A324478 a(n) = (6/((n+1)*(n+2)*(n+3))) * multinomial(4*n;n,n,n,n).

1, 6, 252, 18480, 1801800, 209513304, 27485041584, 3937652896320, 603400560305400, 97512510301206000, 16452310738019476320, 2876570958459008603520, 518262201015698050067520, 95794174581229987212924000, 18101994022606737439599480000

Offset: 0

Author

N. J. A. Sloane, Mar 10 2019, following a suggestion from Luis Fredes and Avelio Sepulveda

Keywords

• nonn

Comments

Theorem (Luis Fredes, Mar 04 2019): (Start)

a(n) is an integer for all n >= 0.

Proof:

a(n) = (6/((n+1)*(n+2)*(n+3)))*multinomial(4*n;n,n,n,n) = multinomial(4*n;n,n,n,n) - multinomial(4*n;n+3,n-3,n,n) + 3*multinomial(4*n;n+2,n-2,n,n) + 15*multinomial(4*n;n+3,n-2,n-1,n) - 18*multinomial(4*n;n+3,n-1,n-1,n-1).

The right hand side is equal (after some manipulation) to

(f(n)/((n+1)*(n+2)*(n+3)))*multinomial(4*n;n,n,n,n)

where f(n) = (n+1)*(n+2)*(n+3) - (n-1)*(n-2)*n + 3*(n-1)*n*(n+3) + 15*(n-1)*n^2 - 18*n^3 = 6. QED (End)

(11!/12)*a(n) is divisible by ((n + 1)*(n + 2)*(n + 3))^2 for all n. More generally, we conjecture that there is a constant C(r) such that C(r)*(4*n)!/(n!*(n+r)!^3) is an integer for all n. Calculation suggests that we may take C(r) = (1/8)*(4*r)!/r! for r >= 1. - Peter Bala, Feb 28 2023

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..400
• Luis Fredes and Avelio Sepulveda, Tree-decorated planar maps, arXiv:1901.04981 [math.CO], 2019. See Remark 4.6.

Crossrefs

Cf. A324152.

Programs

Maple

a:= n-> 6*combinat[multinomial](4*n, n$4)/((n+1)*(n+2)*(n+3)):
seq(a(n), n=0..20);  # Alois P. Heinz, Mar 11 2019

Mathematica

c[m_,n_]:=2m Product[1/(n+i), {i, m}] (Multinomial@@ConstantArray[n, m+1]);{1}~Join~Array[c[3, #]&, 20] (* Vincenzo Librandi, Mar 11 2019 *)
Flatten[{1, Table[6*(4*n)! / ((n!)^3 * (n+3)!), {n, 1, 15}]}] (* Vaclav Kotesovec, Jul 21 2019 *)

Formula

From Vaclav Kotesovec, Jul 21 2019: (Start)

For n>0, a(n) = 6*(4*n)! / ((n!)^3 * (n+3)!).

a(n) ~ 6 * 2^(8*n - 1/2) / (Pi^(3/2) * n^(9/2)). (End)