A084770 Coefficients of 1/(1-4x-16x^2)^(1/2); also, a(n) is the central coefficient of (1+2x+5x^2)^n.
1, 2, 14, 68, 406, 2332, 13964, 83848, 509926, 3118892, 19194724, 118654648, 736365436, 4584612632, 28623792344, 179142212368, 1123532958086, 7059622447052, 44431918660724, 280059644507608, 1767597777222676
Offset: 0
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. G.f. = 1 + 2*x + 14*x^2 + 68*x^3 + 406*x^4 + 2332*x^5 + 13964*x^6 + 83848*x^7 + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
- 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.
- Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
- Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
Programs
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Magma
[n le 2 select 2^(n-1) else (2*(2*n-3)*Self(n-1) + 16*(n-2)*Self(n-2))/(n-1): n in [1..30]]; // G. C. Greubel, May 30 2023
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Mathematica
Table[n!*SeriesCoefficient[E^(2*x)*BesselI[0,2*Sqrt[5]*x],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *) Table[Abs[LegendreP[n, I/2]] 4^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 22 2015 *) a[n_]:= (4/I)^n LegendreP[n, I/2]; (* Michael Somos, Sep 30 2017 *) -
PARI
for(n=0,30,t=polcoeff((1+2*x+5*x^2)^n,n,x); print1(t","))
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PARI
a(n) = 4^n*abs(pollegendre(n, I/2)) \\ after 2nd Mathematica; Michel Marcus, Oct 22 2015
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PARI
{a(n) = (4/I)^n * pollegendre(n, I/2)}; /* Michael Somos, Sep 30 2017 */ -
SageMath
[(-4*i)^n*gen_legendre_P(n, 0, i/2) for n in range(41)] # G. C. Greubel, May 30 2023
Formula
E.g.f.: exp(2*x) * BesselI(0, 2*sqrt(5)*x). More generally, e.g.f.: exp(b*x) * BesselI(0, 2*sqrt(c)*x) yields central coefficients of (1+b*x+c*x^2)^n. - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(2(n-k), n)*4^k. - Paul Barry, Sep 08 2004
Define Q(n, x) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(2(n-k), n)*x^(n-2k). A084770(n) is 2^n*Q(n, 1/2). - Paul Barry, Sep 08 2004
Recurrence: n*a(n) = 2*(2*n-1)*a(n-1) + 16*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ sqrt(50+10*sqrt(5))*(2+2*sqrt(5))^n/(10*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
G.f.: G(0), where G(k)= 1 + 4*x*(1+4*x)*(4*k+1)/(4*k+2 - 4*x*(1+4*x)*(4*k+2)*(4*k+3)/(4*x*(1+4*x)*(4*k+3) + 4*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
a(n) = 2^n * hypergeom([(1-n)/2, -n/2], [1], 5). - Vladimir Reshetnikov, Oct 10 2016
a(n) = (4/i)^(2*n+1) * a(-1-n), and 0 = a(n)*(+256*a(n+1) + 96*a(n+2) - 32*a(n+3)) + a(n+1)*(+32*a(n+1) + 16*a(n+2) - 6*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Sep 30 2017
From Seiichi Manyama, Aug 30 2025: (Start)
a(n) = Sum_{k=0..n} (1-2*i)^k * (1+2*i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit.
a(n) = Sum_{k=0..floor(n/2)} 5^k * 2^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)
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