A078531 Coefficients of power series that satisfies A(x)^2 - 4*x*A(x)^3 = 1, A(0)=1.
1, 2, 10, 64, 462, 3584, 29172, 245760, 2124694, 18743296, 168043980, 1526726656, 14025209100, 130056978432, 1215785268840, 11445014102016, 108401560073190, 1032295389593600, 9877854438949980, 94927710773575680, 915818218696933860, 8866494751734497280
Offset: 0
Keywords
Examples
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.
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.
- Ira M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656 [math.PR], 2014.
- W. Mlotkowski, Karol 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.
Programs
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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
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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 *) -
Maxima
taylor(sqrt(3)/2*(sech(acosh(-sqrt(108)*x)/3)),x,0,10); /* Vladimir Kruchinin, Oct 12 2022 */
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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 */
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 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)
From Akiva Weinberger, Dec 09 2024: (Start)
a(n) = 2*binomial(3*n, 3*n/2)*binomial(3*n/2, n/2)/(2*(n+1)*binomial(n, n/2)).
a(n) = 2^(2*n)/((n+1)*(3*n+1)*(Integral_{t=0..1} (t-t^3)^n dt)). (End)
G.f.: 2*hypergeometric([2/3,1,4/3],[3/2,2],108*x^2)*x + hypergeometric([1/6,5/6],[3/2],108*x^2). - Vladimir Kruchinin, Feb 25 2025
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^4). - Seiichi Manyama, Jun 20 2025
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