A084605 G.f.: 1/(1-2x-15x^2)^(1/2); also, a(n) is the central coefficient of (1+x+4x^2)^n.
1, 1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041, 148331665, 716698425, 3462260265, 16786700865, 81464917185, 396215601225, 1929237099225, 9408084660945, 45928695279345, 224476389327705
Offset: 0
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1433 (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.
- Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 96.
- Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, Some matrix identities on colored Motzkin paths, Discrete Mathematics 340.12 (2017): 3081-3091.
Programs
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Maple
a := n -> simplify(2^n*GegenbauerC(n,-n, -1/4)): seq(a(n), n=0..22); # Peter Luschny, May 08 2016
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Mathematica
Table[n!*SeriesCoefficient[E^x*BesselI[0,4*x],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *) a[n_] := Hypergeometric2F1[1/2 - n/2, -n/2, 1, 16]; Table[a[n], {n, 0, 22}] (* Peter Luschny, Mar 18 2018 *) -
PARI
for(n=0,30,t=polcoeff((1+x+4*x^2)^n,n,x); print1(t",")) for(n=0,20,print1(a(n),", "))
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PARI
{a(n) = sum(k=0,n, (-3)^(n-k)*2^k*binomial(n,k)*binomial(2*k,k))} for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Dec 09 2018
Formula
E.g.f.: exp(x)*BesselI(0, 4*x). - Vladeta Jovovic, Aug 20 2003
a(n) is also the central coefficient of (4+x+x^2)^n; a(n) = Sum_{k=0..n} 3^(n-k) C(n,k) T(k,n), where T(k,n) is the triangle of trinomial coefficients = Coefficient of x^n of (1+x+x^2)^k : A027907. - N-E. Fahssi, Mar 30 2008
a(n) = (1/Pi)*integral(x=-2..2, (2*x+1)^n/sqrt((2-x)*(2+x))). - Peter Luschny, Sep 12 2011
D-finite with recurrence a(n+2) = ((2*n+3)*a(n+1) + 15*(n+1)*a(n))/(n+2); a(0)=a(1)=1 - Sergei N. Gladkovskii, Aug 01 2012
a(n) ~ 5^(n+1/2)/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = 2^n*GegenbauerC(n, -n, -1/4). - Peter Luschny, May 08 2016
a(n) = hypergeom([1/2 - n/2, -n/2], [1], 16). - Peter Luschny, Mar 18 2018
a(n) = Sum_{k=0..n} (-3)^(n-k) * 2^k * binomial(n,k)*binomial(2*k,k). - Paul D. Hanna, Dec 09 2018
a(n) = Sum_{k=0..n} 5^(n-k) * (-2)^k * binomial(n,k)*binomial(2*k,k). - Seiichi Manyama, May 01 2019
a(n) = (1/4)^n * Sum_{k=0..n} (-3)^k * 5^(n-k) * binomial(2*k,k) * binomial(2*(n-k),n-k). - Seiichi Manyama, Aug 18 2025
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