A226974 a(n) = Sum_{k=0..floor(n/3)} binomial(n,3*k)*binomial(4*k,k)/(3*k+1).
1, 1, 1, 2, 5, 11, 25, 64, 169, 443, 1181, 3224, 8897, 24701, 69161, 195255, 554577, 1583109, 4541461, 13086574, 37856437, 109892403, 320034309, 934774902, 2737689189, 8037746691, 23652564261, 69749727716, 206091735797, 610061655665, 1808962146529
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
Programs
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Maple
A226974 := proc(n) hypergeom([-n/3,-n/3+2/3,-n/3+1/3,1/4,1/2,3/4],[1/3,2/3,2/3,1,4/3],-256/27) ; simplify(%) ; end proc: seq(A226974(n),n=0..40) ; # R. J. Mathar, Jan 10 2023
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Mathematica
Table[Sum[Binomial[n,3*k]*Binomial[4*k,k]/(3*k+1), {k,0,Floor[n/3]}],{n,0,20}] (* Vaclav Kotesovec, Jun 28 2013 *) -
Maxima
a(n):=if n<0 then 0 else if n=0 then 1 else sum(sum(sum(a(l)*a(i)*a(j)*a(n-i-j-l-3),l,0,n-3-i-j),j,0,n-3-i),i,0,n-3)+1; /* Vladimir Kruchinin, May 17 2020 */
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PARI
a(n) = sum(k=0, n\3, binomial(n,3*k)*binomial(4*k,k)/(3*k+1)); \\ Michel Marcus, Sep 16 2021
Formula
a(n) = Sum_{k=0..floor(n/3)} binomial(n,3*k)*A002293(k).
Representation in terms of special values of hypergeometric function of type 6F5: a(n) = hypergeom([1/4, 1/2, 3/4, -(1/3)*n, -(1/3)*n+2/3, -(1/3)*n+1/3], [1/3, 2/3, 2/3, 1, 4/3],-4^4/3^3), n>=0.
Recurrence: 27*n*(n+1)*(n-1)*a(n) = 162*n*(n-1)^2*a(n-1) - 81*(5*n^2-15*n+12)*(n-1)*a(n-2) + 4*(199*n^3 - 1098*n^2 + 2043*n - 1296)*a(n-3) - (n-3)*(1173*n^2 - 5097*n + 5584)*a(n-4) + 6*(n-4)*(n-3)*(155*n-401)*a(n-5) - 283*(n-5)*(n-4)*(n-3)*a(n-6). - Vaclav Kotesovec, Jun 28 2013
a(n) ~ (3+4^(1+1/3))^(n+3/2)/(8*3^(n+1)*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 28 2013
G.f. satisfies A(x)=1+x^3*A(x)^4+x/(1-x). - Vladimir Kruchinin, May 17 2020
From Peter Bala, Sep 15 2021: (Start)
O.g.f.: A(x) = (1/x)*series reversion( x*(1 - x^3)/(1 + x*(1 - x^3)) ).
The g.f. of the m-th binomial transform of this sequence is equal to (1/x)*series reversion( x*(1 - x^3)/(1 + (m + 1)*x*(1 - x^3)) ). The case m = -1 gives the sequence [1,0,0,1,0,0,4,0,0,22,0,0,140,...] - an aerated version of A002293. (End)