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.

  • $\begingroup$ 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. $\endgroup$ 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.