OFFSET
0,2
COMMENTS
Inverse binomial transform of A005810.
FORMULA
G.f.: (1/x) * Sum_{k>=0} binomial(4*k,k) * (x/(1 + x))^(k+1).
a(n) = [x^n] (1 + 3*x + 6*x^2 + 4*x^3 + x^4)^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)^4 - x) ). See A317133. - Seiichi Manyama, Apr 17 2025
a(n) ~ 229^(n + 1/2) / (2^(7/2) * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Apr 17 2025
MATHEMATICA
Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[4 k, k], {k, 0, n}], {n, 0, 22}]
Table[(-1)^n HypergeometricPFQ[{1/4, 1/2, 3/4, -n}, {1/3, 2/3, 1}, 256/27], {n, 0, 22}]
nmax = 22; CoefficientList[Series[(1/x) Sum[Binomial[4 k, k] (x/(1 + x))^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
PROG
(PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*binomial(4*k, k)); \\ Seiichi Manyama, Apr 17 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Apr 17 2025
STATUS
approved