A281733 -id:A281733 - cp's OEIS frontend

A078531 Coefficients of power series that satisfies A(x)^2 - 4xA(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

Paul D. Hanna, Nov 28 2002

nonn

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
The even bisection halved gives A176898. The odd bisection halved gives A281733. - Akiva Weinberger, Dec 09 2024

			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.

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.

Cf. A002420, A078532, A078533, A078534, A078535, A214377, A176898, A281733.

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

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 *)

taylor(sqrt(3)/2*(sech(acosh(-sqrt(108)*x)/3)),x,0,10); /* Vladimir Kruchinin, Oct 12 2022 */

{a(n) = if( n<0, 0, n++; polcoeff( serreverse( x * sqrt( 1 - 4*x + O(x^n))), n))}; /* Michael Somos, Feb 05 2004 */

{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 */

{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 */

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*A176898(n/2) for even n and a(n) = 2*A281733((n+1)/2) for odd n.
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

A176898 a(n) = binomial(6n, 3n)binomial(3n, n)/(2(2n+1)binomial(2n, n)).

Original entry on oeis.org

5, 231, 14586, 1062347, 84021990, 7012604550, 607892634420, 54200780036595, 4938927219474990, 457909109348466930, 43057935618181929900, 4096531994713828810686, 393617202432246696493436, 38142088615983865845923052, 3723160004902167033863327592

Offset: 1

Zhi-Wei Sun, Apr 28 2010

During April 26-28, 2010, Zhi-Wei Sun introduced this new sequence and proved that a(n) = binomial(6n,3n)*binomial(3n,n)/(2*(2n+1)*binomial(2n,n)) is a positive integer for every n=1,2,3,... He also observed that a(n) is odd if and only if n is a power of two, and that 3a(n)=0 (mod 2n+3). By Stirling's formula, we have lim_n (8n*sqrt(n*Pi)a(n)/108^n) = 1. It is interesting to find a combinatorial interpretation or recursion for the sequence.
From Tatiana Hessami Pilehrood, Dec 01 2015: (Start)
Zhi-Wei Sun formulated two conjectures concerning a(n) (see Conjectures 1.1 and 1.2 in Z.-W. Sun, "Products and sums divisible by central binomial coefficients" and Conjecture A89 in "Open conjectures on congruences"). The first conjecture states that Sum_{n=1..p-1} a(n)/(108^n) is congruent to 0 or -1 modulo a prime p > 3 depending on whether p is congruent to +-1 or +-5 modulo 12, respectively.
The second conjecture asks about an exact formula for a companion sequence of a(n). Both conjectures as well as many numerical congruences involving a(n) and (2n+1)a(n) were solved by Kh. Hessami Pilehrood and T. Hessami Pilehrood, see the link below. (End)
This is the even bisection of A078531 divided by 2. The odd bisection divided by 2 is A281733. - Akiva Weinberger, Dec 09 2024

			For n=2 we have a(2) = binomial(12,6)*binomial(6,2)/(2*(2*2+1)*binomial(4,2)) = 231.

Indranil Ghosh, Table of n, a(n) for n = 1..450
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi polynomials and congruences involving some higher-order Catalan numbers and binomial coefficients, preprint, arXiv:1504.07944 [math.NT], 2015.
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7
Mark Roger Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
Chen Wang and Hui-Li Han, Supercongruences involving binomial coefficients and Euler polynomials, arXiv:2407.19882 [math.NT], 2024. See p. 2.
Zhi-Wei Sun, Products and sums divisible by central binomial coefficients, preprint, arXiv:1004.4623 [math.NT], 2010.
Zhi-Wei Sun, Open conjectures on congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
Brian Y. Sun and J. X. Meng, Proof of a Conjecture of Z.-W. Sun on Trigonometric Series, arXiv preprint arXiv:1606.08153 [math.CO], 2016.
Chen Wang and Hui-Li Han, Supercongruences involving binomial coefficients and Euler polynomials, arXiv:2407.19882 [math.NT], 2024. See p. 2.
Xiran Zhang and Guo-Shuai Mao, Congruences involving binomial coefficients and Legendre symbol, ResearchGate (2024). See p. 10.

Cf. A000984, A006013, A173774, A176285, A176477, A078531, A281733.

[Binomial(6*n, 3*n)*Binomial(3*n, n)/(2*(2*n+1)*Binomial(2*n, n)): n in [1..15]]; // Vincenzo Librandi, Dec 02 2015

ogf := eval((1-6*s)/((12*s-1)*(8*s-2)) - 1/2, s=RootOf(x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2,s));
series(ogf,x=0,30); # Mark van Hoeij, May 06 2013

S[n_]:=Binomial[6n,3n]Binomial[3n,n]/(2(2n+1)Binomial[2n,n]) Table[S[n],{n,1,50}]

a(n) = binomial(6*n, 3*n) * binomial(3*n, n) / (2*(2*n+1) * binomial(2*n, n)); \\ Indranil Ghosh, Mar 05 2017

import math
f=math.factorial
def C(n,r): return f(n)/f(r)/f(n-r)
def A176898(n): return C(6*n, 3*n) * C(3*n, n) / (2*(2*n+1) * C(2*n, n)) # Indranil Ghosh, Mar 05 2017

G.f.: (1-6*s)/((12*s-1)*(8*s-2)) - 1/2, where x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2 = 0. - Mark van Hoeij, May 06 2013
a(n) ~ 2^(2*n-3) * 3^(3*n) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jan 09 2023
From Peter Bala, Feb 21 2023: (Start)
a(n+1) = 6*(6*n + 1)*(6*n + 5)/((n + 1)*(2*n + 3))*a(n).
a(n) = (2^(2*n-1)) * Product_{1 <= i <= j <= 2*n-1} (2*i + j + 2)/(2*i + j - 1). Cf. A006013. (End)
D-finite with recurrence n*(2*n+1)*a(n) -6*(6*n-1)*(6*n-5)*a(n-1)=0. - R. J. Mathar, Nov 22 2024
a(n) = 2^(4n-1) * binomial(3n-1/2, 2n)/(2n+1). - Akiva Weinberger, Dec 09 2024
a(n) = A078531(2*n)/2. - Akiva Weinberger, Dec 09 2024

cp's OEIS Frontend

A078531 Coefficients of power series that satisfies A(x)^2 - 4xA(x)^3 = 1, A(0)=1.

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A176898 a(n) = binomial(6n, 3n)binomial(3n, n)/(2(2n+1)binomial(2n, n)).

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A284731 Square array read by antidiagonals downwards: T(n,k) = number of linear extensions of the North-East rectangular partial order NE_{n,k} that avoid the pattern 123.

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cp's OEIS Frontend

A078531 Coefficients of power series that satisfies A(x)^2 - 4*x*A(x)^3 = 1, A(0)=1.

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Maple

Mathematica

Maxima

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Formula

A176898 a(n) = binomial(6*n, 3*n)*binomial(3*n, n)/(2*(2*n+1)*binomial(2*n, n)).

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Magma

Maple

Mathematica

PARI

Python

Formula

A284731 Square array read by antidiagonals downwards: T(n,k) = number of linear extensions of the North-East rectangular partial order NE_{n,k} that avoid the pattern 123.

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A078531 Coefficients of power series that satisfies A(x)^2 - 4xA(x)^3 = 1, A(0)=1.

A176898 a(n) = binomial(6n, 3n)binomial(3n, n)/(2(2n+1)binomial(2n, n)).