Ramanujan delivered his [famous congruences](https://en.wikipedia.org/wiki/Ramanujan's_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 $p$.



powered partition function generator: 1/2 of them are zeros?
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).


