A116383 Row sums of number triangle A116382.

1, 1, 4, 6, 19, 33, 93, 175, 460, 910, 2286, 4676, 11388, 23842, 56808, 120926, 283611, 611065, 1416625, 3079635, 7078263, 15490553, 35374519, 77805481, 176813809, 390379483, 883861033, 1957097715, 4418562265, 9805546875, 22090136885

Offset: 0

Author

Paul Barry, Feb 12 2006

Keywords

• easy
• nonn

Comments

Binomial transform is A116387.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Programs

GAP

List([0..40], n-> Sum([0..n], k-> Sum([0..n], j-> (-1)^(n-j)* Binomial(n,j)*Sum([0..j], m-> Binomial(j,m-k)*Binomial(m,j-m) )))); # G. C. Greubel, May 22 2019

Magma

T:= func< n,k | (&+[(-1)^(n-j)*Binomial(n,j)*(&+[Binomial(j,m-k)* Binomial(m,j-m): m in [0..j]]): j in [0..n]]) >;
[(&+[T(n,k): k in [0..n]]): n in [0..40]]; // G. C. Greubel, May 22 2019

Magma

R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( (1+x-4*x^2)/(2*(1-5*x^2)*Sqrt(1-4*x^2)) + (1+x)/(2*(1-5*x^2)) )); // G. C. Greubel, May 22 2019

Mathematica

CoefficientList[Series[(1+x-4*x^2)/(2*(1-5*x^2)*Sqrt[1-4*x^2])+(1+x)/(2*(1-5*x^2)), {x, 0, 40}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

PARI

my(x='x+O('x^40)); Vec((1+x-4*x^2)/(2*(1-5*x^2)*sqrt(1-4*x^2)) + (1+x)/(2*(1-5*x^2))) \\ G. C. Greubel, May 22 2019

Sage

((1+x-4*x^2)/(2*(1-5*x^2)*sqrt(1-4*x^2)) + (1+x)/(2*(1-5*x^2))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 22 2019

Formula

G.f.: (1+x-4*x^2)/(2*(1-5*x^2)*sqrt(1-4*x^2)) + (1+x)/(2*(1-5*x^2)).

a(n) = Sum_{k=0..n} Sum_{j=0..n} (-1)^(n-j)*C(n,j)*Sum_{i=0..j} C(j,i-k) * C(i,j-i).

Conjecture: n*a(n) + (n-2)*a(n-1) + (-13*n+20)*a(n-2) + (-9*n+22)*a(n-3) + 4*(14*n-45)*a(n-4) + 20*(n-3)*a(n-5) + 80*(-n+5)*a(n-6) = 0. - R. J. Mathar, Nov 24 2012

Recurrence: n*(n^2 - 4*n - 1)*a(n) = 4*(2*n-3)*a(n-1) + (9*n^3 - 40*n^2 + 3*n + 48)*a(n-2) - 20*(2*n-3)*a(n-3) - 20*(n-3)*(n^2 - 2*n - 4)*a(n-4). - Vaclav Kotesovec, Feb 12 2014

a(n) ~ (5+sqrt(5))/10 * 5^(n/2). - Vaclav Kotesovec, Feb 12 2014