cp's OEIS Frontend

A384668 a(n) = 12 * (5*n+2)! / ((3*n+1)! * (2*n+2)!).

%I A384668 #13 Jun 12 2025 10:22:39
%S A384668 12,105,1584,29172,596904,13037895,297748800,7023149820,169774618104,
%T A384668 4183919862474,104722807600320,2654939113240050,68033328627480804,
%U A384668 1759318006963275528,45853277234783179392,1203249937243079847660,31764232607604306053400,842982010030680328418706
%N A384668 a(n) = 12 * (5*n+2)! / ((3*n+1)! * (2*n+2)!).
%H A384668 Paolo Xausa, <a href="/A384668/b384668.txt">Table of n, a(n) for n = 0..650</a>
%F A384668 O.g.f.: 12*hypergeom([3/5, 4/5, 1, 6/5, 7/5], [2/3, 4/3, 3/2, 2], (3125*x)/108).
%F A384668 E.g.f.: 12*hypergeom([3/5, 4/5, 6/5, 7/5], [2/3, 4/3, 3/2, 2], (3125*x)/108).
%F A384668 O.g.f. denoted by h(x), satisfies the algebraic equation of order 10:
%F A384668 1889568 - 6141096*x + 10628820*x^2 - 59049*x^3 + (-2834352*x^3 + 4861701*x^2 - 2834352*x - 157464)*h(x) + 13122*x*(14*x^3 - 77*x^2 + 124*x + 30)*h(x)^2 - 4374*x^2*(14*x^2 + 94*x + 99)*h(x)^3 +  729*x^3*(50*x^2 + 32*x + 377)*h(x)^4 - 243*x^4*(11*x^2 - 40*x + 456)*h(x)^5 - 243*x^5*(8*x - 121)*h(x)^6 + 54*x^6*(2*x - 95)*h(x)^7 + 567*x^7*h(x)^8 - 36*x^8*h(x)^9 + x^9*h(x)^10 = 0.
%F A384668 a(n) = Integral_{x=0..3125/108} x^n*W(x)*dx, where W(x) = W1(x)+W2(x)+W3(x)+W4(x), with
%F A384668   W1(x) = (3*sqrt(5)*csc(Pi/5)*sin(Pi/10)*hypergeom([-2/5, 1/10, 4/15, 14/15], [1/5, 2/5, 4/5], (108*x)/3125))/(2*Pi*x^(2/5)),
%F A384668   W2(x) = (6*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([-1/5, 3/10, 7/15, 17/15], [2/5, 3/5, 6/5], (108*x)/3125))/(5*Pi*x^(1/5)),
%F A384668   W3(x) = -(24*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*x^(1/5)*hypergeom([1/5, 7/10, 13/15, 23/15], [4/5, 7/5, 8/5], (108*x)/3125))/(125*Pi), and
%F A384668   W4(x) = -(33*sqrt(5)*csc(Pi/5)*sin(Pi/10)*x^(2/5)*hypergeom([2/5, 9/10, 16/15, 26/15], [6/5, 8/5, 9/5], (108*x)/3125))/(1250*Pi).
%F A384668 This integral representation is unique as it is the solution of the Hausdorff power moment 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 = 3125/108. Therefore a(n) is a positive definite sequence.
%t A384668 A384668[n_] := 12*(5*n + 2)!/((3*n + 1)!*(2*n + 2)!);
%t A384668 Array[A384668, 20, 0] (* _Paolo Xausa_, Jun 12 2025 *)
%Y A384668 Cf. A384585, A002293, A000260, A001450.
%K A384668 nonn
%O A384668 0,1
%A A384668 _Karol A. Penson_, Jun 06 2025