Express $C(x)$ through $A(x)$ and $B(x)$, where $C(x)$, $A(x)$, $B(x)$, are generating functions of sequences $c_{n}$, $a_{n}$, $b_{n}$ respectively I'm trying to solve the following problem:
Given generating functions $A(x)$ for sequence $a_0, a_1, a_2, \dots$ and $B(x)$ for sequence $b_0, b_1, b_2, \dots$ express the generating function $C(x)$ for sequence $c_0, c_1, c_2, \dots$ through $A(x)$ and $B(x)$. The $n$-th term for sequence $c_n = \sum_{k=0}^{[n/3]}{a_kb_{n-3k}}$.
I understand that the answer will be some sort of convolution, however, I'm struggling with getting the modified $B(x)$. I think there will something like modulo, but I can't get my head around it.
Thanks in advance.
 A: Consider that, given
$$
A(z) = \sum\limits_{n\, \geqslant \,0} {\,a_{\,n} \,z^{\,n} } \;\quad \left| {\;a_{\,n\, < \,0}  = 0} \right.
$$
then
$$
\begin{gathered}
  \sum\limits_{0\, \leqslant \,k\, \leqslant \,m - 1} {A(\;e^{\,i\,k\frac{{2\,\pi }}
{m}} )}  = \sum\limits_{n\, \geqslant \,0} {\,a_{\,n} \sum\limits_{0\, \leqslant \,k\, \leqslant \,m - 1} {\left( {e^{\,i\frac{{2k\,\pi }}
{m}} } \right)^{\,n} } \,}  =  \hfill \\
   = \sum\limits_{n\, \geqslant \,0} {\,a_{\,n} \sum\limits_{0\, \leqslant \,k\, \leqslant \,m - 1} {\left( {e^{\,i\frac{{2n\,\pi }}
{m}} } \right)^{\,k} } \,}  = m\,\sum\limits_{n\, \geqslant \,0} {\,a_{\,n\,m} \,}  \hfill \\ 
\end{gathered} 
$$
and
$$
\sum\limits_{0\, \leqslant \,k\, \leqslant \,m - 1} {A(\;z^{\,\frac{1}
{m}} \,e^{\,i\,k\frac{{2\,\pi }}
{m}} )}  = m\,\sum\limits_{n\, \geqslant \,0} {a_{\,n\,m} \,z^{\,n} \,} 
$$
Can you proceed from here?
A: Answer is $С(x)=  B(x) \cdot A(x^3)$:
$$
(a_0 + a_1 \cdot x^3 + a_2 \cdot x^6 + \ldots) \cdot (b_0 + b_1 \cdot x + b_2 \cdot x^2 + b_3 \cdot x^3 + \ldots) = \\
(a_0 \cdot b_0) + (b_1 \cdot a_0) \cdot x + \;\ldots\;\\
 + 
(b_{10} \cdot a_0 + b_7 \cdot a_1 + b_4 \cdot a_2 + b_1 \cdot a_3) \cdot x^{10} 
+ \ldots.
$$
A: \begin{align}
C(x) &= \sum_{n\ge 0} c_n x^n \\
&= \sum_{n\ge 0} \sum_{k=0}^{\lfloor n/3 \rfloor} a_k b_{n-3k} x^n \\
&= \sum_{k\ge 0} a_k x^{3k} \sum_{n\ge 3k} b_{n-3k} x^{n-3k} \\
&= \sum_{k\ge 0} a_k x^{3k} B(x) \\
&= B(x) \sum_{k\ge 0} a_k (x^{3})^k \\
&= B(x) A(x^3)
\end{align}
