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%I #43 Sep 08 2022 08:46:24
%S 1,3,126,9240,900900,104756652,13742520792,1968826448160,
%T 301700280152700,48756255150603000,8226155369009738160,
%U 1438285479229504301760,259131100507849025033760,47897087290614993606462000,9050997011303368719799740000
%N a(0)=1; for n>0, a(n) = (3/((n+1)*(n+2)*(n+3))) * multinomial(4*n;n,n,n,n).
%C It is conjectured that a(n) is always an integer.
%C If all terms except the first are doubled, we get A324478, which IS known to be integral.
%H N. J. A. Sloane, <a href="/A324152/b324152.txt">Table of n, a(n) for n = 0..360</a>
%H Luis Fredes, 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 a(n+1) = a(n)*4*(4*n+1)*(4*n+2)*(4*n+3)/((n+1)^2*(n+4)) for n>0.
%F From _Vaclav Kotesovec_, Jul 21 2019: (Start)
%F For n>0, a(n) = 3*(4*n)! / ((n!)^3 * (n+3)!).
%F a(n) ~ 3 * 2^(8*n - 1/2) / (Pi^(3/2) * n^(9/2)). (End)
%t c[m_, n_] := m Product[1/(n + i), {i, m}] (Multinomial @@ ConstantArray[n, m + 1]); {1}~Join~Array[c[3, #] &, 14] (* _Michael De Vlieger_, Mar 01 2019 *)
%t Flatten[{1, Table[3*(4*n)! / ((n!)^3 * (n+3)!), {n, 1, 15}]}] (* _Vaclav Kotesovec_, Jul 21 2019 *)
%o (Magma) [1] cat [n le 1 select 3 else Self(n-1)*4*(4*n-3)*(4*n-2)*(4*n-1)/((n)^2*(n+3)): n in [1..20]]; // _Vincenzo Librandi_, Mar 11 2019
%Y Cf. A000108, A324151, A324465 (exponent of 2), A324467, A324478.
%K nonn
%O 0,2
%A _Michael De Vlieger_ and _N. J. A. Sloane_, Mar 01 2019
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