I stumbled on the following identity, which has been checked numerically.

Question.Is this true? If so, any proof? $$\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-2k+j}{j,k-2j,n-3k+2j} =\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-k-2j-1}{k-2j}.$$

Here, $\binom{m}{a,b,c}$ is understood as $\frac{m!}{a!\,b!\,c!}$.