cp's OEIS Frontend

Series reversion of x(1-x^2)/(1+x^2)^2 expanded in odd powers of x. [Previous name.]

If x = y*(1-y^2)/(1+y^2)^2 then y = x + 3*x^3 + 22*x^5 + 211*x^7 + 2306*x^9 + ...

G.f. A(x) satisfies x*A(x^2) = (C(x) - C(-x))/(C(x) + C(-x)) where C(x) is g.f. of the Catalan numbers A000108.

a(n) = Sum_{k=0..2n} (-1)^k * A000108(2*n-k) * A000108(k). - David Callan, Aug 16 2006

a(n) = ((2^(4n+2))/Gamma(1/2)) * ((Gamma(n+1/2)/(2*Gamma(n+2))) - Gamma(2n+3/2)/Gamma(2n+3)). [David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009]

G.f.: exp( Sum_{n>=1} C(4n-1,2n)*x^n/n ). - Paul D. Hanna, Dec 30 2010

G.f.: C(sqrt(x))*C(-sqrt(x)) where C(x) is the g.f. for the Catalan numbers A000108. - David Callan, Apr 10 2012

D-finite with recurrence n*(n+1)*(2*n+1)*a(n) -2*n*(32*n^2-32*n+11)*a(n-1) +16*(4*n-5)*(4*n-3)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 29 2012

a(n) ~ (2-sqrt(2))*16^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 20 2013

a(n) = 2^(2*n+1)*Catalan(n) - Catalan(2*n+1) (see Regev). It follows that the 2-adic valuations of a(n) and Catalan(n) are equal. In particular, a(n) is odd iff n is of the form 2^m - 1. - Peter Bala, Aug 02 2016

G.f.: (sqrt(2) * sqrt(1 + sqrt(1-16*x)) - sqrt(1-16*x) - 1)/(4*x). - Vladimir Reshetnikov, Sep 25 2016

G.f. A(x) satisfies A(x^2) = C(x)^2*r(-x*C(x)^2), where C(x) is g.f. of the Catalan numbers A000108, and r(x) is g.f. of the large Schröder numbers A006318. - Alexander Burstein, Nov 21 2019

From Peter Bala, Sep 14 2021: (Start)

A(x) = exp( Sum_{n >= 1} (1/2)*binomial(4*n,2*n)*x^n/n ).

1 + x*A(x) is the o.g.f. of A066357.

The sequence defined by b(n) := [x^n] A(x)^n begins [1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, ...] and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. See A333563. Cf. A060941. (End)

From Peter Bala, Oct 23 2024: (Start)

For integer r and positive integer s, define a sequence {u(n) : n >= 0} by setting u(n) = [x^(s*n)] A(x)^(r*n). We conjecture that the supercongruence u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k.

Let B(x) = 1/x * series_reversion(x*A(x)). Define a sequence {v(n) : n >= 0} by setting v(n) = [x^(s*n)] B(x)^(r*n). We conjecture that the supercongruence v(n*p^k) == v(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k. (End)

For n>0, a(n) = Sum_{r=1..n} C(2*r-1)*a(n-r). Here C(2*r-1) is a Catalan number (A000108). - Paul Boddington, Mar 14 2003

G.f.: 2/(1+4*sqrt(x)/(sqrt(1+4*sqrt(x))-sqrt(1-4*sqrt(x)))).

D-finite with recurrence a(n)*(2*n-1)*(n+1)n = a(n-1)*(32*n^2 - 64*n + 39)*2*n - a(n-2)*(2*n-3)*(4*n-5)*(4*n-7)*16, n>1.

a(0) = 1,a(n) = (1/n)*Sum_{k=0..n} C(4*n,k)*C(3*n-k-2,n-k-1), n>1. - Paul Barry, Apr 09 2007

a(n) = ((2^(4*n))/Gamma(1/2)) * ((6*(2*n+1)*Gamma(2*n+1/2)/Gamma(2*n+3))-2*Gamma(n+1/2)/Gamma(n+2)). - David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009

Convolution of A079489 with itself: (1, 6, 53, 554, ...) = (1, 3, 22, 211, ...)*(1, 3, 22, 211, ...).

Proof. Working with Dyck paths, we must show that Dyck paths of size (semilength) 2n, all of whose components (constituent primitive Dyck paths) have even size, are equinumerous with ordered pairs of nonempty Dyck paths of total size 2n in each of which the first component is of odd size and all other components (if any) are of even size. Given a Dyck path P of the former class, use the first return decomposition to write P (uniquely) as the concatenation of U A_1 A_2 ... A_j O E D Q where U denotes upstep, D denotes downstep, A_1,...,A_j are all primitive Dyck paths of even size with j>=0, O is a primitive Dyck path of odd size, E is a Dyck path of even size, and Q is a Dyck path in which all components are of even size. Then P -> (O A_1 A_2 ... A_j, U E D Q) is the desired bijection. QED - David Callan, Apr 11 2012

a(n) = C(2*n+1) + 2*C(2*n) - 2^(2*n+1)*C(n), where C(n) is the Catalan number A000108. This formula can be obtained by manipulating generating functions. The equivalence of this formula and the Barry (Apr 09 2007) sum can be established by the WZ method with a second-order operator. A combinatorial interpretation of the Barry sum would be nice. - David Callan, Apr 10 2012

a(n) ~ (3-2*sqrt(2)) * 2^(4*n) / (n^(3/2) * sqrt(2*Pi)). - Vaclav Kotesovec, Mar 21 2014

exp( Sum_{n >= 1} binomial(4*n,2*n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... is an o.g.f. for this sequence omitting the initial term. See A001448. - Peter Bala, Oct 02 2015

a(n) = binomial(3*n-2,n-1)*hypergeom([1-n,-4*n],[2-3*n],-1)/n for n>=1. - Peter Luschny, Oct 15 2015

a(n) = 3*(2*n+1) /(2*n+2) /(4*n+1) *binomial(4*n+2,2*n+1) -4^n /(2*n+1) *binomial(2*n+2,n+1) [Merlini et al F_n formula] - R. J. Mathar, Oct 01 2021