# Partitions and Truncations

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!

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

You probably know that for the partition function we have

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

Thus,

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

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

$$\left(1+x^2+x^4+x^6+x^8+x^{10}\right)\left(1+x^4+x^8\right)\left(1+x^6\right)\left(1+x^8\right)\left(1+x^{10}\right)\;.$$

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 $k$.

• 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 $p_n$. – Mike Earnest Nov 5 '16 at 21:27
• @Mike: Good point; thanks! – Brian M. Scott Nov 5 '16 at 21:32
• I got $7 x^{10}+5 x^8+3 x^6+2 x^4+x^2+1,$ is it correct? – Dreamer Nov 8 '16 at 20:38
• @Regina: Yep, looks good. – Brian M. Scott Nov 8 '16 at 21:53