# Combinatorial identity with numerator doubled in binomial coefficient

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.

• Have you tried applying snake oil? This looks pretty similar to some central binomial coefficient identities which that method handles well. – Badam Baplan May 30 at 14:44
• @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 – Ewan Delanoy May 30 at 16:05
• That's just $\frac{1}{(1-t)^{x+1}}$ – Badam Baplan May 30 at 17:10