Motzkin triangles (OEIS A064189) $[T_{n,k}]$ are the Riordan arrays $(M(x),xM(x))$, where $(M(x))$ is the g.f. for the Motzkin numbers(OEIS A001006). The OEIS page shows that $$T_{n,k}=\frac{k}{n}\sum_{j=0}^n \binom{n}{j}\binom{j}{2j-n-k}=\sum_{j=0}^n\binom{n}{j}(\binom{n-j}{j+k}-\binom{n-j}{j+k+2}).$$ Can anyone show some references for this result? Thanks very much.
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$\begingroup$ What is actually in OEIS is the following: $\texttt{T(0,0)=1, T(n,k)=0 if n < k, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1). }$ $\texttt{T(n,k) = Sum_{j=0..n} C(n,j)*(C(n-j,j+k) - C(n-j,j+k+2)). - Paul Barry, Feb 16 2006}$ $\texttt{T(n,k) = (k/n)*Sum_{j=0..n} binomial(n,j)*binomial(j,2*j-n-k). - Vladimir Kruchinin, Feb 12 2011}$ $\texttt{T(n,k) = binomial(n, k)*hypergeom([(k-n)/2, (k-n+1)/2], [k+2], 4). - Peter Luschny, May 19 2021}$ Have you tried contacting Barry, Kruchinin or Luschny? $\endgroup$– SomosCommented Jun 18 at 20:50
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$\begingroup$ Thanks for your advice. I contacted Barry and Kruchinin, and have received their reply. $\endgroup$– xmchenhitCommented Jun 20 at 0:38
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I contacted Barry and Kruchinin, and they kindly provide some information that is very helpful to me. According to their reply, the result can be deduced from the Lagrange inversion formula, by noticing that the g.f. for the $k$-th column of $\{T_{n,k}\}$ is $x^k(M(x))^{k+1}$, and $xM(x)=xg(XM(x))$ with $g(x)=1+x+x^2$. Thanks Barry and Kruchinin.