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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.

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  • $\begingroup$ You seem to be linking to the wrong entry in OEIS -- that seems to be for partitions into odd and distinct parts. $\endgroup$ – Lucia Dec 5 '19 at 17:37
  • $\begingroup$ Thank you, Lucia. $\endgroup$ – T. Amdeberhan Dec 6 '19 at 14:03
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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.

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  • $\begingroup$ 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 $\endgroup$ – Brian Hopkins Dec 6 '19 at 1:49
  • $\begingroup$ @BrianHopkins no: the values of $k$ are not the same as the values of $n$ $\endgroup$ – Fedor Petrov Dec 6 '19 at 8:54
  • $\begingroup$ Thanks @FedorPetrov. And thanks, Lucia, for finishing out the argument. $\endgroup$ – Brian Hopkins Dec 6 '19 at 11:17
  • $\begingroup$ I appreciate all of your comments, suggestions and resolution. $\endgroup$ – T. Amdeberhan Dec 6 '19 at 14:03

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