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

0,2

Links

Table of n, a(n) for n=0..19.

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

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 *)

Prog

(PARI) a(n) = sum(k=0, n, binomial(4*n, k)*binomial(2*n-k-1, n-k));

Crossrefs

Cf. A001448, A352373, A370101, A370102.

Cf. A365754.

Sequence in context: A124267 A124450 A136023 * A328032 A074192 A058806

Adjacent sequences: A370097 A370098 A370099 * A370101 A370102 A370103

Keyword

nonn,easy

Author

Seiichi Manyama, Feb 10 2024

Status

approved