Are the largely composite numbers the same as the fully composite numbers?

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.

edited Nov 7 2010 at 18:22

asked Nov 4 2010 at 19:37

tdnoe
289●1●6

	I see, you do not demand new records, just "at least as good as." Why is every LCN an FCN? – Will Jagy Nov 4 2010 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 2010 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 2010 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 2010 at 21:41
	It appears to be something like c (log x)^2. – tdnoe Nov 4 2010 at 21:53