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

1, 2, 98, 11880, 2432430, 714249900, 275335499824, 131928199603200, 75727786603836510, 50713478000403718500, 38843740303576863755100, 33508462196084294380001040, 32157574295254903735909896240, 33990046387543889224733323929120

Offset

0,2

Links

Alois P. Heinz, Table of n, a(n) for n = 0..250

Formula

a(n) = A253180(2n,n).

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

a(n) = A210029(n) * (4*n)! / (n! * (2*n)! * (2*n + 1)!), for n>0. - Vaclav Kotesovec, Sep 27 2023

Example

a(0) = 1: the empty string.
a(1) = 2: ()(), (()).
a(2) = A000108(4) * (2^3-1) = 14*7 = 98.

Maple

ctln:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
A:= proc(n, k) option remember; k^n*ctln(n) end:
a:= n-> add(A(2*n, n-i)*(-1)^i/((n-i)!*i!), i=0..n):
seq(a(n), n=0..15);

Mathematica

A[n_, k_] := A[n, k] = k^n CatalanNumber[n];
a[n_] := If[n==0, 1, Sum[A[2n, n-i] (-1)^i/((n-i)! i!), {i, 0, n}]];
a /@ Range[0, 15] (* Jean-François Alcover, Jan 01 2021, after Alois P. Heinz *)

Crossrefs

Cf. A000108, A210029, A242446, A253180, A256254, A258426.

Sequence in context: A383734 A223038 A324266 * A212838 A024241 A278684

Adjacent sequences: A258396 A258397 A258398 * A258400 A258401 A258402

Keyword

nonn

Author

Alois P. Heinz, May 28 2015

Status

approved