Chu-Vandermonde Identity for Trinomial Coefficients

Question

It seems as though there is a natural extension to trinomial coefficients. Wikipedia give a variant of the trinomial coefficient as

$$(1+x+x^2)^n=\sum_{k=0}^{2n}\binom{n}{k}_2x^k$$

Given then that you take the convolution of

$$(1+x+x^2)^m(1+x+x^2)^{n-m}=\sum_{k=0}^{2m}\binom{m}{k}_2x^k\sum_{k=0}^{2(n-m)}\binom{n}{k}_2x^k$$

$$=\sum_{k=0}^{2n}\sum_{j=0}^{2k}\binom{m}{j}_2\binom{n-m}{k-j}_2x^k=\sum_{k=0}^n\binom{n}{k}_2x^k=(1+x+x^2)^n$$

Is this right?

$$\sum_{j=0}^{2k}\binom{m}{j}_2\binom{n-m}{k-j}_2=\binom{n}{k}_2$$

Answer 1

Yes.

More generally, if $f(z)=\sum_{k=0}^{\infty} a_kz^k$. Then if we have that:

$$f(z)^n = \sum_{k=0}^{\infty} A_{n,k}z^k$$

then you have that $f(z)^nf(z)^m=f(z)^{n+m}$, which means that:

$$A_{m+n,k} = \sum_{i=0}^{k} A_{m,i}A_{n,k-i}$$