# Size of parities in counting partitions into odd parts

Let $$p_{odd}(n)$$ be the number of partitions of $$n$$ into odd parts (see here). For instance, one has the generating function $$\prod_{k\geq1}\frac1{1-q^{2k-1}}.$$

QUESTION. What is the size of this set $$A_N:=\{n\in\{1,2,\ldots,N\}: \text{p_{odd}(n) is odd}\}$$ for large $$N$$?

Note. I do not expect this to be $$\sim\frac12N$$. Any solution or reference is appreciated.

• You seem to be linking to the wrong entry in OEIS -- that seems to be for partitions into odd and distinct parts. Dec 5 '19 at 17:37
• Thank you, Lucia. Dec 6 '19 at 14:03

Note that the generating function is $$\prod_{k=1}^{\infty} \frac{1}{1-q^{2k-1}} = \prod_{k=1}^{\infty} \frac{1-q^{2k}}{1-q^k} = \prod_{k=1}^{\infty} (1+q^k) \equiv \prod_{k=1}^{\infty} (1-q^k) \mod 2.$$ Now use Euler's pentagonal number theorem, which says that the RHS is $$1+ \sum_{k=1}^{\infty} (-1)^k (q^{k(3k+1)/2} + q^{k(3k-1)/2}).$$ Thus $$p_{\text{odd}}(n)$$ is odd if and only if $$n$$ is of the form $$k(3k\pm 1)/2$$, so that $$A_N$$ has only on the order of $$\sqrt{N}$$ elements.
• Equivalently $k(3k-1)/2$ for all integers $k$, from which it is easier to answer the original question that $A_N$ does, in fact, approach $N/2$ (as every 8 consecutive $k$ values give 4 even and 4 odd values). OEIS entry Dec 6 '19 at 1:49
• @BrianHopkins no: the values of $k$ are not the same as the values of $n$ Dec 6 '19 at 8:54