OFFSET
0,2
COMMENTS
Central coefficients of (1 + 10*x + 24*x^2)^n. 2^n*LegendreP(n,k) yields the central coefficients of (1 + 2*k*x + (k^2-1)*x^2)^n, with g.f. 1/sqrt(1 - 4*k*x + 4*x^2).
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.
Eric Weisstein's World of Mathematics, Legendre Polynomial.
FORMULA
G.f.: 1/sqrt(1-20*x+4*x^2).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n,k)*binomial(2*(n-k), n)*5^(n-2*k).
D-finite with recurrence: n*a(n) +10*(1-2*n)*a(n-1) +4*(n-1)*a(n-2) = 0. - R. J. Mathar, Sep 26 2012
a(n) ~ sqrt(72+30*sqrt(6))*(10+4*sqrt(6))^n/(12*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = A059473(n,n). - Alois P. Heinz, Oct 05 2017
From Peter Bala, Nov 28 2021: (Start)
a(n) = (1/3)*Sum_{k >= n} binomial(k,n)^2*(2/3)^k.
a(n) = (4^n)*Sum_{k = 0..n} binomial(n,k)^2*(3/2)^k.
a(n) = (1/3)*(2/3)^n*hypergeometric2F1([n+1, n+1], [1], 2/3).
a(n) = (4^n)*hypergeometric2F1([-n, -n], [1], 3/2)
a(n) = [x^n] ((2*x - 2)*(3 - 2*x))^n.
a(n) = (2^n)*A006442(n). (End)
MATHEMATICA
Table[SeriesCoefficient[1/Sqrt[1-20*x+4*x^2], {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 14 2012 *)
Table[2^n*LegendreP[n, 5], {n, 0, 40}] (* G. C. Greubel, May 21 2023 *)
PROG
(PARI) a(n)=pollegendre(n, 5)<<n \\ Charles R Greathouse IV, Oct 25 2011
(Sage)
def A098270(n): return 2^n*gen_legendre_P(n, 0, 5)
[A098270(n) for n in (0..16)] # Peter Luschny, Oct 14 2012
(Magma) [2^n*Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // G. C. Greubel, May 21 2023
CROSSREFS
KEYWORD
easy,nonn,changed
AUTHOR
Paul Barry, Sep 01 2004
STATUS
approved