A275206 - OEIS

%I #24 Apr 30 2023 07:10:38

%S 0,0,0,2,6,24,80,286,994,3536,12576,45220,163372,594320,2172768,

%T 7983990,29464010,109174560,405990464,1514797020,5669004692,

%U 21275014800,80047213792,301892460012,1141068949396,4321730134624,16399422014400,62340424959176,237373155238104,905251034394784

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

%C 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.

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

%C Analog of A216785 with Catalan number replacing connected graph counts.

%C From _Petros Hadjicostas_, Jul 27 2020: (Start)

%C 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)).

%C 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.

%C 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)

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

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

%F From _Petros Hadjicostas_, Jul 27 2020: (Start)

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

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

%t A[x_] = (1 - Sqrt[1 - 4x])/(2x) - 1;

%t CoefficientList[(A[x]^2 - A[x^2])/2 + O[x]^30, x] (* _Jean-François Alcover_, Apr 30 2023 *)

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

%K nonn

%O 0,4

%A _R. J. Mathar_, Jul 19 2016