OFFSET
0,3
COMMENTS
Inverse binomial transform of A002293.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000
FORMULA
G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^2 * A(x)^4.
G.f.: Sum_{k>=0} ( binomial(4*k,k) / (3*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) ~ 229^(n + 3/2) / (2048 * sqrt(2*Pi) * n^(3/2) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021
D-finite with recurrence +3*n*(3*n-1)*(3*n+1)*a(n) -74*n*(2*n-1) *(n-1)*a(n-1) -6*(n-1) *(101*n^2 -202*n +105)*a(n-2) -330*(n-1) *(n-2)*(2*n-3) *a(n-3) -229*(n-1)*(n-2) *(n-3)*a(n-4)=0. - R. J. Mathar, Aug 17 2023
MAPLE
A346664 := proc(n)
add( (-1)^(n-k)*binomial(n, k)*binomial(4*k, k)/(3*k+1), k=0..n) ;
end proc:
seq(A346664(n), n=0..80); # R. J. Mathar, Aug 17 2023
MATHEMATICA
Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 23}]
nmax = 23; A[_] = 0; Do[A[x_] = 1/(1 + x) + x (1 + x)^2 A[x]^4 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 23; CoefficientList[Series[Sum[(Binomial[4 k, k]/(3 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
Table[(-1)^n HypergeometricPFQ[{1/4, 1/2, 3/4, -n}, {2/3, 1, 4/3}, 256/27], {n, 0, 23}]
PROG
(PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*binomial(4*k, k)/(3*k+1)); \\ Michel Marcus, Jul 28 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Jul 27 2021
STATUS
approved