Partitions in a generating function $(1-x)(1-x^2)(1-x^3)(1-x^4)...(1-x^{1000})$

Question

I was trying to find the coefficient of each $x^n$ in the expression

$$(1-x)(1-x^2)(1-x^3)(1-x^4)...(1-x^{1000})$$ for $n=1,2,3,4...30$

The answer seems to be equivalent to finding $P_{even}$-$P_{odd}$ where

• $P_{even}$ is the number of partitions of $n$ into an even number of distinct parts and

• $P_{odd}$ is the number of ways of partitioning $n$ into an odd number of distinct parts.

This seems complicated. Do not tell me the answer but tell me if there is an easier method to solve it.

Answer 1

I think what you need is the Euler's pentagonal theorem

There is an interesting video about how Euler solved the problem of the partitions, linked the particular part but I recommend the entire video.

Added:

$\prod_{i=1}^\infty (1-x^i) = \sum_{i=0}^\infty a_ix^i,$

$a_i := \begin{cases}1 & \mbox{ if } i = \frac{1}{2}(3k^2 \pm k) \mbox{ and } k \mbox{ is even}\\ -1 & \mbox{ if } i = \frac{1}{2}(3k^2 \pm k) \mbox{ and } k \mbox{ is odd }\\ 0 & \mbox{ otherwise }\end{cases}$

$P_{even}-P_{odd}=\sum_{i \mathrm{\ even}} p(n{-}g_i)-\sum_{i \mathrm{\ odd}} p(n{-}g_i)=\sum_{i=0}^n p(n{-}i) a_i=0\ $ ($g_k$ the k^th generalized pentagonal number)

Answer 2

Hint: Let $$f(x)=(1-x)(1-x^2)(1-x^3)(1-x^4)...(1-x^{1000})$$ then the coefficient of $x^n$ is $\dfrac{f^{(n)}(0)}{n!}$ and for this purpose consider $\ln f(x)$ and find a formula for $n$-th derivative.