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A384585 a(n) = 6 * (4*n)! / ((n+1)! * (3*n+1)!).

%I A384585 #19 Jun 22 2025 00:51:29
%S A384585 6,3,8,33,168,969,6072,40365,280488,2017356,14914848,112784399,
%T A384585 869046168,6803716710,53997506640,433647466245,3518801467560,
%U A384585 28815074239908,237887596740192,1978246301709540,16558857808956320,139428557033056785,1180350813375438840,10041660963789578955
%N A384585 a(n) = 6 * (4*n)! / ((n+1)! * (3*n+1)!).
%C A384585 Since a(1) < a(0) the sequence is not growing monotonically with n.
%F A384585 O.g.f.: 6*hypergeom([1/4, 1/2, 3/4, 1], [2/3, 4/3, 2], (256*z)/27).
%F A384585 E.g.f.: 6*hypergeom([1/4, 1/2, 3/4], [2/3, 4/3, 2], (256*z)/27).
%F A384585 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.
%F A384585 a(n) = Integral_{x=0..256/27} x^n*W(x)*dx, where W(x) = W1(x)+W2(x)+W3(x), with
%F A384585   W1(x) = 4*sqrt(2)*hypergeom([-3/4, -1/12, 7/12], [1/2, 3/4], (27*x)/256)/(Pi*x^(3/4)),
%F A384585   W2(x) = -3*hypergeom([-1/2, 1/6, 5/6], [3/4, 5/4], (27*x)/256)/(Pi*sqrt(x)), and
%F A384585   W3(x) = -3*sqrt(2)*hypergeom([-1/4, 5/12, 13/12], [5/4, 3/2], (27*x)/256)/(8*Pi*x^(1/4)).
%F A384585 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.
%p A384585 a:= proc(n) option remember; `if`(n<2, 6-3*n,
%p A384585       8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)/(3*(3*n-1)*(3*n+1)*(n+1)))
%p A384585     end:
%p A384585 seq(a(n), n=0..23);  # _Alois P. Heinz_, Jun 04 2025
%t A384585 a[n_]:=6*(4*n)!/((n+1)!*(3*n+1)!); Array[a,24,0] (* _Stefano Spezia_, Jun 04 2025 *)
%Y A384585 Cf. A002293, A000260.
%K A384585 nonn
%O A384585 0,1
%A A384585 _Karol A. Penson_, Jun 04 2025