A227035 a(n) = Sum_{k=0..floor(n/4)} binomial(n,4*k)*binomial(5*k,k)/(4*k+1).

1, 1, 1, 1, 2, 6, 16, 36, 76, 172, 436, 1156, 3006, 7606, 19202, 49466, 130156, 345356, 915196, 2421532, 6427001, 17163581, 46087911, 124133531, 334850208, 904691576, 2449891276, 6651540676, 18100561856, 49344295152, 134719523056, 368350942416, 1008680051756

Offset: 0

Author

Vaclav Kotesovec, Jun 28 2013

Keywords

• nonn
• easy

Comments

Generally, Sum(binomial(n,p*k)*binomial((p+1)*k,k)/(p*k+1), k=0..floor(n/p)) is asymptotic to (p+(p+1)^(1+1/p))^(n+3/2)/(p^(n+1)*(p+1)^(1+3/(2*p))*n^(3/2)*sqrt(2*Pi)).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A002294, A007317 (p=1), A049130 (p=2), A226974 (p=3), A226910 (p=5).

Programs

Mathematica

Table[Sum[Binomial[n,4*k]*Binomial[5*k,k]/(4*k+1),{k,0,Floor[n/4]}],{n,0,20}]

PARI

a(n)=sum(k=0,n\4,binomial(n,4*k)*binomial(5*k,k)/(4*k+1)) \\ Charles R Greathouse IV, Jun 28 2013

Formula

Recurrence: -2869*(n-7)*(n-6)*(n-5)*(n-4)*a(n-8) + 2*(n-6)*(n-5)*(n-4)*(5226*n-17267)*a(n-7) - (n-5)*(n-4)*(11582*n^2-55156*n+50139)*a(n-6) - 3*(n-4)*(612*n^3 - 18926*n^2 + 102684*n - 155665)*a(n-5) + 5*(n-4)*(2959*n^3 - 26172*n^2 + 77408*n - 76800)*a(n-4) - 1024*(n-2)*(2*n-5)*(7*n^2-35*n+48)*a(n-3) + 1024*(n-2)*(n-1)*(7*n^2-28*n+30)*a(n-2) - 1024*(n-2)*(n-1)*n*(2*n-3)*a(n-1) + 256*(n-2)*(n-1)*n*(n+1)*a(n) = 0.

a(n) ~ (4+5^(1+1/4))^(n+3/2)/(4^(n+1)*5^(1+3/8)*n^(3/2)*sqrt(2*Pi)).

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x^4 * A(x)^5. - Ilya Gutkovskiy, Jul 25 2021

From Peter Bala, Sep 15 2021: (Start)

O.g.f.: A(x) = (1/x)*series reversion ( x*(1 - x^4)/(1 + x*(1 - x^4) )).

The g.f. of the m-th binomial transform of this sequence is equal to (1/x)*series reversion ( x*(1 - x^4)/(1 + (m + 1)*x*(1 - x^4)) ). The case m = -1 gives the sequence [1,0,0,0,1,0,0,0,5,0,0,0,35,0,0,0,285,...] - an aerated version of A002294. (End)