G.f.: (1 - x - (1 - 6*x + x^2)^(1/2))/(2*x).
For n > 0, a(n) = (1/n)*Sum_{k = 0..n} 2^k*C(n, k)*C(n, k-1). -
Benoit Cloitre, May 10 2003
The g.f. satisfies (1 - x)*A(x) - x*A(x)^2 = 1. -
Ralf Stephan, Jun 30 2003
a(n) = Sum_{k = 0..n} C(n+k, n)*C(n, k)/(k+1). (End)
With offset 1: a(1) = 1, a(n) = a(n-1) + Sum_{i = 1..n-1} a(i)*a(n-i). -
Benoit Cloitre, Mar 16 2004
a(n) = (CentralDelannoy(n+1) - 3 * CentralDelannoy(n))/(2*n) = (-CentralDelannoy(n+1) + 6 * CentralDelannoy(n) - CentralDelannoy(n-1))/2 for n >= 1, where CentralDelannoy is
A001850. -
David Callan, Aug 16 2006
and 2*
A123164(n) = (n+1)*a(n) - (n-1)*a(n-1) (n > 0). (End)
Define the general Delannoy numbers d(i, j) as in
A001850. Then a(k) = d(2*k, k) - d(2*k, k-1) and a(0) = 1, Sum_{j=0..n} ((-1)^j * (d(n, j) + d(n-1, j-1)) * a(n-j)) = 0. -
Peter E John, Oct 19 2006
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, Phi([1]) is the Catalan numbers
A000108. The present sequence is (essentially) Phi([2]). -
Gary W. Adamson, Oct 27 2008
G.f.: 1/(1-2x/(1-x/(1-2x/(1-x/(1-2x/(1-x/(1-2x/(1-x/(1-2x/(1-x.... (continued fraction). -
Paul Barry, Dec 08 2008
G.f.: 1/(1 - x - x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ... (continued fraction). -
Paul Barry, Jan 29 2009
a(n) ~ ((3 + 2*sqrt(2))^n)/(n*sqrt(2*Pi*n)*sqrt(3*sqrt(2) - 4))*(1-(9*sqrt(2) + 24)/(32*n) + ...). - G. Nemes (nemesgery(AT)gmail.com), Jan 25 2009
a(n) = the upper left term in M^(n+1), M = the production matrix:
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
2, 2, 1, 1, 0, 0, ...
4, 4, 2, 1, 1, 0, ...
8, 8, 8, 2, 1, 1, ...
a(n) is the sum of top row terms in Q^n, Q = an infinite square production matrix as follows:
1, 1, 0, 0, 0, 0, ...
1, 1, 2, 0, 0, 0, ...
1, 1, 1, 2, 0, 0, ...
1, 1, 1, 1, 2, 0, ...
1, 1, 1, 1, 1, 2, ...
With F(x) = (1 - 3*x - sqrt(1 - 6*x + x^2))/(2*x) an o.g.f. (nulling the n = 0 term) for
A006318, G(x) = x/(2 + 3*x + x^2) is the compositional inverse.
Consequently, with H(x) = 1/ (dG(x)/dx) = (2 + 3*x + x^2)^2 / (2 - x^2),
a(n) = (1/n!)*[(H(x)*d/dx)^n] x evaluated at x = 0, i.e.,
F(x) = exp[x*H(u)*d/du] u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)). (End)
a(n-1) = number of ordered complete binary trees with n leaves having k internal vertices colored black, the remaining n - 1 - k internal vertices colored white, and such that each vertex and its rightmost child have different colors ([Drake, Example 1.6.7]). For a refinement of this sequence see
A175124. -
Peter Bala, Sep 29 2011
D-finite with recurrence: (n-2)*a(n-2) - 3*(2*n-1)*a(n-1) + (n+1)*a(n) = 0. -
Vaclav Kotesovec, Oct 05 2012
G.f.: A(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) = (1 - G(0))/x; G(k) = 1 + x - 2*x/G(k+1); (continued fraction, 1-step). -
Sergei N. Gladkovskii, Jan 04 2012
G.f.: A(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) = (G(0) - 1)/x; G(k) = 1 - x/(1 - 2/G(k+1)); (continued fraction, 2-step). -
Sergei N. Gladkovskii, Jan 04 2012
G.f.: 1/Q(0) where Q(k) = 1 + k*(1 - x) - x - x*(k+1)*(k+2)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Mar 14 2013
G.f.: 1/x - 1 - U(0)/x, where U(k) = 1 - x - x/U(k+1); (continued fraction). -
Sergei N. Gladkovskii, Jul 16 2013
G.f.: (2 - 2*x - G(0))/(4*x), where G(k) = 1 + 1/( 1 - x*(6 - x)*(2*k - 1)/(x*(6 - x)*(2*k - 1) + 2*(k + 1)/G(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Jul 16 2013
a(n) = 1/(n + 1) * (Sum_{j=0..n} C(n+j, j)*C(n+j+1, j+1)*(Sum_{k=0..n-j} (-1)^k*C(n+j+k, k))). -
Graham H. Hawkes, Feb 15 2015
a(n) = hypergeom([-n, n+1], [2], -1). -
Peter Luschny, Mar 23 2015
a(n) = sqrt(2) * LegendreP(n, -1, 3) where LegendreP is the associated Legendre function of the first kind (in Maple's notation). -
Robert Israel, Mar 23 2015
G.f. A(x) satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*A(x)^k. -
Ilya Gutkovskiy, Apr 11 2019
a(n) = 2 * Sum_{k = 0..floor(n/2)} binomial(n, 2*k)*binomial(2*n-2*k, n)/(n-2*k+1) for n >= 1.
a(n) = Integral_{x = 0..1} Legendre_P(n, 2*x+1) dx. (End)
G.f. A(x) = 1/(1 - x) * c(x/(1-x)^2), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers
A000108. -
Peter Bala, Aug 29 2024
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