Let $p_3(n)$ be the number of partitions of $n$ into parts that are not multiples of 3. What is the generating function for the sequence $\{p_3(n)\}$

Question

I believe that in order to solve this Eulers theorem must be used, where the generating function $P(x)$ of $(p(n))_{n\geq0}$ is
$$P(x) = \prod_{k=1}^{\infty}\frac{1}{1-x^k}$$ I am not sure, however, how to remove all factors that are multiples of 3. How can this be solved? Thanks.

Answer 1

$$ \sum_{n\geq 0} p_{3}(n) x^{n} = (1+x+x^{2}+x^{3}+\cdots)(1+x^{2}+x^{4}+\cdots)(1+x^{4}+x^{8}+\cdots)\cdots = \prod_{3\nmid k}\frac{1}{1-x^{k}} = \frac{\prod_{k\geq 1}\frac{1}{1-x^{k}}}{\prod_{3|k}\frac{1}{1-x^{k}}} = \frac{P(x)}{P(x^{3})} $$