The parities of the number of partitions of $n!/6$ and $n!/2$ appear to be non-random initially, as follows — is there an explanation for this other than chance? With $p$ being the partition function and $\lfloor \rfloor$ the floor,
$$ \begin{align} & p(n!/6) \equiv \lfloor n/3\rfloor\ (\text{mod}\ 2), && \text{for}\ 3\leq n\leq 19\\ & p(n!/2) \equiv 1\ (\text{mod}\ 2), && \text{for}\ 2\leq n\leq 16 \end{align} $$
However, neither statement is true for all $n$: $p(20!/6) \equiv 1\ (\text{mod}\ 2)$ and $p(n!/2) \equiv 0\ (\text{mod}\ 2)$ for $17\leq n\leq 19$. This is as far as I have computed (using SageMath with the "flint" algorithm).
Also, is there a faster algorithm for computing the parity of the number of partitions than first computing the number of partitions with the "flint" algorithm (which uses the Hardy-Ramanujan-Rademacher formula in an especially efficient way, as I understand)?
Any way to use the special form $n!/k!$ to speed evaluation, e.g., of a recurrence relation for $p$?
