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A324478 a(n) = (6/((n+1)*(n+2)*(n+3))) * multinomial(4*n;n,n,n,n). 2
1, 6, 252, 18480, 1801800, 209513304, 27485041584, 3937652896320, 603400560305400, 97512510301206000, 16452310738019476320, 2876570958459008603520, 518262201015698050067520, 95794174581229987212924000, 18101994022606737439599480000 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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. QED

(End)

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..400

Luis Fredes, Avelio Sepulveda, Tree-decorated planar maps, arXiv:1901.04981 [math.CO], 2019. See Remark 4.6.

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)

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 *)

CROSSREFS

Cf. A324152.

Sequence in context: A056238 A221822 A184424 * A230881 A041853 A168476

Adjacent sequences:  A324475 A324476 A324477 * A324479 A324480 A324481

KEYWORD

nonn

AUTHOR

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

STATUS

approved

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Last modified October 18 05:18 EDT 2019. Contains 328146 sequences. (Running on oeis4.)