14
$\begingroup$

Ramanujan delivered his famous congruences $$p(5n+4)\equiv_50, \qquad p(7n+5)\equiv_70, \qquad p(11n+6)\equiv_{11}0$$ for the integer partitions with generating function $F(x)=\prod_{k=0}^{\infty}\frac1{1-x^k}=\sum_{n\geq0}p(n)x^n$.

Let $p\geq5$ be a prime number, and consider the series $$F(x)^{p-1}=\sum_{n\geq0}a_p(n)x^n.$$ Experimental evidence suggests (curiously) that, modulo $p$, exactly half of the $p-1$ rows $$\begin{cases} a_p(pn+1): n\geq0 \\ a_p(pn+2): n\geq0 \\ a_p(pn+3): n\geq0 \\ \qquad\dots\dots\dots \\ a_p(pn+p-1): n\geq0 \end{cases}$$ are identically zero.

Question. Is this true? Why? Or, is it known?

Question. Why does this fail to be true for the prime $p=3$?

For example, $a_5(5n+3)\equiv_50$ and $a_5(5n+4)\equiv_50$ for any $n\geq0$.

Notation. $\equiv_p$ means congruent modulo $p$.

$\endgroup$
21
$\begingroup$

This is true. First, note that by the Pentagonal number theorem due to Euler, $$\frac{1}{F(x)} = \sum_{k \in \mathbb{Z}} (-1)^k x^{\frac{k}{2}(3k-1)}.$$

For a given prime $p \ge 5$, the function $f(k)=\frac{k}{2}(3k-1),f:\mathbb{Z}\to \mathbb{F}_p$ has only $\frac{p+1}{2}$ values in its image, one of which is $0$. This follows by completing the square: $$ \frac{k}{2}(3k-1) = \frac{3}{2}\left( (k-\frac{1}{6})^2 - \frac{1}{6^2} \right).$$ We deduce that there are $\frac{p-1}{2}$ non-zero distinct values $a_1,\cdots,a_{\frac{p-1}{2}}$ modulo $p$ such that the $n$'th coefficient of $\frac{1}{F(x)}$ is zero if $n \equiv a_i \bmod p$ for some $i$. We conclude by noting that modulo $p$, $$F(x)^{p-1} \equiv_p \frac{F(x^p)}{F(x)}.$$ In particular, $a_p(n) = 0$ if $n \equiv a_i \bmod p$.

The result does not hold for $p=3$ because $f$ behaves differently then (it is surjective in that case).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.