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For any given number n $\in N$, how would I find the number of factorizations of n? For example, the number 8 has three combinations* of factorizations ($2^3, 2^2\cdot2,$ and $2\cdot2\cdot2$) so we can say the Factor Combination function $FC(8) = 3$. How would I find $FC(n)$ for an arbitrary $n$?

By the Fundamental Theorem of Arithmetic, every number can be expressed as a unique prime factorization in the form $p_1^{\alpha_1}\cdotp_2^{\alpha_2}\cdot...p_n^{\alpha_n}$. I know that the $FC(n)$ doesn't depend on the particular prime factors, but only on the different $\alpha$'s (e.g. $FC(5^3) = FC(2^3) = FC(p^3) = 3$, for any arbitrary prime $p$.

*My application doesn't care about order, so I would like to find combinations instead of permutations. If it helps, I am trying to find the number of Abelian groups of a given order up to isomorphsim.

Bill Dubuque
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Jafego
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    You may need to use partitions for the exponents of each prime and then combine them. – insipidintegrator Aug 10 '22 at 10:47
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    If we have the prime factorization, this is straight forward. – Peter Aug 10 '22 at 10:54
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    See also https://mathoverflow.net/questions/230960/asymptotics-for-the-number-of-abelian-groups-of-order-at-most-x – lhf Aug 10 '22 at 10:54
  • @Peter, I know how to find the prime factorization for any given number. How can I leverage that to find all of the factorizations? – Jafego Aug 10 '22 at 10:56
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    @Peter If you are only looking for the number of ways to write it as a product of 2 numbers, then it is essentially just finding the number of divisors and this is easy. Here however it is about the number of factorizations into any number of factors, ignoring order. – Jaap Scherphuis Aug 10 '22 at 11:17
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    @JaapScherphuis I know, therefore I called the solution "straight forward" anot not "easy" or even "trivial". – Peter Aug 10 '22 at 11:21
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    @Jafego I don't think factorizations is exactly what you need to be counting. For example, 30=235=215=310=56 but there is only one group of order 30, the cyclic group. But the number of Abelian groups is a multiplicative function, so you do have NA(30)=NA(2)NA(3)NA(5)=11*1=1. – Jaap Scherphuis Aug 10 '22 at 11:25
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    @JaapScherphuis thank you for your clarification. The mod linked to a post that answered the question I should have been asking. – Jafego Aug 10 '22 at 11:56

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