A084772 Coefficients of 1/sqrt(1 - 12*x + 16*x^2); also, a(n) is the central coefficient of (1 + 6*x + 5*x^2)^n.
1, 6, 46, 396, 3606, 33876, 324556, 3151896, 30915046, 305543556, 3038019876, 30354866856, 304523343996, 3065412858696, 30946859111256, 313206733667376, 3176825392214406, 32284147284682596, 328643023505612596
Offset: 0
Keywords
Examples
G.f.: 1/sqrt(1 - 2*b*x + (b^2-4*c)*x^2) yields central coefficients of (1 + b*x + c*x^2)^n.
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.
- H. A. Verrill, Sums of squares of binomial coefficients, with applications to Picard-Fuchs equations, arXiv:math/0407327 [math.CO], 2008, Theorem 8.
Crossrefs
Cf. A001850.
Programs
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GAP
List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*5^k)); # Muniru A Asiru, Jul 29 2018
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Magma
[4^n*Evaluate(LegendrePolynomial(n), 3/2) : n in [0..40]]; // G. C. Greubel, May 31 2023
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Mathematica
Table[n! SeriesCoefficient[E^(6 x) BesselI[0, 2 Sqrt[5] x], {x, 0, n}], {n, 0, 20}] (* Vincenzo Librandi, May 10 2013 *) CoefficientList[Series[1/Sqrt[1-12x+16x^2],{x,0,30}],x] (* Harvey P. Dale, Apr 17 2015 *) Table[4^n*LegendreP[n,3/2], {n,0,40}] (* G. C. Greubel, May 31 2023 *) a[n_] := 4^n*HypergeometricPFQ[{-n, n + 1}, {1}, -1/4]; Flatten[Table[a[n], {n,0,18}]] (* Detlef Meya, May 22 2024 *) -
PARI
for(n=0,30,t=polcoeff((1+6*x+5*x^2)^n,n,x); print1(t","))
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SageMath
[4^n*gen_legendre_P(n, 0, 3/2) for n in range(41)] # G. C. Greubel, May 31 2023
Formula
a(n) = Sum_{k=0..n} 5^k*C(n,k)^2. - Benoit Cloitre, Oct 26 2003
E.g.f.: exp(6*x)*BesselI(0, 2*sqrt(5)*x). - Paul Barry, Sep 20 2004
Asymptotic: a(n) ~ (1+sqrt(5))^(2*n+1)/(2*5^(1/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Sep 11 2012
D-finite with recurrence: n*a(n) = 6*(2*n-1)*a(n-1) - 16*(n-1)*a(n-2). - R. J. Mathar, Nov 09 2012
a(n) = 4^n * LegendreP(n, 3/2). - G. C. Greubel, May 31 2023
a(n) = 4^n*hypergeom([-n, n + 1], [1], -1/4). - Detlef Meya, May 22 2024
Comments