Let $P(x)=\sum_{n=0}^{\infty} p_nx^n$ be the partition generating function, and let $P^*(x)=\sum_{n=0}^{\infty} p^*_nx^n$, where $$p^*_n = \binom{\text{number of partitions of }n}{\text{into an even number of parts}} - \binom{\text{number of partitions of }n}{\text{into an odd number of parts}}.$$ For example, $p^*_4=3-2=1$, because there are $3$ partitions of $4$ into an even number of parts $(3+1,\ 2+2,\ 1+1+1+1)$ and $2$ partitions of $4$ into an odd number of parts $(4,\ 2+1+1)$.

Compute the truncation of $P(x)P^*(x)$ to degree $10$; that is, determine the polynomial consisting of all terms in the power series expansion of $P(x)P^*(x)$ with degree less than or equal to $10$.

(As an example, the truncation of $\frac 1{1-x}$ to degree $3$ is $1+x+x^2+x^3$.)

I have no idea on how to even start this problem, I am stuck. Solutions are greatly appreciated!


HINT: This answer shows that

$$P^*(x)=\sum_{k\ge 0}p_k^*x^k=\prod_{k\ge 1}\frac1{1+x^k}\;.$$

You probably know that for the partition function we have

$$P(x)=\sum_{k\ge 0}p_kx^k=\prod_{k\ge 1}\frac1{1-x^k}\;.$$


$$P(x)P^*(x)=\left(\prod_{k\ge 1}\frac1{1+x^k}\right)\left(\prod_{k\ge 1}\frac1{1-x^k}\right)=\prod_{k\ge 1}\frac1{1-x^{2k}}=\prod_{k\ge 1}\sum_{n\ge 0}x^{2kn}\;.$$

It’s a little tedious if you do the calculation directly, but you can now get the truncation to degree $10$ by looking only at


As Mike Earnest points out in a comment below, you can be a little cleverer and realize that $P(x)P^*(x)=P(x^2)$; then you can get the coefficients from the known values of $p_k$ for small $k$.

  • 1
    $\begingroup$ It doesn't have to be tedious: $P(x)P^*(x)=P(x^2)$, so the $x^n$ coefficient of $P(x)P^*(x)$ is $p_{n/2}$ when $n$ is even and 0 otherwise. It's then pretty easy to compute the first 5 values of $p_n$. $\endgroup$ – Mike Earnest Nov 5 '16 at 21:27
  • $\begingroup$ @Mike: Good point; thanks! $\endgroup$ – Brian M. Scott Nov 5 '16 at 21:32
  • $\begingroup$ I got $7 x^{10}+5 x^8+3 x^6+2 x^4+x^2+1,$ is it correct? $\endgroup$ – Dreamer Nov 8 '16 at 20:38
  • $\begingroup$ @Regina: Yep, looks good. $\endgroup$ – Brian M. Scott Nov 8 '16 at 21:53

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.