This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A275206 #27 Feb 12 2025 12:56:09
%S A275206 0,0,0,2,6,24,80,286,994,3536,12576,45220,163372,594320,2172768,
%T A275206 7983990,29464010,109174560,405990464,1514797020,5669004692,
%U A275206 21275014800,80047213792,301892460012,1141068949396,4321730134624,16399422014400,62340424959176,237373155238104,905251034394784
%N A275206 Expansion of (A(x)^2 - A(x^2))/2 where A(x) = A000108(x) - 1.
%C A275206 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 A275206 If one of the groups is allowed to have no parentheses, we arrive at A000150 (with a different offset).
%C A275206 Analog of A216785 with Catalan number replacing connected graph counts.
%C A275206 From _Petros Hadjicostas_, Jul 27 2020: (Start)
%C A275206 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 A275206 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 A275206 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 A275206 a(2*n+1) = A007223(2*n+1).
%F A275206 Conjecture: a(n) = 2*A050182(n-2) for n >= 2.
%F A275206 From _Petros Hadjicostas_, Jul 27 2020: (Start)
%F A275206 a(n) = (-2*c(n) + c(n+1) - [(n mod 2) == 0]*c(n/2))/2 for n >= 1, where c = A000108.
%F A275206 a(n) = (1/n)*(binomial(2*n, n-2) - [(n mod 2) == 0]*binomial(n, (n/2) - 1)) for n >= 2. (End)
%F A275206 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
%t A275206 A[x_] = (1 - Sqrt[1 - 4x])/(2x) - 1;
%t A275206 CoefficientList[(A[x]^2 - A[x^2])/2 + O[x]^30, x] (* _Jean-François Alcover_, Apr 30 2023 *)
%Y A275206 Cf. A000108, A000150, A007223, A050182, A216785.
%K A275206 nonn
%O A275206 0,4
%A A275206 _R. J. Mathar_, Jul 19 2016