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What can be said about the set of positive integers representable as a product of the form $p(n_1)\cdots p(n_r)$ for the partition function $p(n)$ ? Such numbers $k$ arise as the number of distinct abelian groups of order $n$ for some $n$, see here.
For example, $k=10^6$ is representable as $p(2)^6p(4)^6=2^6\cdot 5^6$, but $k=10^6+1=101\cdot 9901$ is not: we have $101=p(13)$, but no $k$ with $p(k)=9901$, or $p(k)=10^6+1$.

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Dietrich Burde
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1 Answers1

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Very few integers can be so represented, basically because $p(n)$ grows extremely fast.

Let $x$ be a large real number. Suppose that $m$ can be represented as $p(2)^{a_2} p(3)^{a_3} \cdots$ for some nonnegative integers $a_j$; then $\log m = a_2p(2) + a_3p(3) + \cdots$. Therefore the number of $m\le x$ that can be so represented is at most the number of solutions to $a_2p(2) + a_3p(3) + \cdots \le \log x$ in nonnegative integers $a_j$.

We know that $\log p(n) \sim c\sqrt n$, where $c=\pi\sqrt{\frac23}$. So roughly we are interested in the number of solutions to $a_2\sqrt2 + a_3\sqrt3 + \cdots \le \frac1c\log x$ in nonnegative integers $a_2,a_3,\dots$. In fact we can stop at the term $a_B\sqrt B$ where $B \sim \frac1{c^2}\log^2x$.

Even if we insisted only that each individual $a_j\sqrt j$ is at most $\frac1c\log x$, the number of solutions would be about $$ \sum_{j=2}^B \frac{\frac1c\log x}{\sqrt j} \sim \frac1c\log x \cdot 2\sqrt B \sim \frac2{c^2} \log^2x. $$ So the number of integers up to $x$ that can be so represented is at most a constant times $\log^2x$, and probably even rather smaller.

Greg Martin
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