I'm under the impression that if $n$ is even then the number of partitions of $n$ into an even number of parts exceeds the number of partitions of $n$ into an odd number of parts. And the opposite if $n$ is odd.
I feel like this is clear by looking at small examples, but I can't give a rigorous proof. I think it follows from how you make the partitions out of $n$ ones.
Would anyone be able to help me gain some more intuition about this or provide me with a proof? Thanks!
Your conjecture is correct for $n>2$.
If $n>2$, $n$ has at least one self-conjugate partition; see the answer to this question. If $c(n)$ is the number of self-conjugate partitions of $n$, $$c(n)=(-1)^n(p_E(n)-p_O(n))\,,$$ where $p_E(n)$ and $p_O(n)$ are the numbers of partitions of $n$ with an even and an odd number of parts, respectively; this is proved in the answers to this question. Thus, $c(n)$ is positive for $n>2$, so $p_E(n)-p_O(n)$ must be positive for even $n>2$ and negative for odd $n>2$.
