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

3, 9, 280, 17325, 1513512, 162954792, 20193091776, 2768662192725, 409716429837000, 64358256798795960, 10605621798062141760, 1817833036248401270280, 321997225483126007438400, 58649494641569379926280000, 10941649720331183519046796800, 2084191938036600263793119045925

Offset: 0

Author

Peter Bala, Mar 01 2023

Keywords

• nonn
• easy

Comments

Row 0 of A361032.

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?

Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.

Crossrefs

Cf. A000108, A007226, A007228, A008977, A361028, A361032, A361034, A361035.

Programs

Maple

seq(3*(4*n)!/(n!*(n+1)!^3), n = 0..20);

Mathematica

Table[3 (4n)!/(n! ((n+1)!)^3),{n,0,15}] (* Harvey P. Dale, Jul 30 2024 *)

Formula

a(n) = 3*A008977(n)/(n+1)^3.

a(n) = (3/4)*A008977(n+1)/((4*n+1)*(4*n+2)*(4*n+3)).

a(n) = (1/2)*A007228(n)*A007226(n)*A000108(n).

P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+1)^3 * a(n-1) with a(0) = 3.

The o.g.f. A(x) satisfies the differential equation

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.

a(n) ~ 3*sqrt(1/(2*Pi^3)) * 2^(8*n)/n^(9/2).