Counting the number of partitions such that every part is divisible by $k$.

Question

I would like to do this using generating functions. I'm comfortable with the generating function for the number of partitions of $n$: $$(1+x+x^2+\dots)(1+x^2+x^4+\dots)(1+x^3+x^6+\dots)\cdot\dots,$$ although I'm not entirely sure how to adapt it to deal with certain restrictions, such as asking that every part is divisible by $k$, or divisible by some pair of numbers, say $k$ and $j$.

I was thinking something like $$(1+x^k+x^{2k}+\dots)(1+x^{2k}+x^{4k}+\dots)(1+x^{3k}+x^{6k}+\dots)\cdot\dots.$$

Any advice or hints to get be going would be greatly appreciated!

Answer 1

What you have is correct. If $$f(x)=\sum_{n\ge 0} p(n) x^n= \prod_{j\ge 1} \frac{1}{1-x^j}$$ is the generating function for all partitions, then $$\prod_{\substack{j\ge 1:\\ j\pmod k=0}}\frac{1}{1-x^j} =\prod_{m\ge 1}\frac{1}{1-x^{km}} = f(x^k) $$ is your desired generating function.

Answer 2

A partition of $n$ with every part divisible by $k$ correspond bijectively to partitions of $n/k.$