A084774 Coefficients of 1/sqrt(1-14*x+9*x^2); also, a(n) is the central coefficient of (1+7x+10x^2)^n.
1, 7, 69, 763, 8881, 106407, 1298949, 16065483, 200630241, 2524253767, 31947470149, 406281388443, 5187375332881, 66454791792487, 853788052488069, 10996378059281643, 141934540736139201, 1835494145265388167, 23776671158743933509, 308463567293772941883
Offset: 0
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Paul Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), #13.5.1.
- 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.
- Vaclav Kotesovec, Asymptotic of a sums of powers of binomial coefficients * x^k, 2012.
Programs
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GAP
List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*2^k*5^(n-k))); # Muniru A Asiru, Jul 29 2018
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Magma
[3^n*Evaluate(LegendrePolynomial(n), 7/3) : n in [0..40]]; // G. C. Greubel, May 31 2023
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Mathematica
Table[Sum[Binomial[n,k]^2*2^k*5^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *) Table[n! SeriesCoefficient[E^(7 x) BesselI[0, 2 Sqrt[10] x], {x,0,n}], {n,0,20}] (* Vincenzo Librandi, May 10 2013 *) Table[3^n*LegendreP[n, 7/3], {n,0,40}] (* G. C. Greubel, May 31 2023 *) a[n_] := 3^n*HypergeometricPFQ[{-n, n + 1}, {1}, -2/3]; Flatten[Table[a[n], {n,0,19}]] (* Detlef Meya, May 22 2024 *) -
PARI
for(n=0,30,t=polcoeff((1+7*x+10*x^2)^n,n,x); print1(t","))
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PARI
{a(n)=sum(k=0, n, binomial(n, k)^2*2^k*5^(n-k))} \\ Paul D. Hanna, Sep 28 2012 -
SageMath
[3^n*gen_legendre_P(n, 0, 7/3) for n in range(41)] # G. C. Greubel, May 31 2023
Formula
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 2^k * 5^(n-k). - Paul D. Hanna, Sep 28 2012
D-finite with recurrence: n*a(n) = 7*(2*n-1)*a(n-1) - 9*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ sqrt(200 + 70*sqrt(10))*(7 + 2*sqrt(10))^n/(20*sqrt(Pi*n)) = (sqrt(2) + sqrt(5))^(2*n+1)/(2*10^(1/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = 3^n * LegendreP(n, 7/3). - G. C. Greubel, May 31 2023
a(n) = 3^n*hypergeom([-n, n + 1], [1], -2/3). - Detlef Meya, May 22 2024
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