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If d(n) is a function that gives the number of divisors (positive) of n, then is there any relation between d(n1), d(n2), d(n1*n2) and GCD(n1,n2)? I was wondering if any mathematical formula existed that related the above elements.

yobro97
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    A similar question but does not explain the case where GCD!=1 https://math.stackexchange.com/questions/1198639/is-there-any-formula-for-number-of-divisors-of-a-times-b – yobro97 Apr 14 '17 at 17:35
  • It's a slightly different question, but note that $d(n_1)d(n_2) = d(\mathop{\rm lcm}(n_1,n_2))d(\gcd(n_1,n_2))$. – Greg Martin Apr 14 '17 at 17:51
  • But no: given values for $d(n_1)$, $d(n_2)$, and $\gcd(n_1,n_2)$, there can be multiple possible values for $d(n_1n_2)$. An example is $d(n_1)=4$, $d(n_2)=4$, and $\gcd(n_1,n_2)=2$: each of $n_1$ and $n_2$ must be either $8$ or $2p$ for an odd prime $p$, but their product could be either $64$ or $16p$ or $4p^2$ or $4pp'$, which have $7$ or $8$ or $9$ or $12$ divisors, respectively. – Greg Martin Apr 14 '17 at 17:55

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If the prime factorization of $n_1$ and $n_2$ are $$ n_1 = 2^{a_0}3^{a_1}5^{a_2}7^{a_3}\cdots \qquad\qquad n_2 = 2^{b_0}3^{b_1}5^{b_2}7^{b_3}\cdots $$ then the divisor counts are the products $$ d(n_1) = \prod_{i\ge 0}(a_i+1) \qquad d(n_2) = \prod_{i\ge 0}(b_i+1) \qquad d(\gcd(n_1,n_2)) = \prod_{i\ge 0}\min(a_i+1,b_i+1) $$ (each of these products is finite because only finitely many of the $a_i$s and $b_i$s can be nonzero).

This does not seem to lead to any particularly useful relation between the divisor counts. We cannot choose $d(n_1)$, $d(n_2)$ and $d(\gcd(n_1,n_2))$ completely independently from each other, but there's still a lot of freedom.

hmakholm left over Monica
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