A098270 a(n) = 2^n*P_n(5), 2^n times the Legendre polynomial of order n at 5.
1, 10, 148, 2440, 42256, 752800, 13660480, 251113600, 4660568320, 87140108800, 1638884021248, 30970912737280, 587599919386624, 11185644310405120, 213540626285805568, 4086692369433395200, 78378887309200261120
Offset: 0
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.
Crossrefs
Programs
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Magma
[2^n*Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // G. C. Greubel, May 21 2023
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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 *) -
PARI
a(n)=pollegendre(n,5)<
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
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)
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