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A387248 a(n) = 3/(n + 1) * Catalan(2*n).

%I A387248 #16 Sep 02 2025 11:00:24
%S A387248 3,3,14,99,858,8398,89148,1002915,11785890,143291610,1790214660,
%T A387248 22870640910,297670187844,3935861372604,52749590350072,
%U A387248 715309969142307,9800129095949682,135490673691621794,1888389218820071604,26510079418051005210,374589577468070301260,5324240442532424176260,76082624294738699098440
%N A387248 a(n) = 3/(n + 1) * Catalan(2*n).
%C A387248 Bisection of A007054.
%C A387248 Compare with Catalan(n) = 1/(n + 1) * binomial(2*n, n).
%C A387248 For r >= 2, there is a constant K_r such that K_r/(n + 1) * Catalan(r*n) is integral for all n.
%H A387248 Paolo Xausa, <a href="/A387248/b387248.txt">Table of n, a(n) for n = 0..800</a>
%F A387248 a(n) = 6/((2*n + 1)*(2*n + 2)) * binomial(4*n, 2*n).
%F A387248 a(n) = 4*Catalan(2*n) - Catalan(2*n+1) (showing a(n) to be an integer)
%F A387248 G.f.: A(x) = ((2 - f(x))*sqrt(2 + 2*f(x)) - 2)/(4*x), where f(x) = sqrt(1 - 16*x).
%F A387248 a(n) = 2*(4*n - 1)*(4*n - 3)/((n + 1)*(2*n + 1)) * a(n-1) with a(0) = 3.
%F A387248 a(n) ~ 3/(2*sqrt(2*Pi)) * 16^n/n^(5/2).
%F A387248 a(n) is odd iff n = 2^k - 1 for some k, so a(n) has the same parity as Catalan(n).
%F A387248 E.g.f.: 3*hypergeom([1/4, 3/4], [3/2, 2], 16*x). - _Stefano Spezia_, Aug 27 2025
%p A387248 seq( 6/((2*n+1)*(2*n+2)) * binomial(4*n, 2*n), n = 0..22);
%t A387248 A387248[n_] := 3*CatalanNumber[2*n]/(n + 1); Array[A387248, 25, 0] (* _Paolo Xausa_, Sep 02 2025 *)
%Y A387248 Cf. A000108, A000984, A001448, A007054, A387249, A387250.
%K A387248 nonn,easy,new
%O A387248 0,1
%A A387248 _Peter Bala_, Aug 24 2025