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Symmetric square table of coefficients, read by antidiagonals, where T(n,k) is the coefficient of x^n*y^k in f(x,y) that satisfies: f(x,y) = g(x,y) + xy*f(x,y)^4 and where g(x,y) satisfies: 1 + (x+y-1)*g(x,y) + xy*g(x,y)^2 = 0.
3

%I #15 Jun 07 2016 10:27:57

%S 1,1,1,1,4,1,1,10,10,1,1,20,48,20,1,1,35,162,162,35,1,1,56,441,841,

%T 441,56,1,1,84,1036,3314,3314,1036,84,1,1,120,2184,10786,18004,10786,

%U 2184,120,1,1,165,4236,30460,77952,77952,30460,4236,165,1,1,220,7689,77044

%N Symmetric square table of coefficients, read by antidiagonals, where T(n,k) is the coefficient of x^n*y^k in f(x,y) that satisfies: f(x,y) = g(x,y) + xy*f(x,y)^4 and where g(x,y) satisfies: 1 + (x+y-1)*g(x,y) + xy*g(x,y)^2 = 0.

%C Explicitly, g(x,y) = ((1-x-y)+sqrt((1-x-y)^2-4xy))/(2xy) = sum(n>=0, sum(k>=0, N(n,k)*x^n*y^k), where N(n,k) are the Narayana numbers: N(n,k) = C(n+k,k)*C(n+k+2,k+1)/(n+k+2). This array is directly related to sequence A002293, which has a g.f. h(x) that satisfies h(x) = 1 + x*h(x)^4. The inverse binomial transform of the rows grows by three terms per row.

%e Rows begin:

%e [1, 1, 1, 1, 1, 1, 1, 1, ...];

%e [1, 4, 10, 20, 35, 56, 84, 120, ...];

%e [1, 10, 48, 162, 441, 1036, 2184, 4236, ...];

%e [1, 20, 162, 841, 3314, 10786, 30460, 77044, ...];

%e [1, 35, 441, 3314, 18004, 77952, 284880, 912042, ...];

%e [1, 56, 1036, 10786, 77952, 435654, 2007456, 7951674, ...];

%e [1, 84, 2184, 30460, 284880, 2007456, 11427992, 55009548, ...];

%e [1, 120, 4236, 77044, 912042, 7951674, 55009548, 317112363, ...];

%e [1, 165, 7689, 178387, 2624453, 27870393, 231114465, 1576219474, ...]; ...

%o (PARI) {L=10; T=matrix(L,L,n,k,1); for(n=1,L-1, for(k=1,L-1, T[n+1,k+1]=binomial(n+k,k)*binomial(n+k+2,k+1)/(n+k+2)+ sum(j3=1,k,sum(i3=1,n,T[n-i3+1,k-j3+1]* sum(j2=1,j3,sum(i2=1,i3,T[i3-i2+1,j3-j2+1]* sum(j1=1,j2,sum(i1=1,i2,T[i2-i1+1,j2-j1+1]*T[i1,j1])); )); )); )); T}

%Y Cf. A089448 (diagonal), A089449 (antidiagonal sums), A086617, A088925, A002293.

%K nonn,tabl

%O 0,5

%A _Paul D. Hanna_, Nov 02 2003