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%I #23 Mar 13 2023 07:22:21
%S 1,6,252,18480,1801800,209513304,27485041584,3937652896320,
%T 603400560305400,97512510301206000,16452310738019476320,
%U 2876570958459008603520,518262201015698050067520,95794174581229987212924000,18101994022606737439599480000
%N a(n) = (6/((n+1)*(n+2)*(n+3))) * multinomial(4*n;n,n,n,n).
%C Theorem (Luis Fredes, Mar 04 2019): (Start)
%C a(n) is an integer for all n >= 0.
%C Proof:
%C 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).
%C The right hand side is equal (after some manipulation) to
%C (f(n)/((n+1)*(n+2)*(n+3)))*multinomial(4*n;n,n,n,n)
%C 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)
%C (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
%H Vincenzo Librandi, <a href="/A324478/b324478.txt">Table of n, a(n) for n = 0..400</a>
%H Luis Fredes and Avelio Sepulveda, <a href="https://arxiv.org/abs/1901.04981">Tree-decorated planar maps</a>, arXiv:1901.04981 [math.CO], 2019. See Remark 4.6.
%F From _Vaclav Kotesovec_, Jul 21 2019: (Start)
%F For n>0, a(n) = 6*(4*n)! / ((n!)^3 * (n+3)!).
%F a(n) ~ 6 * 2^(8*n - 1/2) / (Pi^(3/2) * n^(9/2)). (End)
%p a:= n-> 6*combinat[multinomial](4*n, n$4)/((n+1)*(n+2)*(n+3)):
%p seq(a(n), n=0..20); # _Alois P. Heinz_, Mar 11 2019
%t 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 *)
%t Flatten[{1, Table[6*(4*n)! / ((n!)^3 * (n+3)!), {n, 1, 15}]}] (* _Vaclav Kotesovec_, Jul 21 2019 *)
%Y Cf. A324152.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Mar 10 2019, following a suggestion from Luis Fredes and Avelio Sepulveda.
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