A370100 a(n) = Sum_{k=0..n} binomial(4*n,k) * binomial(2*n-k-1,n-k).
1, 5, 47, 500, 5615, 65005, 767396, 9183144, 110995695, 1351922495, 16566597047, 204010570296, 2522556212228, 31298015910140, 389458822888280, 4858487926378000, 60742838865326319, 760901358321592611, 9547848458062427405, 119990407515367475700
Offset: 0
Programs
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Mathematica
Table[Sum[Binomial[4*n, k]*Binomial[2*n - k - 1, n - k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 12 2024 *) -
PARI
a(n) = sum(k=0, n, binomial(4*n, k)*binomial(2*n-k-1, n-k));
Formula
a(n) = [x^n] ( (1+x)^4/(1-x) )^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)/(1+x)^4 ). See A365754.
From Peter Bala, Jun 08 2024: (Start)
2*n*(n - 1)*(2*n - 1)*(51*n^2 - 144*n + 100)*a(n) = -(n - 1)*(5457*n^4 - 20865*n^3 + 26366*n^2 - 12172*n + 1560)*a(n-1) + 64*(2*n - 3)*(4*n - 5)*(4*n - 7)*(51*n^2 - 42*n + 7)*a(n-2) with a(0) = 1 and a(1) = 5.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. See A352373 for a more general conjecture. (End)
a(n) ~ sqrt(3 + 5/sqrt(17)) * (51*sqrt(17) - 107)^n / (sqrt(Pi*n) * 2^(3*n + 3/2)). - Vaclav Kotesovec, Jun 12 2024
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(2*n+1) * (1-2*x)^n).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k) * binomial(3*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(n+k-1,k) * binomial(3*n-k,n-k). (End)