OFFSET
0,2
COMMENTS
Radius of convergence of g.f. A(x) is r = 1/(2*3^(3/2)) where A(r) = sqrt(3).
If A(x)=sum_{k=1..inf} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(k)=n^(2k)*binomial(k/n+1/n+k-1,k)/(k+1) and, consequently, a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2). - Emeric Deutsch, Dec 10 2002
A generalization of the Catalan sequence (A000108) since for n = 1 the equation A(x)^n -(n^2)*x*A(x)^(n+1) = 1 reduces to A(x)=1+xA(x)^2. - Emeric Deutsch, Dec 10 2002
Number of symmetric non-crossing connected graphs on 2n+1 equidistant nodes on a circle (it is assumed that the axis of symmetry is a diameter of the circle passing through a given node). Example: a(1)=2 because on the nodes A,B,C (axis of symmetry through A) the only symmetric non-crossing connected graphs are {AB,AC} and {AB,AC,BC}. - Emeric Deutsch, Dec 03 2003
LINKS
Robert Israel, Table of n, a(n) for n = 0..900
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq. 14 (2011) # 11.4.3, example 10.
Gi-Sang Cheon, S.-T. Jin, and L. W. Shapiro, A combinatorial equivalence relation for formal power series, Linear Algebra and its Applications, Available online 30 March 2015.
P. Flajolet and M. Noy, Analytic combinatorics of non-crossing configurations, Discrete Math., 204, 203-229, 1999.
Loïc Foissy, Free quadri-algebras and dual quadri-algebras, arXiv preprint, 2015.
I. M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656, 2014
W. Mlotkowski, K. A. Penson and K. Zyczkowski, Densities of the Raney distributions, arXiv preprint arXiv:1211.7259, 2012. - From N. J. A. Sloane, Jan 03 2013
V. U. Pierce, Continuum limits of Toda lattices for map enumeration, in Algebraic and Geometric Aspects of Integrable Systems and Random Matrices, edited by Anton Dzhamay, Ken'ichi Maruno, Virgil U. Pierce; Contemporary Mathematics, Vol. 593, 2013.
Vincent Pilaud, Pebble trees, arXiv:2205.06686 [math.CO], 2022.
M. R. Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
FORMULA
a(n) = 2*(sum_{i=0..n-2} binomial(3n-3, i)*binomial(2n-2-i, n))/(n-1) for n>1. - Emeric Deutsch, Nov 29 2002
G.f.: (12x)^(-1) + (6x)^(-1)*sin(arcsin(216x^2-1)/3). - Emeric Deutsch, Nov 30 2002
a(n) = 2^(2n)*binomial(3n/2-1/2, n)/(n+1). - Emeric Deutsch, Dec 10 2002
G.f. A(x) = y satisfies y' * (6*x*y - 1) + 2*y^2 = 0, y' * (y^2 - 3) + 4*y^4 = 0. - Michael Somos, Feb 05 2004
Sequence with offset 1 is expansion of reversion of g.f. x*sqrt(1-4x). - Ralf Stephan, Mar 22 2004
G.f. satisfies: A(x) = 1 / sqrt(1 - 4*x*A(x)).
G.f. satisfies: A(x) = Sum_{n>=0} ((2*n)!/n!^2)*x^n*A(x)^n. - Paul D. Hanna, Mar 03 2011
Self-convolution yields A214377, where A214377(n) = 4^n*binomial(3/2*n,n)*2/(n+2). - Paul D. Hanna, Jul 14 2012
D-finite with recurrence n*(n+1)*a(n) + n*(n-1)*a(n-1) - 12*(3*n-1)*(3*n-5)*a(n-2) - 12*(3*n-4)*(3*n-8)*a(n-3) = 0. - R. J. Mathar, Jun 07 2013
REVERSION transform of A002420 (both offsets 1). - Michael Somos, Jun 18 2014
0 = a(n)*(16*a(n+1) - 10*a(n+2)) + a(n+1)*(2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jun 18 2014
a(n) ~ 2^(n-1/2) * 3^(3*n/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
G.f. satisfies: 1-2*x*A(x)*C(x*A(x)) = 1/A(x), where C is g.f. of A000108. - Werner Schulte, Aug 07 2015
G.f.: (sqrt(3)/2)*(sech(arccosh(-sqrt(108)*x)/3)). - Vladimir Kruchinin, Oct 11 2022
From Karol A. Penson, Oct 28 10 2024: (Start)
G.f.: ((i*sqrt(3)-1)*g1(z)-(i*sqrt(3)+1)*g2(z)+2)/(24*z), with g1(z) = (sqrt(-108*z^2 + 1) - 6*i*sqrt(3)*z)^(2/3), and g2(z) = (sqrt(-108*z^2 + 1) + 6*i*sqrt(3)*z)^(2/3), where i = sqrt(-1), the imaginary unit.
a(n) = Integral_{x=0..sqrt(108)} x^n*W(x), where W(x) = (3^(1/6)/(24*Pi*x^(2/3)))* (W1(x) - W2(x)), with W1(x) = (18 + sqrt(-3*x^2 + 324))^(2/3) and
W2(x) = (18 - sqrt(-3*x^2 + 324))^(2/3).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, sqrt(108)). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-2/3), and for x > 0 is monotonically decreasing to zero at x = sqrt(108). (End)
EXAMPLE
G.f. = 1 + 2*x + 10*x^2 + 64*x^3 + 462*x^4 + 3584*x^5 + 29172*x^6 + ...
A(x)^2 - 4x*A(x)^3 = 1 since A(x)^2 = 1 + 4x + 24x^2 + 148x^3 + 1280x^4 + 10296x^5 + ... and A(x)^3 = 1 + 6x + 42x^2 + 320x^3 + 2574x^4 + ... also a(1)=2^1, a(3)=2^6.
MAPLE
S:= series(RootOf(Z^2 - 4*x*Z^3-1, Z, 1), x, 101):
seq(coeff(S, x, j), j=0..100); # Robert Israel, Aug 07 2015
MATHEMATICA
a[n_] := 2^(2n)*Binomial[3n/2-1/2, n]/(n+1); Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jan 21 2013, after Emeric Deutsch *)
a[ n_] := With[ {m = n + 1}, If[ m < 1, 0, SeriesCoefficient[ InverseSeries @ Series[ x Sqrt[1 - 4 x], {x, 0, m}], {x, 0, m}]]]; (* Michael Somos, Jun 18 2014 *)
PROG
(PARI) {a(n) = if( n<0, 0, n++; polcoeff( serreverse( x * sqrt( 1 - 4*x + O(x^n))), n))}; /* Michael Somos, Feb 05 2004 */
(PARI) {a(n) = if( n<1, n==0, polcoeff( serreverse( x * (2 + x) / (4 * (1 + x)^3) + x * O(x^n)), n))}; /* Michael Somos, Feb 05 2004 */
(PARI) {a(n)=local(B=sum(m=0, n, binomial(2*m, m)*x^m+x*O(x^n))); polcoeff(1/x*serreverse(x/B), n)} /* Paul D. Hanna, Mar 03 2011 */
(Maxima)
taylor(sqrt(3)/2*(sech(acosh(-sqrt(108)*x)/3)), x, 0, 10); /* Vladimir Kruchinin Oct 12 2022 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 28 2002
STATUS
approved