A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9*x + 20*x^2)^n.
1, 9, 121, 1809, 28401, 458649, 7544041, 125700129, 2114588641, 35836273449, 610897146201, 10463745263409, 179939616743121, 3104680678772409, 53721299280288201, 931852905510160449, 16198821321758152641
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.
- Vaclav Kotesovec, Asymptotic of a sums of powers of binomial coefficients * x^k, 2012.
- Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
Crossrefs
Programs
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Magma
[Evaluate(LegendrePolynomial(n), 9) : n in [0..40]]; // G. C. Greubel, May 17 2023
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Mathematica
Table[SeriesCoefficient[1/Sqrt[1-18*x+x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *) LegendreP[Range[0, 40], 9] (* G. C. Greubel, May 17 2023 *) a[n_] := 4^n*Sum[(5/4)^k*Binomial[n, k]^2, {k, 0, n}]; Table[a[n], {n, 0, 16}] (* Detlef Meya, May 22 2024 *)
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PARI
for(n=0,30,print1(subst(pollegendre(n),x,9)","))
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PARI
{a(n)=sum(k=0, n, binomial(n, k)^2*4^k*5^(n-k))} \\ Paul D. Hanna, Sep 29 2012
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SageMath
[gen_legendre_P(n,0,9) for n in range(41)] # G. C. Greubel, May 17 2023
Formula
G.f.: 1/sqrt(1-18*x+x^2).
Also a(n) = (n+1)-th term of the binomial transform of 1/(1-4x)^(n+1).
From Paul Barry, May 25 2005: (Start)
E.g.f.: exp(9*x) * Bessel_I(0, 2*sqrt(20)*x).
a(n) = Sum_{k=0..n} C(n, k)*C(n+k, k)4^k. (End)
D-finite with recurrence: n*a(n) + 9*(1-2*n)*a(n-1) + (n-1)*a(n-2) = 0. - R. J. Mathar, Sep 27 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 4^k * 5^(n-k). - Paul D. Hanna, Sep 29 2012
a(n) ~ sqrt(200 + 90*sqrt(5))*(9 + 4*sqrt(5))^n/(20*sqrt(Pi*n)) = (2 + sqrt(5))^(2*n+1)/(5^(1/4)*2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = hypergeom([-n, n+1], [1], -4). - Peter Luschny, May 23 2014
x*exp(Sum_{n >= 1} a(n)*x^n/n) = x + 9*x^2 + 101*x^3 + 1269*x^4 + ... is an integral power series, the o.g.f. for A269732. - Peter Bala, Jan 25 2018
a(n) = (-1)^n * Sum_{k=0..n} (1/18)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 28 2025
a(n) = Sum_{k=0..floor(n/2)} 20^k * 9^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, Aug 30 2025