Number of partitions of $n$ formed by combinations of $2$ and $4$

Question

I'm trying to find the number of partitions of a natural number that are a combination of $2$ and $4$.

For example: $$6 = 2+2+2 = 2+4 \Rightarrow p_6 = 2$$

So I start by defining $p_n$ as the coefficient of $x^n$ in the product

$(1+x^2+x^4+\cdots)(1+x^4+x^8+\cdots)$ = $\frac{1}{(1-x^2)}$$\frac{1}{(1-x^4)}$

From here it seems to me that the partial fraction expansion of this last product will give something in terms of complex roots when translated to the generating function of the number of partitions.

But I don't know how to go from here on to determine the value of each coefficient.

Thanks!

Answer 1

Here a simple way to solve it without involving generating functions. You have just to choose the number of $4$s in the partition, and then you can uniquely complete the partition with $2$s. The maximum number of $4$s that can appear in the partition of the (even) positive integer $n$ is $\lfloor \frac{n}{4}\rfloor$.

Hence for every $k=0,\ldots,\lfloor\frac{n}{4}\rfloor$ there is a unique partition involving $k$ $4$s. So $$ p_n=\left\lfloor\dfrac{n}{4}\right\rfloor+1$$