A202476 G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (1 + k*x + x^2).

1, 1, 2, 5, 11, 28, 74, 206, 601, 1826, 5766, 18851, 63676, 221678, 793958, 2920292, 11014653, 42543773, 168074091, 678403932, 2794920078, 11742254750, 50266213000, 219085792538, 971543475593, 4380664101448, 20071848941411, 93403455862117, 441206005123701

Offset: 0

Author

Paul D. Hanna, Dec 19 2011

Keywords

• nonn

Examples

			The coefficients in Product_{k=1..n} (1+k*x+x^2), n>=0, form the triangle:
[1];
[1, 1, 1];
[1, 3, 4, 3, 1];
[1, 6, 14, 18, 14, 6, 1];
[1, 10, 39, 80, 100, 80, 39, 10, 1];
[1, 15, 90, 285, 539, 660, 539, 285, 90, 15, 1];
[1, 21, 181, 840, 2339, 4179, 5038, 4179, 2339, 840, 181, 21, 1];
[1, 28, 329, 2128, 8400, 21392, 36630, 43624, 36630, 21392, 8400, 2128, 329, 28, 1]; ...
the antidiagonal sums of which form this sequence.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A201951, A201826, A202474.

Programs

PARI

{a(n)=sum(k=0,n,polcoeff(prod(j=1,n-k,1+j*x+x^2),k))}

PARI

{a(n)=local(CF=1+x+x*O(x^n)); for(k=1, n-1, CF=(1+(n-k)*x+x^2)/(1 + x*(1+(n-k)*x+x^2) - x*CF+x*O(x^n))); polcoeff(1/(1-x*CF), n)}

Formula

Antidiagonal sums of the irregular triangle in which row n is defined by the g.f.: Product_{k=1..n} (1 + k*x + x^2) for n>=0.

G.f.: 1/(1 - x*(1+x+x^2)/(1 + x*(1+x+x^2) - x*(1+2*x+x^2)/(1 + x*(1+2*x+x^2) - x*(1+3*x+x^2)/(1 + x*(1+3*x+x^2) - x*(1+4*x+x^2)/(1 + x*(1+4*x+x^2) -...))))), a continued fraction.