cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A089447 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.

Original entry on oeis.org

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, 441, 56, 1, 1, 84, 1036, 3314, 3314, 1036, 84, 1, 1, 120, 2184, 10786, 18004, 10786, 2184, 120, 1, 1, 165, 4236, 30460, 77952, 77952, 30460, 4236, 165, 1, 1, 220, 7689, 77044
Offset: 0

Views

Author

Paul D. Hanna, Nov 02 2003

Keywords

Comments

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.

Examples

			Rows begin:
[1,   1,     1,      1,       1,        1,         1,          1, ...];
[1,   4,    10,     20,      35,       56,        84,        120, ...];
[1,  10,    48,    162,     441,     1036,      2184,       4236, ...];
[1,  20,   162,    841,    3314,    10786,     30460,      77044, ...];
[1,  35,   441,   3314,   18004,    77952,    284880,     912042, ...];
[1,  56,  1036,  10786,   77952,   435654,   2007456,    7951674, ...];
[1,  84,  2184,  30460,  284880,  2007456,  11427992,   55009548, ...];
[1, 120,  4236,  77044,  912042,  7951674,  55009548,  317112363, ...];
[1, 165,  7689, 178387, 2624453, 27870393, 231114465, 1576219474, ...]; ...
		

Crossrefs

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

Programs

  • 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}

A089449 Antidiagonal sums of square table A089447, which lists the coefficients 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.

Original entry on oeis.org

1, 2, 6, 22, 90, 396, 1837, 8870, 44186, 225628, 1175322, 6222788, 33392644, 181216728, 992829379, 5483790870, 30502513970, 170705626308, 960498281302, 5430200987260, 30830681187480, 175715526842056, 1004931956037782
Offset: 0

Views

Author

Paul D. Hanna, Nov 02 2003

Keywords

Crossrefs

Cf. A089447 (table), A089448 (diagonal), A002293.

Formula

G.f.: A(x) = sum(n>=0, Catalan(n+1)*x^n) + x^2*A(x)^4, where Catalan(n)=(2n)!/(n!*(n+1)!).
From Vaclav Kotesovec, Oct 10 2020: (Start)
G.f.: A(x) = (1 - Sqrt[1-4*x] - 2*x)/(2*x^2) + x^2*A(x)^4.
a(n) ~ sqrt(11) * 3^(15/2 + 3*n) / ((8 + 3*sqrt(3) + 4*sqrt(4 + 3*sqrt(3))) * sqrt((2519 + 528*sqrt(3) + 2*sqrt(1484692 + 881529*sqrt(3))) * Pi) * n^(3/2) * 2^(2*n + 5/2) * (-32 - 18*sqrt(3) + sqrt(1996 + 1233*sqrt(3)))^(n+2)). (End)
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