A275206 Expansion of (A(x)^2 - A(x^2))/2 where A(x) = A000108(x) - 1.

0, 0, 0, 2, 6, 24, 80, 286, 994, 3536, 12576, 45220, 163372, 594320, 2172768, 7983990, 29464010, 109174560, 405990464, 1514797020, 5669004692, 21275014800, 80047213792, 301892460012, 1141068949396, 4321730134624, 16399422014400, 62340424959176, 237373155238104, 905251034394784

Offset: 0

Author

R. J. Mathar, Jul 19 2016

Keywords

• nonn

Comments

Number of ways to distribute n pairs of parentheses into 2 distinct patterns where each pattern represents a Catalan ordering (A000108), and each pattern must contain at least one pair of parentheses.

If one of the groups is allowed to have no parentheses, we arrive at A000150 (with a different offset).

Analog of A216785 with Catalan number replacing connected graph counts.

From Petros Hadjicostas, Jul 27 2020: (Start)

It is proved in A050182 that A050182(n) = 1/(2*n + 4)*(binomial(2*n + 4, n) - [(n mod 2) == 0]*binomial(n + 2, n/2)).

Let C(x) = A(x) + 1 = Sum_{n >= 0} c(n)*x^n be the g.f. of the Catalan numbers A000108. Then C(x)^2 = (C(x) - 1)/x. Then (A(x) + 1)^2 = A(x)/x, and thus, A(x)^2 = -2*A(x) - 1 + A(x)/x. Thus, (A(x)^2 - A(x^2))/2 = (-2*A(x) - 1 + A(x)/x - A(x^2))/2.

Substituting A(x) = Sum_{n >= 1} c(n)*x^n in the above expression, we get (after some algebra) that a(n) = (-2*c(n) + c(n+1) - [(n mod 2) == 0]*c(n/2))/2 for n >= 1. It is then easy to prove that a(n) = 2*A050182(n-2) = (1/n)*(binomial(2*n, n-2) - [(n mod 2) == 0]*binomial(n, (n/2) - 1)) for n >= 2, thus proving the conjecture below. (End)

Crossrefs

Cf. A000108, A000150, A007223, A050182, A216785.

Programs

Mathematica

A[x_] = (1 - Sqrt[1 - 4x])/(2x) - 1;
CoefficientList[(A[x]^2 - A[x^2])/2 + O[x]^30, x] (* Jean-François Alcover, Apr 30 2023 *)

Formula

a(2*n+1) = A007223(2*n+1).

Conjecture: a(n) = 2*A050182(n-2) for n >= 2.

From Petros Hadjicostas, Jul 27 2020: (Start)

a(n) = (-2*c(n) + c(n+1) - [(n mod 2) == 0]*c(n/2))/2 for n >= 1, where c = A000108.

a(n) = (1/n)*(binomial(2*n, n-2) - [(n mod 2) == 0]*binomial(n, (n/2) - 1)) for n >= 2. (End)

D-finite with recurrence: 5*(n-3)*(n+2)*(n+1)*a(n) + 32*(n-5)*(n-6)*(2*n-11)*a(n-6) + 80*(n-5)*(n-4)*(n-3)*a(n-5) - 40*(n-4)*(7*n^2-35*n+39)*a(n-4) + (120*n^3 - 600*n^2 + 660*n + 60)*a(n-3) + 2*(n-5)*(23*n^2-44*n+6)*a(n-2) - 5*(n+1)*(7*n^2-25*n+6)*a(n-1) = 0. - Georg Fischer, Feb 12 2025