The following has strong experimental evidence.



> 
> **Question.** For $n\geq3k$, is this identity true? Proof?
>  $$\sum\_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-2k+j}{j,k-2j,n-3k+2j}=\sum\_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-k-2j-1}{n-2k-1,k-2j}.$$
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In search of a binomial identity proof
I use a common notation $[x^ay^b]f(x,y)$ for a coefficient of $x^ay^b$ in the series $f(x,y)$.


LHS equals $[t^kz^{n-2k}]\sum\_{j=0}^{k/2} (t^2+tz+z)^{n-2k+j}$, and we may extend the summation for $j$ from $2k-n$ to $\infty$ as our monomial $t^kz^{n-2k}$ does not appear in other summands. That is, 
$$LHS=[t^kz^{n-2k}]\sum\_{s=0}^{\infty} (t^2+tz+z)^{s}=[t^kz^{n-2k}]\frac1{1-t^2-tz-z^2}=\\
=[t^kz^{n-2k}]\frac1{2-z}\left(\frac1{1+t}+\frac1{1-t-z}\right)=[z^{n-2k}]\left(\frac{(-1)^k}{2-z}+\frac1{(1-z)^{k+1}(2-z)}\right).
$$
As for RHS, it equals, where $m=[k/2]$,
$$
[x^{n-2k-1}]\sum\_{j=0}^{m}(1+x)^{n-k-2j-1}=[x^{n-2k-1}](1+x)^{n-k-1}\sum\_{j=0}^{m} (1+x)^{-2j}=\\
[x^{n-2k-1}](1+x)^{n-k-1}\frac{1-(1+x)^{-2m-2}}{1-(1+x)^{-2}}=[x^{n-2k}]\frac{(1+x)^{n-k+1}-(1+x)^{n-k-1-2m}}{2+x}.
$$
Looks similar, but still bit different. Well, do not give up and prove that two expressions are equal. Denoting $z=-x$ in the expression for LHS we have to prove that 
$$
(1)\,\,\,\,[x^{n-2k}]\frac{(-1)^{n+k}-(1+x)^{n-k+1}+(-1)^n(1+x)^{-k-1}+(1+x)^{n-k-1-2m}}{2+x}=0.
$$
Expand a geometric progression $$\frac{(-1)^{n+k}+(-1)^n(1+x)^{-k-1}}{2+x}=\sum\_{s=0}^k(-1)^{n+s}(1+x)^{s-k-1}.$$
We may add if necessary a summand corresponding to $s=k+1$ (it has zero coefficient of $x^{n-2k}$ in any case) and get
$$
(-1)^{n+1}x\sum\_{s=0}^m (1+x)^{2s-k-1}.
$$
A coefficient of $x^{n-2k}$ equals $$(-1)^{n+1}\sum\_{s=0}^m\binom{2s-k-1}{n-2k-1}=\sum\_{s=0}^m\binom{n-k-2s-1}{n-2k-1}.$$
But two other terms of (1) give $$\frac{-(1+x)^{n-k+1}+(1+x)^{n-k-1-2m}}{2+x}=-x(1+x)^{n-k-1-2m}\sum\_{t=0}^{m}(1+x)^{2t}.$$
We see that a coefficient of $x^{n-2k-1}$ succesfully cancels with the previous expression. 


