Let $A_n=\prod_{j=1}^{n}(3j-2)$, $B_n=\prod_{j=1}^{n}(3j-1)$, $C_n=\prod_{j=1}^{n}(3j)=3^n n!$.
For $A_0=B_0=C_0=1$.
Show that $C_n=\sum_{k=0}^{n}$${n}\choose{k}$$A_k B_{n-k}$.
I've tried to prove this inductively with no result. I've also tried to find the generating function for the $(C_n)_{n\in\Bbb{N}}$ sequence but I wasn't able to find anything, not even sure if one exists.
Any hints?
On
Observe that $ \frac{ (-1)^k A_k } { 3^k k!} = \frac{ (\frac{2}{3} - 1)(\frac{2}{3} - 2 ) \ldots ( \frac{ 2}{3} - j ) } { k!} = { -\frac{1}{3} \choose k } $.
Likewise, $ \frac{ (-1)^{n-k} A_{n-k} } { 3^{n-k} (n-k)!} = { - \frac{2}{3} \choose n-k } $.
Show via Vandermonde's identity (which holds for negative non-integer binomial coefficients too), that
$$\sum_{k=0}^n \frac{ (-1)^n A_nB_{n-k} } {3^n k! (n-k)!} = \sum_{k=0}^n { -\frac{1}{3}\choose k } { - \frac{2}{3} \choose n-k } = { -1 \choose n } = (-1)^n. $$
Hence, we are done.
The case $n=0$ is clearly true, let $n\in\mathbb{N}$ such that $$ C_n=\sum_{k=0}^n\binom{n}{k}A_k B_{n-k} $$ $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$ for all $k\in\mathbb{N}^*$ thus $$ \begin{aligned}\sum_{k=0}^{n+1}\binom{n+1}{k}A_k B_{n+1-k}&=B_{n+1}+\sum_{k=1}^{n+1}\binom{n}{k}A_k B_{n+1-k}+\sum_{k=1}^{n+1}\binom{n}{k-1}A_kB_{n+1-k} \\ &=\sum_{k=0}^{n}\binom{n}{k}A_kB_{n+1-k}+\sum_{k=0}^n\binom{n}{k}A_{k+1}B_{n-k} \end{aligned}$$ However $A_{p+1}=(3p+1)A_p$ and $B_{p+1}=(3p+2)B_p$ so that $$ \sum_{k=0}^{n}\binom{n}{k}A_kB_{n+1-k}=\sum_{k=0}^{n}\binom{n}{k}(3n-3k+2)A_kB_{n-k} $$ and $$ \sum_{k=0}^n\binom{n}{k}A_{k+1}B_{n-k}=\sum_{k=0}^n\binom{n}{k}(3k+1)A_kB_{n-k} $$ Finally, $$ \sum_{k=0}^{n+1}\binom{n+1}{k}A_kB_{n+1-k}=(3n+3)\sum_{k=0}^n\binom{n}{k}A_kB_{n-k}=3(n+1)C_n=C_{n+1} $$