Take the 2-minute tour ×

Let $d(n)$ be the number of divisors of $n$. Let $p(n)$ be the product of the divisors of $n$. Ramanujan called a number $n$ largely composite if $d(n) \ge d(m)$ for $m < n$. Let's call $n$ fully composite if $p(n) \ge p(m)$ for $m < n$. It is conjectured that largely composite numbers (LCN) are the same as fully composite numbers (FCN). It is easy to show that an LCN is an FCN. Is every FCN an LCN? In the OEIS, these are sequences http://oeis.org/classic/A067128 and http://oeis.org/classic/A034287. These two sequences are the same for the 105834 terms less than 10^150.

This question is interesting because it connects the number of divisors to the product of divisors.

share|improve this question
I see, you do not demand new records, just "at least as good as." Why is every LCN an FCN? –  Will Jagy Nov 4 '10 at 19:46
Actually, it can be proved that $p(n) \ne p(m)$ for $n \ne m$. Using the identity $p(n) = n^{(d(n)/2)}$, it is easy to show that $d(n) \ge d(m)$ implies $p(n) \ge p(m)$. –  tdnoe Nov 4 '10 at 20:10
I'm not getting that $p(n)$ is multiplicative. Instead, if $\gcd(m,n) = 1,$ I get $$ p(mn) = p(m)^{d(n)} \; p(n)^{d(m)}.$$ Screws up the only method I had in mind. –  Will Jagy Nov 4 '10 at 20:31
About how frequent are these numbers, either list? Roughly how many up to some large real $x,$ at least statistically from the data you have? –  Will Jagy Nov 4 '10 at 21:41
It appears to be something like c (log x)^2. –  tdnoe Nov 4 '10 at 21:53
add comment


Your Answer



Browse other questions tagged or ask your own question.