A078531 -id:A078531 - cp's OEIS frontend

A224884 Expansion of x / Series_Reversion(xsqrt(1 + 4x)).

1, 2, -6, 32, -210, 1536, -12012, 98304, -831402, 7208960, -63740820, 572522496, -5209363380, 47915728896, -444799488600, 4161823309824, -39209074920090, 371626340253696, -3541117629057540, 33902753847705600, -325969196485349340, 3146175557067079680, -30471769822097981160

Offset: 0

Paul D. Hanna, Aug 21 2013

sign

Signed version of A206300. - Peter Bala, Mar 05 2020

			G.f.: A(x) = 1 + 2*x - 6*x^2 + 32*x^3 - 210*x^4 + 1536*x^5 - 12012*x^6 + ..
The coefficients in the powers A(x)^n of the g.f. begin:
n= 1: [1,  2,  -6,   32,  -210,  1536,-12012,  98304, -831402, ...];
n= 2: [1,  4,  -8,   40,  -256,  1848,-14336, 116688, -983040, ...];
n= 3: [1,  6,  -6,   32,  -210,  1536,-12012,  98304, -831402, ...];
n= 4: [1,  8,   0,   16,  -128,  1008, -8192,  68640, -589824, ...];
n= 5: [1, 10,  10,    0,   -50,   512, -4620,  40960, -364650, ...];
n= 6: [1, 12,  24,   -8,     0,   168, -2048,  20592, -196608, ...];
n= 7: [1, 14,  42,    0,    14,     0,  -588,   8192,  -90090, ...];
n= 8: [1, 16,  64,   32,     0,   -32,     0,   2112,  -32768, ...];
n= 9: [1, 18,  90,   96,   -18,     0,    84,      0,   -7722, ...];
n=10: [1, 20, 120,  200,     0,    24,     0,   -240,       0, ...];
n=11: [1, 22, 154,  352,   110,     0,   -44,      0,     726, ...];
n=12: [1, 24, 192,  560,   384,   -48,     0,     96,       0, ...];
n=13: [1, 26, 234,  832,   910,     0,    52,      0,    -234, ...];
n=14: [1, 28, 280, 1176,  1792,   392,     0,    -80,       0, ...];
n=15: [1, 30, 330, 1600,  3150,  1536,  -140,      0,     150, ...];
n=16: [1, 32, 384, 2112,  5120,  4032,     0,    128,       0, ...];
n=17: [1, 34, 442, 2720,  7854,  8704,  1428,      0,    -170, ...];
n=18: [1, 36, 504, 3432, 11520, 16632,  6144,   -432,       0, ...];
n=19: [1, 38, 570, 4256, 16302, 29184, 17556,      0,     342, ...];
n=20: [1, 40, 640, 5200, 22400, 48048, 40960,   5280,       0, ...]; ...
which illustrates the property [x^n] A(x)^(n+2*k) = 0 for k=1..n-1:
[x^2] A(x)^4 = 0;
[x^3] A(x)^5 = 0, [x^3] A(x)^7 = 0;
[x^4] A(x)^6 = 0, [x^4] A(x)^8 = 0, [x^4] A(x)^10 = 0; ...
[x^5] A(x)^7 = 0, [x^5] A(x)^9 = 0, [x^5] A(x)^11 = 0, [x^5] A(x)^13 = 0; ...
Related series:
sqrt(1+4*x) = 1 + 2*x - 2*x^2 + 4*x^3 - 10*x^4 + 28*x^5 - 84*x^6 + 264*x^7 - 858*x^8 + ... + (-1)^(n-1)*2*A000108(n-1)*x^n + ...

G. C. Greubel, Table of n, a(n) for n = 0..980

Cf. A000108, A206300.

Cf. A000984, A048990, A078531, A182122, A245112, A245113.

CoefficientList[Series[x/InverseSeries[Series[x*Sqrt[1+4*x],{x,0,20}],x],{x,0,20}],x] (* Vaclav Kotesovec, Aug 22 2013 *)

{a(n)=polcoeff(x/serreverse(x*sqrt(1+4*x +x^2*O(x^n))),n)}
for(n=0,25,print1(a(n),", "))

G.f. A(x) satisfies:
(1) A(x) = A(x)^3 - 4*x.
(2) A(x) = sqrt(1 + 4*x/A(x)).
(3) A(x*sqrt(1+4*x)) = sqrt(1+4*x).
(4) [x^n] A(x)^(n+2*k) = 0 for k=1..n-1, for n >= 2.
From Vaclav Kotesovec, Aug 22 2013: (Start)
a(n) = (-1)^(n+1) * 3^(3*n/2-1) * 4^(n-1) * GAMMA(n/2 - 1/6) * GAMMA(n/2 + 1/6)/(Pi*n!).
|a(n)| ~ 6^(n-1)*3^(n/2)/(sqrt(Pi/2)*n^(3/2)).
D-finite with recurrence: (n-1)*n*a(n) = 12*(3*n-7)*(3*n-5)*a(n-2). (End)
G.f.: (2/sqrt(3))*cosh(1/3*arccosh(sqrt(108)*x)). - Vladimir Kruchinin, Oct 11 2022
G.f. A(x) satisfies A(x) = 1/A(-x/A(x)^4). - Seiichi Manyama, Jun 20 2025

A245113 G.f. A(x) satisfies A(x)^2 = 1 + 4xA(x)^6.

Original entry on oeis.org

1, 2, 22, 340, 6118, 120060, 2492028, 53798888, 1195684230, 27175425004, 628705751828, 14756641134872, 350529497005532, 8410852483002200, 203561027031883320, 4963404936414528720, 121810229481173225670, 3006555636255509030220, 74585744314812449403300, 1858695101618327423328312

Offset: 0

Paul D. Hanna, Jul 31 2014

Radius of convergence of g.f. A(x) is r = 1/27 where A(r) = sqrt(3/2).

			G.f.: A(x) = 1 + 2*x + 22*x^2 + 340*x^3 + 6118*x^4 + 120060*x^5 + ...
where A(x)^2 = 1 + 4*x*A(x)^6:
A(x)^2 = 1 + 4*x + 48*x^2 + 768*x^3 + 14080*x^4 + 279552*x^5 + ...
A(x)^6 = 1 + 12*x + 192*x^2 + 3520*x^3 + 69888*x^4 + 1462272*x^5 + ...
Related series:
A(x)^5 = 1 + 10*x + 150*x^2 + 2660*x^3 + 51750*x^4 + 1068012*x^5 + ...
A(x)^10 = 1 + 20*x + 400*x^2 + 8320*x^3 + 179200*x^4 + 3969024*x^5 + ...
where A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5.

Cf. A245114.

Cf. A000984, A048990, A078531, A182122, A245112.

A245113:=n->4^n*binomial((6*n-1)/2,n)/(4*n+1): seq(A245113(n), n=0..30); # Wesley Ivan Hurt, Aug 11 2015

Table[4^n*Binomial[(6 n - 1)/2, n]/(4 n + 1), {n, 0, 20}] (* Wesley Ivan Hurt, Aug 11 2015 *)

/* From A(x)^2 = 1 + 4*x*A(x)^6 : */
{a(n) = local(A=1+x);for(i=1,n,A=sqrt(1 + 4*x*A^6 +x*O(x^n)));polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

{a(n) = 4^n * binomial((6*n - 1)/2, n) / (4*n + 1)}
for(n=0,20,print1(a(n),", "))

/* From A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5 : */
{a(n) = local(A=1+x);for(i=1,n,A = sqrt(1 + 4*x^2*A^10 +x*O(x^n)) + 2*x*A^5);polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

a(n) = 4^n * binomial((6*n - 1)/2, n) / (4*n + 1).
G.f. A(x) satisfies A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5.
G.f.: A(x) = sqrt(D(4*x)), where D(x) is the g.f. of A001764. - Werner Schulte, Aug 10 2015
From Karol A. Penson, Mar 19 2024: (Start)
a(n) = 4^n*binomial(3*n+1/2,n)/(6*n+1).
G.f.: 3F2([1/6, 1/2, 5/6], [3/4, 5/4], 27*x).
G.f.: sqrt(2)*sqrt((-(sqrt(1 - 27*x) + 3*i*sqrt(3)*sqrt(x))^(1/3) + (sqrt(1 - 27*x) - 3*i*sqrt(3)*sqrt(x))^(1/3))*i)*3^(3/4)/(6*x^(1/4)), where i is the imaginary unit.
a(n) = Integral_{x=0...27} x^n*W(x), where W(x) = h1(x) + h2(x) + h3(x) and
h1(x) = 2^(2/3)*3F2([-1/12, 1/6, 5/12], [1/3, 2/3], x/27)/(4*Pi*x^(5/6));
h2(x) = -3F2([1/4, 1/2, 3/4], [2/3, 4/3], x/27)/(12*Pi*sqrt(x));
h3(x) = -2^(1/3)*3F2([7/12, 5/6, 13/12], [4/3, 5/3], x/27)/(576*Pi*x^(1/6)).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 27). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0 and for x > 0 is monotonically decreasing to zero at x = 27. For x -> 27, W'(x) tends to -infinity. (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^10). - Seiichi Manyama, Jun 20 2025
a(n) ~ 27^n / (4 * n^(3/2) * sqrt(2*Pi)). - Amiram Eldar, Sep 04 2025

A281733 Positive integers T_i such that Sum_{k >= 0} (S_k * x^(2k+1)) + 1/24 - Sum_{k >= 1} (T_k x^(2k)) = (cos((2/3) arccos(6 * sqrt(3) * x)))/12 for all real x with |x| <= 1/(6 * sqrt(3)), where S_k = A176898(k).

Original entry on oeis.org

1, 32, 1792, 122880, 9371648, 763363328, 65028489216, 5722507051008, 516147694796800, 47463855386787840, 4433247375867248640, 419423751734223175680, 40109816011998942461952, 3870915577031009050296320, 376519953782381735485374464, 36874663860751966094632157184

Offset: 1

Felix Fröhlich, Jan 31 2017

nonn

The terms are given on page 3 in Sun (2013).
Conjecture: T_p == -2 (mod p) for any prime p (cf. Sun (2013), Conjecture 4).
This is the odd bisection of A078531 divided by 2. The even bisection divided by 2 is A176898. - Akiva Weinberger, Dec 09 2024

Davin Park, Table of n, a(n) for n = 1..100
K. H. Pilehrood and T. H. Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, The Journal of Integer Sequences, 18 (2015), Article 15.11.7.
Z. W. Sun, Products and sums divisible by central binomial coefficients, The Electronic Journal of Combinatorics, 20(1) (2013), #P9.

Cf. A176898, A078531.

CoefficientList[Series[(1/24)(1 - Cos[(2/3) ArcSin[6 Sqrt[3x]]]), {x, 0, 20}], x] // Rest (* Davin Park, Feb 06 2017, updated by Jean-François Alcover, Mar 21 2020 *)
CoefficientList[Series[(1-HypergeometricPFQ[{-1/3,1/3},{1/2,1},108x])/24,{x,0,16}],x]*Table[n!,{n,0,16}] (* Stefano Spezia, Mar 21 2020 *)

a(n) = 16^(n-1) * binomial(3*n-2, 2*n-1)/n. - Sarah Selkirk, Feb 11 2020
From Stefano Spezia, Feb 11 2020: (Start)
O.g.f.: (1/24)*(1 - cos((2/3) * arcsin(6 * sqrt(3*x)))).
E.g.f.: (1/24)*(1 - F([-1/3, 1/3], [1/2, 1], 108*x)), where F is the generalized hypergeometric function. (End)
a(n) = binomial(6n-3, 3n-3/2)*binomial(3n-3/2, n-1/2)/(4*n*binomial(2*n-1, n-1/2)). - Akiva Weinberger, Dec 09 2024
a(n) = A078531(2*n-1)/2. - Akiva Weinberger, Dec 09 2024

Extended by Davin Park, Feb 06 2017

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

A385208 G.f. A(x) satisfies A(x) = ( 1 + 49xA(x)^8 )^(1/7).

Original entry on oeis.org

1, 7, 245, 11319, 593047, 33429123, 1977326743, 121034349975, 7601257418678, 487008549508481, 31705597390195820, 2091361378163375955, 139468121325692304390, 9387480337647754305649, 636914947847207765431080, 43512658997082838985965655, 2990750175103769856729417627

Offset: 0

Seiichi Manyama, Jun 21 2025

nonn

Cf. A078531, A078532, A078533, A078534, A078535.

Cf. A385206, A385207.

a(n) = 49^n*binomial(8*n/7+1/7, n)/(8*n+1);

a(n) = 49^n * binomial(8*n/7+1/7,n)/(8*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^9).
G.f.: ( (1/x) * Series_Reversion(x/(1+49*x)^(8/7)) )^(1/8).

A307489 G.f. A(x) satisfies: A(x) = 1/(1 - 2xA(x) - xA(x)/(1 - xA(x)/(1 - x*A(x)/(1 - ...)))), a continued fraction.

Original entry on oeis.org

1, 3, 19, 152, 1367, 13195, 133556, 1398696, 15029311, 164764985, 1835614027, 20722066612, 236524088612, 2725081792932, 31649837891768, 370161223462480, 4355751419996559, 51532214460643957, 612604251998847641, 7313900470316335280, 87659840436181657215

Offset: 0

Ilya Gutkovskiy, Apr 10 2019

nonn

			G.f.: A(x) = 1 + 3*x + 19*x^2 + 152*x^3 + 1367*x^4 + 13195*x^5 + 133556*x^6 + 1398696*x^7 + 15029311*x^8 + 164764985*x^9 + 1835614027*x^10 + ...

Cf. A001700, A001764, A078531, A246432, A307490.

terms = 21; CoefficientList[1/x InverseSeries[Series[x (1 - 4 x + Sqrt[1 - 4 x])/2, {x, 0, terms}], x], x]
terms = 21; A[] = 0; Do[A[x] = 1/(1 - 2 x A[x] + ContinuedFractionK[-x A[x], 1, {k, 1, j}]) + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]
terms = 20; A[] = 1; Do[A[x] = 2/(1 - 4 x A[x] + Sqrt[1 - 4 x A[x]]) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]
terms = 21; A[] = 1; Do[A[x] = Sum[Binomial[2 k + 1, k + 1] x^k A[x]^k, {k, 0, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = 2/(1 - 4*x*A(x) + sqrt(1 - 4*x*A(x))).
G.f. A(x) satisfies: A(x) = Sum_{k>=0} binomial(2*k+1,k+1)*x^k*A(x)^k.
G.f.: A(x) = (1/x)*Series_Reversion(x*(1 - 4*x + sqrt(1 - 4*x))/2).
a(n) ~ 2^(12*n + 5) / (sqrt(Pi) * 17^(1/4) * n^(3/2) * (107 + 51*sqrt(17))^(n + 1/2)). - Vaclav Kotesovec, Sep 16 2021

A307490 G.f. A(x) satisfies: A(x) = 1/(1 - xA(x) - x^2A(x)^2/(1 - x^2A(x)^2/(1 - x^2A(x)^2/(1 - ...)))), a continued fraction.

Original entry on oeis.org

1, 1, 3, 10, 39, 161, 700, 3144, 14495, 68167, 325787, 1577654, 7724612, 38176620, 190196440, 954182528, 4816268367, 24441691827, 124633707865, 638272397350, 3281390284623, 16929034639153, 87617665434328, 454796186669800, 2367025393846724, 12349738834935876

Offset: 0

Ilya Gutkovskiy, Apr 10 2019

nonn

			G.f.: A(x) = 1 + x + 3*x^2 + 10*x^3 + 39*x^4 + 161*x^5 + 700*x^6 + 3144*x^7 + 14495*x^8 + 68167*x^9 + 325787*x^10 + ...

Cf. A001405, A001764, A078531, A307489.

terms = 26; CoefficientList[1/x InverseSeries[Series[x (1 - 2 x + Sqrt[1 - 4 x^2])/2, {x, 0, terms}], x], x]
terms = 26; A[] = 0; Do[A[x] = 1/(1 - x A[x] + ContinuedFractionK[-x^2 A[x]^2, 1, {k, 1, j}]) + O[x]^j, {j, 1,terms}]; CoefficientList[A[x], x]
terms = 25; A[] = 1; Do[A[x] = 2/(1 - 2 x A[x] + Sqrt[1 - 4 x^2 A[x]^2]) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]
terms = 26; A[] = 1; Do[A[x] = Sum[Binomial[k, Floor[k/2]] x^k A[x]^k, {k, 0, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = 2/(1 - 2*x*A(x) + sqrt(1 - 4*x^2*A(x)^2)).
G.f. A(x) satisfies: A(x) = Sum_{k>=0} binomial(k,floor(k/2))*x^k*A(x)^k.
G.f.: A(x) = (1/x)*Series_Reversion(x*(1 - 2*x + sqrt(1 - 4*x^2))/2).
a(n) ~ 1/(n^(3/2)*sqrt(Pi*(103/6 + (1/6)*sqrt(36037)*cos((1/3)*(4*Pi + arccos(6832781/(36037*sqrt(36037))))))) * (-7/48 + (1/24)*sqrt(203/2) * cos((1/3)*arccos(-(1849/(203*sqrt(406))))))^n). - Vaclav Kotesovec, Sep 16 2021

A379383 G.f. A(x) satisfies A(x) = sqrt( (1 + 2xA(x))/(1 - 2xA(x)^3) ).

Original entry on oeis.org

1, 2, 10, 80, 750, 7680, 83252, 939008, 10905942, 129548288, 1566565452, 19220267008, 238662840780, 2993651974144, 37876206019560, 482802294325248, 6194365014836582, 79930063134392320, 1036640587694252380, 13505632613590630400, 176673045664669396132, 2319654465118014537728

Offset: 0

Seiichi Manyama, Dec 22 2024

nonn

Cf. A138020, A379382.

Cf. A078531, A379328, A379331.

a(n) = 2^n*sum(k=0, n, binomial(n/2+2*k-1/2, k)*binomial(n/2+k+1/2, n-k)/(n+2*k+1));

a(n) = 2^n * Sum_{k=0..n} binomial(n/2+2*k-1/2,k) * binomial(n/2+k+1/2,n-k)/(n+2*k+1).

a(n) = 2^n * Sum_{k=0..n} binomial(n,k) * binomial(n/2+2*k+1/2,n)/(n+4*k+1).

A089073 Number of symmetric non-crossing connected graphs on n equidistant nodes on a circle.

Original entry on oeis.org

1, 1, 2, 5, 10, 32, 64, 231, 462, 1792, 3584, 14586, 29172, 122880, 245760, 1062347, 2124694, 9371648, 18743296, 84021990, 168043980, 763363328, 1526726656, 7012604550, 14025209100, 65028489216, 130056978432, 607892634420

Offset: 1

Emeric Deutsch, Dec 04 2003

nonn

Number of symmetric non-crossing connected graphs on n equidistant nodes on a circle (it is assumed that the axis of symmetry is a diameter of the circle passing through a given node).

			a(4)=5 because on the nodes A,B,C,D (axis of symmetry through A) the only symmetric non-crossing connected graphs are (AB,AC,AD), (AC,BC,DC), (AB,BC,CD,DA), (AB,BC,CD,DA,BD), (AB,BC,CD,DA,AC).

P. Flajolet and M. Noy, Analytic combinatorics of non-crossing configurations, Discrete Math., 204, 203-229, 1999.
M. R. Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.

Cf. A078531.

a := proc(n) if n mod 2 = 0 then 4^(n/2)*binomial((3*(n/2)-1)/2,n/2)/2/(n/2+1) else 2*4^((n-1)/2)*binomial((3*((n-1)/2)-1)/2,(n-1)/2)/2/((n-1)/2+1) fi end; seq(a(n), n=1..30);

a[n_] := If[EvenQ[n], 2^n Binomial[(3n-2)/4, n/2]/(n+2), 2^n Binomial[ (3n-5)/4, (n-1)/2]/(n+1)];
Array[a, 28] (* Jean-François Alcover, Jul 29 2018 *)

a(2k) = 4^k*binomial((3k-1)/2, k)/[2(k+1)], a(2k+1) = 2a(2k).
a(2k) = (1/2)A078531(k), a(2k+1) = A078531(k).
Conjecture D-finite with recurrence n*(n+2)*(23*n^2-162*n+199) *a(n) +12*(27*n^2-47*n-10) *a(n-1) +24*(-27*n^2+47*n+10) *a(n-2) +48 *(27*n^2-47*n-10) *a(n-3) -12*(3*n-13)*(3*n-5)*(23*n^2-116*n+60) *a(n-4)=0. - R. J. Mathar, Jul 22 2022

Name edited by Michel Marcus, Jul 30 2018

A360636 Triangle read by rows. T(n, m) = (1/(n + 1)) * C(n + 1, m) * 4^n * C((3n - m + 1)/2 - 1, n) if n is odd, otherwise (1/(n + 1)) C(n + 1, m) * C((3n - m)/2, n) C(3n - m, (3n - m)/2) / C(n - m, (n - m)/2).

Original entry on oeis.org

1, 2, 2, 10, 16, 6, 64, 140, 96, 20, 462, 1280, 1260, 512, 70, 3584, 12012, 15360, 9240, 2560, 252, 29172, 114688, 180180, 143360, 60060, 12288, 924, 245760, 1108536, 2064384, 2042040, 1146880, 360360, 57344, 3432

Offset: 0

Vladimir Kruchinin, Feb 14 2023

			Triangle T(n, m) begins:
[0]     1;
[1]     2,      2;
[2]    10,     16,      6;
[3]    64,    140,     96,     20;
[4]   462,   1280,   1260,    512,    70;
[5]  3584,  12012,  15360,   9240,  2560,   252;
[6] 29172, 114688, 180180, 143360, 60060, 12288, 924;

Cf. A078531, A000984, A151403 (row sums).

T := (n, k) -> ifelse(n mod 2 = 1, 4^n*((3*n - k - 1)/2)! / (k!*(n + 1 - k)! * ((n - k - 1)/2)!), binomial(n + 1, k) * ((n - k)/2)! * (3*n - k)! / (((3*n - k)/2)! * (n + 1)! * (n - k)!)): for n from 0 to 6 do seq(simplify(T(n, k)), k=0..n) od;
# Alternative:
gf := ((1 - 4*x*y)*sin(arcsin((216*x^2) / (1 - 4*x*y)^3 - 1)/3))/(6*x) + (1 - 4*x*y) / (12*x): assume(x > 0); serx := series(gf, x, 9): poly := n -> simplify(coeff(serx, x, n)): seq(print(seq(coeff(poly(n), y, k), k = 0..n)), n = 0..6); # Peter Luschny, Feb 15 2023

Maxima
```
T(n,m):=if n
				
```

G.f.: ((1 - 4*x*y)*sin(arcsin((216*x^2) / (1 - 4*x*y)^3 - 1)/3))/(6*x) + (1 - 4*x*y) / (12*x).

cp's OEIS Frontend

A224884 Expansion of x / Series_Reversion(x*sqrt(1 + 4*x)).

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A245113 G.f. A(x) satisfies A(x)^2 = 1 + 4*x*A(x)^6.

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A281733 Positive integers T_i such that Sum_{k >= 0} (S_k * x^(2*k+1)) + 1/24 - Sum_{k >= 1} (T_k * x^(2*k)) = (cos((2/3) * arccos(6 * sqrt(3) * x)))/12 for all real x with |x| <= 1/(6 * sqrt(3)), where S_k = A176898(k).

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

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A385208 G.f. A(x) satisfies A(x) = ( 1 + 49*x*A(x)^8 )^(1/7).

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A307489 G.f. A(x) satisfies: A(x) = 1/(1 - 2*x*A(x) - x*A(x)/(1 - x*A(x)/(1 - x*A(x)/(1 - ...)))), a continued fraction.

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A307490 G.f. A(x) satisfies: A(x) = 1/(1 - x*A(x) - x^2*A(x)^2/(1 - x^2*A(x)^2/(1 - x^2*A(x)^2/(1 - ...)))), a continued fraction.

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A379383 G.f. A(x) satisfies A(x) = sqrt( (1 + 2*x*A(x))/(1 - 2*x*A(x)^3) ).

A224884 Expansion of x / Series_Reversion(xsqrt(1 + 4x)).

A245113 G.f. A(x) satisfies A(x)^2 = 1 + 4xA(x)^6.

A281733 Positive integers T_i such that Sum_{k >= 0} (S_k * x^(2k+1)) + 1/24 - Sum_{k >= 1} (T_k x^(2k)) = (cos((2/3) arccos(6 * sqrt(3) * x)))/12 for all real x with |x| <= 1/(6 * sqrt(3)), where S_k = A176898(k).

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

A385208 G.f. A(x) satisfies A(x) = ( 1 + 49xA(x)^8 )^(1/7).

A307489 G.f. A(x) satisfies: A(x) = 1/(1 - 2xA(x) - xA(x)/(1 - xA(x)/(1 - x*A(x)/(1 - ...)))), a continued fraction.

A307490 G.f. A(x) satisfies: A(x) = 1/(1 - xA(x) - x^2A(x)^2/(1 - x^2A(x)^2/(1 - x^2A(x)^2/(1 - ...)))), a continued fraction.

A379383 G.f. A(x) satisfies A(x) = sqrt( (1 + 2xA(x))/(1 - 2xA(x)^3) ).

A360636 Triangle read by rows. T(n, m) = (1/(n + 1)) * C(n + 1, m) * 4^n * C((3n - m + 1)/2 - 1, n) if n is odd, otherwise (1/(n + 1)) C(n + 1, m) * C((3n - m)/2, n) C(3n - m, (3n - m)/2) / C(n - m, (n - m)/2).