Let $F_m(x)=\binom{x+m}{m}$. Does anyone know how to prove or disprove that

$$ F_{m-1}(2x)=2^{m-1}F_m(x)+\sum_{d=2}^{\lfloor \frac{m}{2} \rfloor +1} \frac{(-1)^{d-1}}{2} \frac{m2^m}{(d-1)2^{d-1}} \binom{m-d}{d-2}F_{m-d}(x) $$

What I did : I have the checked the identity for $m\leq 10$.

Context : If true, the identity would help me a lot into turning a recent partial answer of mine into a full answer.

  • $\begingroup$ Have you tried applying snake oil? This looks pretty similar to some central binomial coefficient identities which that method handles well. $\endgroup$ – Badam Baplan May 30 at 14:44
  • $\begingroup$ @BadamBaplan Interesting idea, but I don't see how to implement it. $\sum_{m\geq 0}\binom{x+m}{x}t^m$ is not a value I know $\endgroup$ – Ewan Delanoy May 30 at 16:05
  • $\begingroup$ That's just $\frac{1}{(1-t)^{x+1}}$ $\endgroup$ – Badam Baplan May 30 at 17:10

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