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

%I A361028 #18 Mar 13 2023 07:21:11
%S A361028 2,3,20,210,2772,42042,700128,12471030,233716340,4557468630,
%T A361028 91752013080,1896208270320,40055997189600,862021408906800,
%U A361028 18849534808095360,417929529573239310,9379553386892837940,212776905535994934750,4873239487455972633000,112571832160232967822300
%N A361028 a(n) = 2*(3*n)!/(n!*(n+1)!^2).
%C A361028 Row 0 of square array A361027.
%C A361028 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?
%F A361028 a(n) = (2/(n+1)^2) * (3*n)!/n!^3.
%F A361028 a(n) = (2/3) * A006480(n+1)/((3*n + 1)*(3*n + 2)), where A006480(n) = (3*n)!/n!^3.
%F A361028 a(n) = (1/3)*27^(n+1)*binomial(1/3, n+1)*binomial(2/3, n+1).
%F A361028 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.
%F A361028 a(n) = A007226(n) * A000108(n).
%F A361028 a(n) ~ sqrt(3)*27^n/(Pi*n^3).
%F A361028 P-recursive: (n + 1)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 2.
%F A361028 The o.g.f. A(x) satisfies the differential equation
%F A361028 x^2*(1 - 27*x)*A''(x) + x*(3 - 54*x)*A'(x) + (1 - 6*x)*A(x) - 2 = 0, with
%F A361028 A(0) = 2 and A'(0) = 3.
%p A361028 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:
%p A361028 seq(a(n), n = 0..20);
%Y A361028 Row 0 of A361027. Cf. A000108, A006480, A007226, A361029, A361030, A361031, A361033.
%K A361028 nonn,tabl,easy
%O A361028 0,1
%A A361028 _Peter Bala_, Feb 28 2023