A361028 a(n) = 2*(3*n)!/(n!*(n+1)!^2).

2, 3, 20, 210, 2772, 42042, 700128, 12471030, 233716340, 4557468630, 91752013080, 1896208270320, 40055997189600, 862021408906800, 18849534808095360, 417929529573239310, 9379553386892837940, 212776905535994934750, 4873239487455972633000, 112571832160232967822300

Offset

0,1

Comments

Row 0 of square array A361027.

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 de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 2*A006480(n) is divisible by (n+1)^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) = (2/(n+1)^2) * (3*n)!/n!^3.

a(n) = (2/3) * A006480(n+1)/((3*n + 1)*(3*n + 2)), where A006480(n) = (3*n)!/n!^3.

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

a(n) = 2*C(2*n,n)*C(3*n,n) - 3*C(2*n,n+1)*C(3*n,2*n) + 2*C(2*n,n)*C(3*n,2*n+2) + 2*C(2*n,n+1)*C(3*n,2*n+2). This formula shows that a(n) is an integer for all n.

a(n) = A007226(n) * A000108(n).

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

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

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

x^2*(1 - 27*x)*A''(x) + x*(3 - 54*x)*A'(x) + (1 - 6*x)*A(x) - 2 = 0, with

A(0) = 2 and A'(0) = 3.

Maple

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

Crossrefs

Row 0 of A361027. Cf. A000108, A006480, A007226, A361029, A361030, A361031, A361033.

Sequence in context: A012416 A261317 A353921 * A177946 A006246 A349600

Adjacent sequences: A361025 A361026 A361027 * A361029 A361030 A361031

Keyword

nonn,tabl,easy

Author

Peter Bala, Feb 28 2023

Status

approved