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A361033 a(n) = 3*(4*n)!/(n!*(n+1)!^3).

%I A361033 #14 Jul 30 2024 11:23:10
%S A361033 3,9,280,17325,1513512,162954792,20193091776,2768662192725,
%T A361033 409716429837000,64358256798795960,10605621798062141760,
%U A361033 1817833036248401270280,321997225483126007438400,58649494641569379926280000,10941649720331183519046796800,2084191938036600263793119045925
%N A361033 a(n) = 3*(4*n)!/(n!*(n+1)!^3).
%C A361033 Row 0 of A361032.
%C A361033 The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that A000984(n) is divisible by n + 1 and the result (2*n)!/(n!*(n+1)!) is the n-th Catalan number A000108(n). Similarly, the  numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 3*A008977(n) is divisible by (n + 1)^3, leading to the present sequence. Cf. A361028. Do these numbers have a combinatorial interpretation?
%C A361033 Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.
%F A361033 a(n) = 3*A008977(n)/(n+1)^3.
%F A361033 a(n) = (3/4)*A008977(n+1)/((4*n+1)*(4*n+2)*(4*n+3)).
%F A361033 a(n) = (1/2)*A007228(n)*A007226(n)*A000108(n).
%F A361033 P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+1)^3 * a(n-1) with a(0) = 3.
%F A361033 The o.g.f. A(x) satisfies the differential equation
%F A361033 x^3*(1 - 256*x)*A(x)''' + x^2*(6 - 1152*x)*A(x)'' + x*(7 - 816*x)*A(x)' + (1 - 24*x)*A(x) - 3 = 0 with A(0) = 3, A'(0) = 9 and A''(0) = 560.
%F A361033 a(n) ~ 3*sqrt(1/(2*Pi^3)) * 2^(8*n)/n^(9/2).
%p A361033 seq(3*(4*n)!/(n!*(n+1)!^3), n = 0..20);
%t A361033 Table[3 (4n)!/(n! ((n+1)!)^3),{n,0,15}] (* _Harvey P. Dale_, Jul 30 2024 *)
%Y A361033 Cf. A000108, A007226, A007228, A008977, A361028, A361032, A361034, A361035.
%K A361033 nonn,easy
%O A361033 0,1
%A A361033 _Peter Bala_, Mar 01 2023