OFFSET
0,1
COMMENTS
Since a(1) < a(0) the sequence is not growing monotonically with n.
FORMULA
O.g.f.: 6*hypergeom([1/4, 1/2, 3/4, 1], [2/3, 4/3, 2], (256*z)/27).
E.g.f.: 6*hypergeom([1/4, 1/2, 3/4], [2/3, 4/3, 2], (256*z)/27).
O.g.f. = h(z) satisfies algebraic equation of order 4: -6 - 39*z + 4096*z^2 + (1 - 12*z - 768*z^2)*h(z) - 3*z*(2*z - 1)*h(z)^2 + 3*z^2*h(z)^3 + z^3*h(z)^4 = 0.
a(n) = Integral_{x=0..256/27} x^n*W(x)*dx, where W(x) = W1(x)+W2(x)+W3(x), with
W1(x) = 4*sqrt(2)*hypergeom([-3/4, -1/12, 7/12], [1/2, 3/4], (27*x)/256)/(Pi*x^(3/4)),
W2(x) = -3*hypergeom([-1/2, 1/6, 5/6], [3/4, 5/4], (27*x)/256)/(Pi*sqrt(x)), and
W3(x) = -3*sqrt(2)*hypergeom([-1/4, 5/12, 13/12], [5/4, 3/2], (27*x)/256)/(8*Pi*x^(1/4)).
This integral representation is unique as it is the solution of the Hausdorff power moment problem of the function W(x). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0 and for x > 0 is monotonically decreasing to zero at x = 256/27. Therefore a(n) is a positive definite sequence.
MAPLE
a:= proc(n) option remember; `if`(n<2, 6-3*n,
8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)/(3*(3*n-1)*(3*n+1)*(n+1)))
end:
seq(a(n), n=0..23); # Alois P. Heinz, Jun 04 2025
MATHEMATICA
a[n_]:=6*(4*n)!/((n+1)!*(3*n+1)!); Array[a, 24, 0] (* Stefano Spezia, Jun 04 2025 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Karol A. Penson, Jun 04 2025
STATUS
approved
