Blogrolling On II

Recall that binary trees satisfied the Catalan relation $T=1+T^2$, whereas Schroeder numbers, or Motzkin numbers, satisfy the substitution rule $X=1+X+X^2$. In this paper, Cossali studies a generalisation of the generating function for the Catalan numbers $C_n$. Let

$L(x,z)=\sum _n\sum _m{C}_{m}B(2m+n,n)x^m{z}^n=J(x,z/x)$

where $B(i,j)$ is the binomial coefficient. The substitution rules above are all united by the function $J(x,y)$ which satisfies the rule

$x{J}^2+xyJ+1=J$

When $x=1$ and $y=0$ this reduces to $J=1+J^2$, the rule for counting the number of vertices on an associahedron, but when $x=y=1$ it reduces to the rule $J^2+J+1=J$. In general, the generating function takes the form

$J(x,y)=\frac{1}{2x}\left[\right(1-yx)\pm \sqrt{(1-yx{)}^2-4x}]$

Note that the coefficients $g(m,n)=C_{m}B(2m+n,n)$ of the double series $L(x,z)$ could only give the Schroeder numbers when $x=z$, so by summing the diagonals of the table on page 4 (that is, summing the $g(m,n)$ of weight $i$, starting with $g(0,0)=1$ for $i=0$) the Schroeder numbers are obtained. This proves that the Schroeder numbers $S_i$ are given by

$S_i=\sum _{m+n=i}{C}_{m}B(2m+n,n)$

which is a little nicer than the hypergeometric expression $2F1(1-i,2+i;2;-1)$. By setting $y=-1$ the coefficients are alternating sums of the $g(m,n)$ and this leads to the sequence $1,0,0,0,0,0,...$ The Motzkin rule follows from setting $y={\sqrt{x}}^{-1}$ and then working with $z=\sqrt{x}$, which one may verify by looking at the generating function. Can we generate more cute Abel sums using $J(x,y)$?