We refer to this article by Ira Gessel. In section 6, page 10, equation (28), the Super Catalan numbers are defined as $$S(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}.$$ On page 12, equation (31), there goes the identity $$S(m,n)=\sum_{k=0}^m(-1)^k\binom{2n+2k}{n+k}\binom{m}k4^{m-k}. \tag1$$ In another article by Ira Gessel and Guoce Xin, the authors introduced the notation $$T(m,n)=\frac12\frac{(2m)!(2n)!}{m!n!(m+n)!}.$$ On pages 1 and 2, in their attempt at a combinatorial proof they mentioned the two relations $$T(2,n)=4C_n-C_{n+1} \qquad \text{and} \qquad T(3,n)=16C_n-8C_{n+1}+C_{n+2}$$ where $C_n=\frac1{n+1}\binom{2n}n$ are the Catalan numbers. These became my motivation to experiment further and I came up with another version of (1) in the form $$S(m,n)=\frac12\sum_{j=0}^{m-1}(-1)^jC_{n+j}\binom{m-1}j 4^{m-j}. \tag2$$
QUESTION. It appears that (1) and (2) are very nearly the same, can one be directly transformed to the other without employing $S(m,n)$ as a "negotiator"?
Remark. After seeing the above two identities, Gessel showed me a generalization of them both (private communication).
