A115864 Legendre_P(n,2)*4^n.

1, 8, 88, 1088, 14176, 190208, 2600704, 36030464, 504047104, 7104278528, 100726755328, 1435037302784, 20526579564544, 294599134674944, 4240277467168768, 61183611081064448, 884741809748967424

Offset: 0

Author

Paul Barry, Feb 01 2006

Keywords

• easy
• nonn

Comments

Central coefficients of (1+8x+12x^2)^n. In general, Jacobi_P(n,0,0,sqrt(m))(k*sqrt(m))^n=Legendre_P(n,sqrt(m))(k*sqrt(m))^n has g.f. 1/sqrt(1-2*k*m*x+k^2*x^2), e.g.f. exp(k*m*x)Bessel_I(0,k*sqrt(m(m-1))*x) and gives the central coefficients of (1+k*m*x+k^2*(m(m-1)/4)*x^2)^n.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
• Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

Programs

Mathematica

CoefficientList[Series[1/Sqrt[1-16*x+16*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

PARI

a(n)=pollegendre(n,2)<<(2*n) \\ Charles R Greathouse IV, Mar 18 2017

Formula

G.f.: 1/sqrt(1-16x+16x^2);

E.g.f.: exp(8x)Bessel_I(0,2*sqrt(12)x);

a(n)=Jacobi_P(n,0,0,sqrt(4))*(2*sqrt(4))^n; a(n)=2^n*A069835(n).

D-finite with recurrence: n*a(n) +8*(1-2*n)*a(n-1) +16*(n-1)*a(n-2) =0. - R. J. Mathar, Nov 16 2011

a(n) ~ sqrt(18+12*sqrt(3))*(8+4*sqrt(3))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 19 2012

a(n) = 2^n*A069835(n). - R. J. Mathar, Jan 20 2020