A370101 a(n) = Sum_{k=0..n} binomial(4*n,k) * binomial(4*n-k-1,n-k).
1, 7, 97, 1519, 25089, 427007, 7408897, 130287871, 2313945089, 41409732607, 745530884097, 13488086405119, 245014271688705, 4465915098890239, 81637668328243201, 1496095489290731519, 27477504726883368961, 505627095685486608383, 9320167322334416338945
Offset: 0
Keywords
Programs
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Mathematica
Table[Sum[Binomial[4n,k]Binomial[4n-k-1,n-k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Dec 09 2024 *) Table[Sum[2^k*(-1)^(n-k)*Binomial[4*n, k], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 31 2025 *) -
PARI
a(n) = sum(k=0, n, binomial(4*n, k)*binomial(4*n-k-1, n-k));
Formula
a(n) = [x^n] ( (1+x)^4/(1-x)^3 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^3/(1+x)^4 ). See A365846.
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k). - Seiichi Manyama, Jul 31 2025
a(n) ~ 2^(9*n + 3/2) / (7 * sqrt(Pi*n) * 3^(3*n - 1/2)). - Vaclav Kotesovec, Jul 31 2025
a(n) = Sum_{k=0..n} 2^k * binomial(3*n+k-1,k). - Seiichi Manyama, Aug 01 2025
a(n) = [x^n] 1/((1-x) * (1-2*x)^(3*n)). - Seiichi Manyama, Aug 09 2025