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A258399 Number of 4n-length strings of balanced parentheses of exactly n different types that are introduced in ascending order.

%I A258399 #22 Sep 27 2023 16:10:08
%S A258399 1,2,98,11880,2432430,714249900,275335499824,131928199603200,
%T A258399 75727786603836510,50713478000403718500,38843740303576863755100,
%U A258399 33508462196084294380001040,32157574295254903735909896240,33990046387543889224733323929120
%N A258399 Number of 4n-length strings of balanced parentheses of exactly n different types that are introduced in ascending order.
%H A258399 Alois P. Heinz, <a href="/A258399/b258399.txt">Table of n, a(n) for n = 0..250</a>
%F A258399 a(n) = A253180(2n,n).
%F A258399 a(n) ~ c * d^n * n! / n^(5/2), where d = A256254 = -64/(LambertW(-2*exp(-2))*(2 + LambertW(-2*exp(-2)))) = 98.8248737517356857317..., c = 1/(2^(5/2) * Pi^(3/2) * sqrt(1 + LambertW(-2*exp(-2)))) = 0.0412044746356859529237459292541572856326... . - _Vaclav Kotesovec_, Jun 01 2015, updated Sep 27 2023
%F A258399 a(n) = A210029(n) * (4*n)! / (n! * (2*n)! * (2*n + 1)!), for n>0. - _Vaclav Kotesovec_, Sep 27 2023
%e A258399 a(0) = 1: the empty string.
%e A258399 a(1) = 2: ()(), (()).
%e A258399 a(2) = A000108(4) * (2^3-1) = 14*7 = 98.
%p A258399 ctln:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
%p A258399 A:= proc(n, k) option remember; k^n*ctln(n) end:
%p A258399 a:= n-> add(A(2*n, n-i)*(-1)^i/((n-i)!*i!), i=0..n):
%p A258399 seq(a(n), n=0..15);
%t A258399 A[n_, k_] := A[n, k] = k^n CatalanNumber[n];
%t A258399 a[n_] := If[n==0, 1, Sum[A[2n, n-i] (-1)^i/((n-i)! i!), {i, 0, n}]];
%t A258399 a /@ Range[0, 15] (* _Jean-François Alcover_, Jan 01 2021, after _Alois P. Heinz_ *)
%Y A258399 Cf. A000108, A210029, A242446, A253180, A256254, A258426.
%K A258399 nonn
%O A258399 0,2
%A A258399 _Alois P. Heinz_, May 28 2015