I am trying to prove the following relation. For $k \in \mathbb{N},k \geq 2$ and $i=1,\ldots,k-1$ \begin{equation} {2k-2-i \choose k-1-i}=\sum \limits_{l=0}^{k-1-i} 2^{2l} T(k-2-l,k-1-i-l)-2 \sum \limits_{l=0}^{k-2-i} 2^{2l} T(k-2-l,k-2-i-l) \end{equation} where \begin{equation} T(w,u):=\frac{(w+u)!(w-u+1)}{u!(w+1)!} \end{equation} are the entries in the Catalan triangle for $w \geq 0$ and $ w \geq u \geq 0$ (else T(w,u)=0).
I tried induction over $i$ or use the fact that the 2nd sum is (without the "$2$" factor) the 1st sum evaluated at $i+1$.
Could you please help me? Thank you.
Edit: If you write the rhs above in terms of factorials, one obtains: \begin{align*} \sum \limits_{l=0}^{k-1-i} 2^{2l} \frac{(2k-i-3-2l)! i}{(k-1-i-l)!(k-1-l)!} -2 \sum \limits_{l=0}^{k-2-i} 2^{2l} \frac{(2k-i-4-2l)! (i+1)}{(k-2-i-l)!(k-1-l)!}. \end{align*} One can also write $2^{2l}$ in terms of binomial coefficients. \begin{align*} &\sum \limits_{l=0}^{k-1-i} \sum \limits_{j=0}^{l} \binom{2l-j}{l-j} 2^{j} \frac{(2k-i-3-2l)! i}{(k-1-i-l)!(k-1-l)!} -2 \sum \limits_{l=0}^{k-2-i} \sum \limits_{j=0}^{l} \binom{2l-j}{l-j} 2^{j}\frac{(2k-i-4-2l)! (i+1)}{(k-2-i-l)!(k-1-l)!}\\ &=\sum \limits_{l=0}^{k-1-i} \sum \limits_{j=0}^{l} \sum \limits_{r=0}^{j} \binom{2l-j}{l-j} \binom{j}{r} \frac{(2k-i-3-2l)! i}{(k-1-i-l)!(k-1-l)!} -2 \sum \limits_{l=0}^{k-2-i} \sum \limits_{j=0}^{l} \sum \limits_{r=0}^{j} \binom{2l-j}{l-j} \binom{j}{r}\frac{(2k-i-4-2l)! (i+1)}{(k-2-i-l)!(k-1-l)!}. \end{align*} The 1st equation comes from the question "partial sum involving factorials" in this forum and the 2nd one from \begin{equation} \sum \limits_{k=0}^{n} \binom{n}{k}=2^{k}.\\ \end{equation} However I have no clue how to continue.