What is the number of unordered triples $a,b,c$ such that $abc = n$?

Question

If $abc = n$

where $a,b,c,n\in \mathbb{N}$

then can you derive a formula to find the total number of triples of a,b,c as such?

eg : $abc = 12$ has $4$ such triples, $\{(1,1,12),(1,2,6),(1,3,4),(2,2,3)\}$

Answer 1

Write $$\Large N=\prod^m p_i^{\alpha_i}$$ Where $p_i$ are all prime numbers then if we need to find for this: $$\Large N=\prod^nx_j$$ We can write: $$\Large x_i=\prod^m p_i^{\alpha_{x_j}}$$where $\large 0\le\alpha_{x_j}\le\alpha_i$ Then it can be sais that: $$\Large \sum^n \alpha_{x_j}=\alpha_i$$ So total solutions[ordered] to these are: $$\Large t=\prod^{m}\binom{\alpha_i+n-1}{n-1}$$

Answer 2

Let $N$ be a positive integer with prime factorisation $\displaystyle \prod^m p_i^{\alpha_i}$. Then the number of unordered triples $(a,b,c)$ such that $N = abc$ is $$\frac{1}{6}\prod^{m}\binom{\alpha_i+2}{2} + \frac{1}{2}\prod^{m}\left(\left\lfloor\frac{\alpha_i}{2}\right\rfloor + 1\right)+\frac{1}{3}\prod^m [3\mid\alpha_i]$$ where $[3\mid\alpha_i]$ is an Iverson bracket, i.e. $[3\mid\alpha_i] = 1$ if $3\mid\alpha_i$ and is $0$ otherwise.

Since this number is an integer and the last summand is either $0$ or $1/3$ we could also write this as $$\left\lceil\frac{1}{6}\prod^{m}\binom{\alpha_i+2}{2} + \frac{1}{2}\prod^{m}\left(\left\lfloor\frac{\alpha_i}{2}\right\rfloor + 1\right)\right\rceil.$$

Proof

We prove this using Burnside's lemma. Let the symmetric group $S_3$ act on the set of ordered triples $(a_1,a_2,a_3)$ such that $a_1a_2a_3 = N$ by $(a_1,a_2,a_3)^\pi = (a_{\pi(1)},a_{\pi(2)},a_{\pi(3)})$ for all $\pi \in S_3$. The number of unordered triples is the number of orbits under this action which, by Burnside's lemma, is the average number of triples fixed by an element of $S_3$.

As Aditya has shown, the number of ordered triples is $\displaystyle \prod^{m}\binom{\alpha_i+2}{2}$, so this is the number of ordered triples fixed by the identity.

The number of triples fixed by each transposition is equal to the number of square divisors of $N$, since $(ij)$ fixes an ordered triple $(a_1,a_2,a_3)$ iff $a_i=a_j$. The number of square divisors of $N$ is $\displaystyle \prod^{m}\left(\left\lfloor\frac{\alpha_i}{2}\right\rfloor + 1\right)$.

The $3$-cycles $(123)$ and $(132)$ fix an ordered triple $(a_1,a_2,a_3)$ iff $a_1=a_2=a_3$, so the number of ordered triples fixed by a $3$-cycle is $1$ if $N$ is a cube and $0$ otherwise. $N$ is a cube iff $\alpha_i$ is divisible by $3$ for all $i$, so we can write the number of triples fixed by a $3$-cycle as $\displaystyle \prod^m[3\mid\alpha_i]$.

Hence the average number of triples fixed by an element of $S_3$ is $$\frac{1}{6}\left(\prod^{m}\binom{\alpha_i+2}{2} + 3\prod^{m}\left(\left\lfloor\frac{\alpha_i}{2}\right\rfloor + 1\right)+2\prod^m [3\mid\alpha_i]\right).$$