OFFSET
0,2
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
From Peter Bala, Oct 11 2025: (Start)
a(n) = binomial(4*n-1, n)*hypergeom([-4*n, -n], [1 - 4*n], -1).
3*n*(3*n-1)*(3*n-2)*(77*n^3-343*n^2+504*n-244)*a(n) = (37345*n^6-223412*n^5+523453*n^4-606562*n^3+360148*n^2-101400*n+10080)*a(n-1) + 8*(4*n-5)*(4*n-6)*(4*n-7)*(77*n^3-112*n^2+49*n-6)*a(n-2) with a(0) = 1 and a(1) = 7.
The Gauss congruences a(n*p^r) == a(n*p^(r-1)) ( mod p^r ) hold for all primes p and all positive integers n and r.
We conjecture that the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) hold for all primes p >= 5 and all positive integers n and r. (End)
MAPLE
seq(simplify(binomial(4*n-1, n)*hypergeom([-4*n, -n], [1 - 4*n], -1)), n = 0..20); # Peter Bala, Oct 11 2025
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 *)
PROG
(PARI) a(n) = sum(k=0, n, binomial(4*n, k)*binomial(4*n-k-1, n-k));
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Seiichi Manyama, Feb 10 2024
STATUS
approved