A361029 a(n) = 120*(3*n)!/(n!*(n+2)!^2).

30, 20, 75, 504, 4620, 51480, 656370, 9237800, 140229804, 2259901800, 38230005450, 673210036800, 12262039956000, 229872375708480, 4417859720647350, 86767376381987400, 1736954330906081100, 35364582637561485000, 730985923118395894950, 15315895532004485418000

Offset

0,1

Comments

Row 1 of the square array A361027.

The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result 6*(2n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 120*A006480(n) is divisible by ((n + 1)*(n + 2))^2, leading to the present sequence. Do these numbers have a combinatorial interpretation?

Links

Table of n, a(n) for n=0..19.

Formula

a(n) = 120/((n+1)*(n+2))^2 * (3*n)!/n!^3.

a(n) = (1/3)*(1*2*4*5) * A006480(n+2)/((3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n +

5)), where A006480(n) = (3*n)!/n!^3.

a(n) = (10*binomial(3*n,n) - 7*binomial(3*n,n+1) + binomial(3*n,n+2)) * (3*binomial(2*n,n) - 4*binomial(2*n,n+1) + binomial(2*n,n+2)), shows that a(n) is an integer for all n.

a(n) = (1/3)*27^(n+2)*binomial(4/3, n+2)*binomial(5/3, n+2).

a(n) ~ sqrt(3)*60*27^n/(Pi*n^5).

P-recursive: (n + 2)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 30.

The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(5 - 54*x)*A'(x) + (4 - 6*x)*A(x) - 120 = 0, with A(0) = 30 and A'(0) = 20.

Maple

a := proc(n) option remember; if n = 0 then 30 else 3*(3*n-1)*(3*n-2)/(n+2)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Mathematica

Table[120 (3n)!/(n!(n+2)!^2), {n, 0, 20}] (* Harvey P. Dale, Jul 02 2023 *)

Crossrefs

Row 1 of A361027. Cf. A006480, A007054, A361028, A361030, A361031, A361037.

Sequence in context: A033971 A193220 A040872 * A268855 A117749 A283621

Adjacent sequences: A361026 A361027 A361028 * A361030 A361031 A361032

Keyword

nonn,easy

Author

Peter Bala, Mar 01 2023

Status

approved