Let $p\_{odd}(n)$ be the number of partitions of $n$ into odd parts ([see here](https://oeis.org/A000700)). 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.



Size of parities in counting partitions into odd parts
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.


