A098269 a(n) = 2^n*P_n(4), 2^n times the Legendre polynomial of order n at 4.

1, 8, 94, 1232, 16966, 240368, 3468844, 50712992, 748553926, 11131168688, 166498969924, 2502416381792, 37759888297756, 571681667171168, 8679980422677784, 132116085646644032, 2015249400937940806

Offset: 0

Author

Paul Barry, Sep 01 2004

Keywords

• easy
• nonn

Comments

Central coefficients of (1+8x+15x^2)^n. 2^n*LegendreP(n,k) yields the central coefficients of (1+2kx+(k^2-1)x^2)^n, with g.f. 1/sqrt(1-4kx+4x^2).

16th binomial transform of 2^n*LegendreP(n,-4) = (-1)^n*A098269(n). - Paul Barry, Sep 03 2004

Diagonal of rational functions 1/(1 + x + 3*y + x*z - 2*x*y*z), 1/(1 - x + y + 3*x*z - 2*x*y*z), 1/(1 - x - x*y - 3*y*z - 2*x*y*z). - Gheorghe Coserea, Jul 03 2018

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.

Crossrefs

Cf. A069835, A084773.

Programs

Mathematica

Table[SeriesCoefficient[1/Sqrt[1-16*x+4*x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
a[n_] := 3^n*HypergeometricPFQ[{-n, -n}, {1}, 5/3]; Flatten[Table[a[n], {n,0,16}]] (* Detlef Meya, May 21 2024 *)

PARI

a(n)=pollegendre(n,4)<Charles R Greathouse IV, Oct 24 2011

PARI

{a(n)=sum(k=0, n, binomial(n, k)^2*3^k*5^(n-k))} \\ Paul D. Hanna, Sep 29 2012

Formula

G.f.: 1/sqrt(1-16x+4x^2).

a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n, k)*binomial(2(n-k), n)*4^(n-2k).

E.g.f.: exp(8*x)*BesselI(0, 2*sqrt(15)*x), cf. A084770. - Vladeta Jovovic, Sep 01 2004

a(n) = Sum_{k=0..n} binomial(n,k)^2 * 3^k * 5^(n-k). - Paul D. Hanna, Sep 29 2012

D-finite with recurrence: n*a(n) = 8*(2*n-1)*a(n-1) - 4*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 14 2012

a(n) ~ sqrt(450+120*sqrt(15))*(8+2*sqrt(15))^n/(30*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012

a(n) = 3^n*hypergeom([-n, -n], [1], 5/3) = 5^n*hypergeom([-n, -n], [1], 3/5). - Detlef Meya, May 21 2024