if $\sum_0^\infty a_n x^n = (\sum_0^\infty x^n )(\sum_0^\infty x^{2n})$ what is $a_n$? if $$\sum_0^\infty a_n x^n = (\sum_0^\infty x^n )(\sum_0^\infty x^{2n})$$
what is $a_n$?
Here is my approach 
let $b_n= 1$ and $c_n= x^n$
Then by forming / relating to cauchy product we can conlude that the product is equal to : 
$$\sum_0^\infty a_n x^n$$
where $a_n = \sum_{k=0}^{n} b_k c_{n-k} = \sum x^{n-k}= x^n+x^{n-1}+...+x^0 = \frac{1-x^{n+1}}{1-x}$
What do you think about my approach? I feel like there is something wrong with it. If not - could you provide a different approach? By the way depending on the way I solve - I keep finding different answers. I also ended up with $a_n=\frac{3+2n+(-1)^n}{4}$ without using cauchy product.

Let us write the first sum on the RHS as a geometrical series:
$$\sum_{n=0}^\infty a_n x^n =\dfrac{1}{1-x}\sum_{n=0}^\infty x^{2n} \ \ \iff$$
$$(1-x)\sum_{k=0}^\infty a_k x^k =\sum_{n=0}^\infty x^{2n} \ \ \iff$$
$$\tag{1}\sum_{k=0}^\infty a_{k}x^k - \sum_{k=0}^\infty a_{k}x^{k+1} = \sum_{n=0}^\infty x^{2n}.$$
Setting $K:=k+1 \ \iff \ k=K-1$, (1) becomes:
$$\sum_{k=0}^\infty a_{k}x^k - \sum_{K=1}^\infty a_{K-1}x^{K} = \sum_{n=0}^\infty x^{2n}$$
$$a_0+\sum_{k=1}^\infty (a_{k}-a_{k-1}) x^k =\sum_0^\infty x^{2n} $$
An immediate consequence is that $a_0=1$ ; then, for positive indices, identitifying the coefficients in the LHS and RHS, two cases occur, depending if they correspond to odd or even exponents:
$$\begin{cases}k&=&2n+1& \ \implies \ & a_{2n+1}-a_{2n}=0\\k&=&2n& \ \implies \ & a_{2n}-a_{2n-1}=1 \end{cases}$$
In other words: $$\begin{cases}a_{2n+1}=a_{2n}\\a_{2n}=a_{2n-1}+1 \end{cases}.$$
Summarizing, we have a unit increase at even steps, and no increase at odd steps.
Hence, taking into account value $a_0=1$, we get the following explicit formula for the general term:

$$a_n=\left\lfloor \frac{n}{2} \right\rfloor+1$$

