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 $p$.