OFFSET
0,3
COMMENTS
Inverse binomial transform of A002294.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..958
FORMULA
G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^3 * A(x)^5.
G.f.: Sum_{k>=0} ( binomial(5*k,k) / (4*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) ~ 2869^(n + 3/2) / (78125 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)). - Vaclav Kotesovec, Jul 30 2021
D-finite with recurrence 8*n*(4*n+1)*(2*n-1)*(4*n-1)*a(n) -(n-1) *(1845*n^3 -1333*n^2 -238*n +240)*a(n-1) -4*(n-1) *(2485*n^3 -7263*n^2 +7388*n -2580) *a(n-2) -2*(n-1) *(n-2) *(8095*n^2 -24029*n +18924) *a(n-3) -4*(n-1) *(n-2) *(n-3) *(2805*n -5578) *a(n-4) -2869*(n-1) *(n-2) *(n-3) *(n-4) *a(n-5)=0. - R. J. Mathar, Aug 17 2023
MAPLE
A346665 := proc(n)
add((-1)^(n-k)*binomial(n, k)*binomial(5*k, k)/(4*k+1), k=0..n) ;
end proc:
seq(A346665(n), n=0..80); # R. J. Mathar, Aug 17 2023
MATHEMATICA
Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 22}]
nmax = 22; A[_] = 0; Do[A[x_] = 1/(1 + x) + x (1 + x)^3 A[x]^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 22; CoefficientList[Series[Sum[(Binomial[5 k, k]/(4 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
Table[(-1)^n HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5, -n}, {1/2, 3/4, 1, 5/4}, 3125/256], {n, 0, 22}]
PROG
(PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*binomial(5*k, k)/(4*k + 1)); \\ Michel Marcus, Jul 28 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Jul 27 2021
STATUS
approved