7}{2}}+{\frac {\sqrt {-15}}{2}}\right)}$. But that would be wrong. Each number in the latter two factorizations has nontrivial non-unit divisors and thus probably don't correspond to distinct factorizations any more than 4 2 does. First notice that the "half-integer" ${\displaystyle \left({\frac {1}{2}}+{\frac {\sqrt {-15}}{2}}\right)}$ is fully an algebraic integer, with minimal polynomial ${\displaystyle x^{2}-x+4}$, and therefore an integer of this domain. Then verify that ${\displaystyle 2\left({\frac {1}{2}}+{\frac {\sqrt {-15}}{2}}\right)=1+{\sqrt {-15}}}$ and ${\displaystyle \left({\frac {1}{2}}+{\frac {\sqrt {-15}}{2}}\right)^{2}=-{\frac {7}{2}}+{\frac {\sqrt {-15}}{2}}}$. With adjustments of signs, the other "composite" factors are addressed.

This raises the question of whether ${\displaystyle 2^{2}\left({\frac {1}{2}}-{\frac {\sqrt {-15}}{2}}\right)\left({\frac {1}{2}}+{\frac {\sqrt {-15}}{2}}\right)}$ is a distinct factorization. If it is distinct, it is rather inelegant.

As for ${\displaystyle \mathbb {Z} [{\sqrt {15}}]}$, the of the MORE REMARKS GO HERE

 ${\displaystyle n}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-15}})}}$ ${\displaystyle \mathbb {Z} [{\sqrt {15}}]}$ 2 Irreducible 3 4 2 2 OR ${\displaystyle \left({\frac {1}{2}}-{\frac {\sqrt {-15}}{2}}\right)\left({\frac {1}{2}}+{\frac {\sqrt {-15}}{2}}\right)}$ 2 2 5 Irreducible 6 2 × 3 2 × 3 OR ${\displaystyle (-1)(3-{\sqrt {15}})(3+{\sqrt {15}})}$ 7 Prime Irreducible 8 2 3 9 3 2 10 2 × 5 OR ${\displaystyle \left({\frac {5}{2}}-{\frac {\sqrt {-15}}{2}}\right)\left({\frac {5}{2}}+{\frac {\sqrt {-15}}{2}}\right)}$ 2 × 5 OR ${\displaystyle (5-{\sqrt {15}})(5+{\sqrt {15}})}$ 11 Prime ${\displaystyle (-1)(2-{\sqrt {15}})(2+{\sqrt {15}})}$ 12 2 2 × 3 13 Prime 14 2 × 7 2 × 7 OR ${\displaystyle (-1)(1-{\sqrt {15}})(1+{\sqrt {15}})}$ 15 3 × 5 OR ${\displaystyle (-1)({\sqrt {-15}})^{2}}$ 3 × 5 OR ${\displaystyle ({\sqrt {15}})^{2}}$ 16 2 4 OR ${\displaystyle \left({\frac {1}{2}}\pm {\frac {\sqrt {-15}}{2}}\right)^{2}}$ 2 4 17 Irreducible 18 2 × 3 2 19 ${\displaystyle (2-{\sqrt {-15}})(2+{\sqrt {-15}})}$ Prime 20 2 2 × 5

Ideals can help us make sense of the lack of unique factorization in these two rings.

 ${\displaystyle p}$ Factorization of ${\displaystyle \langle p\rangle }$ In ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-15}})}}$ In ${\displaystyle \mathbb {Z} [{\sqrt {15}}]}$ 2 ${\displaystyle \left\langle 2,{\frac {1}{2}}-{\frac {\sqrt {-15}}{2}}\right\rangle \left\langle 2,{\frac {1}{2}}+{\frac {\sqrt {-15}}{2}}\right\rangle }$ ${\displaystyle \langle 2,1+{\sqrt {15}}\rangle ^{2}}$ 3 ${\displaystyle \langle 3,{\sqrt {-15}}\rangle ^{2}}$ ${\displaystyle \langle 3,{\sqrt {15}}\rangle ^{2}}$ 5 ${\displaystyle \langle 5,{\sqrt {-15}}\rangle ^{2}}$ ${\displaystyle \langle 5,{\sqrt {15}}\rangle ^{2}}$ 7 Prime ${\displaystyle \langle 7,1-{\sqrt {15}}\rangle \langle 7,1+{\sqrt {15}}\rangle }$ 11 Prime ${\displaystyle \langle 2-{\sqrt {15}}\rangle \langle 2+{\sqrt {15}}\rangle }$ 13 Prime 17 ${\displaystyle \langle 17,6-{\sqrt {-15}}\rangle \langle 17,6+{\sqrt {-15}}\rangle }$ ${\displaystyle \langle 17,7-{\sqrt {15}}\rangle \langle 17,7+{\sqrt {15}}\rangle }$ 19 ${\displaystyle \langle 2-{\sqrt {-15}}\rangle \langle 2+{\sqrt {-15}}\rangle }$ Prime 23 29 31 37 41 43 47

## Factorization of 15 in some quadratic integer rings

In ${\displaystyle \mathbb {Z} }$, 15 is composite, being the product of two primes. It is of course also composite in any quadratic integer rings, but it has different factorizations in some of those.

 ${\displaystyle \mathbb {Z} [i]}$ ${\displaystyle 3(2\pm i)}$ ${\displaystyle \mathbb {Z} [{\sqrt {-2}}]}$ ${\displaystyle (1\pm {\sqrt {-2}})5}$ ${\displaystyle \mathbb {Z} [{\sqrt {2}}]}$ 3 × 5 ${\displaystyle \mathbb {Z} [\omega ]}$ ${\displaystyle (-1)(1+2\omega )^{2}5}$ ${\displaystyle \mathbb {Z} [{\sqrt {3}}]}$ ${\displaystyle ({\sqrt {3}})^{2}5}$ ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ ${\displaystyle (-1)3({\sqrt {-5}})^{2}}$ ${\displaystyle \mathbb {Z} [\phi ]}$ ${\displaystyle 3(-1+2\phi )^{2}}$ ${\displaystyle \mathbb {Z} [{\sqrt {-6}}]}$ 3 × 5 OR ${\displaystyle (3-{\sqrt {-6}})(3+{\sqrt {-6}})}$ ${\displaystyle \mathbb {Z} [{\sqrt {6}}]}$ ${\displaystyle (-1)(3\pm {\sqrt {6}})(1\pm {\sqrt {6}})}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-7}})}}$ 3 × 5 ${\displaystyle \mathbb {Z} [{\sqrt {7}}]}$ ${\displaystyle (-1)(2\pm {\sqrt {7}})5}$ ${\displaystyle \mathbb {Z} [{\sqrt {-10}}]}$ ${\displaystyle \mathbb {Z} [{\sqrt {10}}]}$ 3 × 5 OR ${\displaystyle (5-{\sqrt {10}})(5+{\sqrt {10}})}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-11}})}}$ ${\displaystyle \left({\frac {1}{2}}\pm {\frac {\sqrt {-11}}{2}}\right)\left({\frac {3}{2}}\pm {\frac {\sqrt {-11}}{2}}\right)}$ ${\displaystyle \mathbb {Z} [{\sqrt {11}}]}$ ${\displaystyle 3(4\pm {\sqrt {11}})}$ ${\displaystyle \mathbb {Z} [{\sqrt {-13}}]}$ 3 × 5 ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {13}})}}$ ${\displaystyle (-1)\left({\frac {1}{2}}\pm {\frac {\sqrt {13}}{2}}\right)5}$ ${\displaystyle \mathbb {Z} [{\sqrt {-14}}]}$ ${\displaystyle \mathbb {Z} [{\sqrt {14}}]}$ ${\displaystyle (-1)3(3\pm {\sqrt {14}})}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-15}})}}$ 3 × 5 OR ${\displaystyle (-1)({\sqrt {-15}})^{2}}$ ${\displaystyle \mathbb {Z} [{\sqrt {15}}]}$ 3 × 5 OR ${\displaystyle ({\sqrt {15}})^{2}}$ ${\displaystyle \mathbb {Z} [{\sqrt {-17}}]}$ 3 × 5 ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {17}})}}$ 3 × 5 ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-19}})}}$ ${\displaystyle 3\left({\frac {1}{2}}\pm {\frac {\sqrt {-19}}{2}}\right)}$ ${\displaystyle \mathbb {Z} [{\sqrt {19}}]}$ ${\displaystyle (-1)(4\pm {\sqrt {19}})(9\pm 2{\sqrt {19}})}$

## Representation of 15 in various bases

 Base 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 through 36 Representation 1111 120 33 30 23 21 17 16 15 14 13 12 11 10 F

We see that 15 is a Harshad number in bases 3, 5, 6, 7, 11, 12, 13 and trivially in base 15.

 ${\displaystyle -1}$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 1729