A078532 -id:A078532 - cp's OEIS frontend

A004987 a(n) = (3^n/n!)Product_{k=0..n-1} (3k + 1). 3-central binomial coefficients.

1, 3, 18, 126, 945, 7371, 58968, 480168, 3961386, 33011550, 277297020, 2344420260, 19927572210, 170150808870, 1458435504600, 12542545339560, 108179453553705, 935434098376155, 8107095519260010, 70403724246205350, 612512400941986545, 5337608065351597035, 46582761297613937760

Offset: 0

Joe Keane (jgk(AT)jgk.org)

Diagonal of rational function R(x,y) = (1 - 9*x*y) / (1 - 2*x - 3*y + 3*y^2 + 9*x^2*y). - Gheorghe Coserea, Jul 01 2016

This is the k = 3 variant of the k-central binomial coefficients c(n,k) with g.f. (1 - k^2*x)^(-1/k), which yield the usual central binomial coefficients A001405 for k = 2. - M. F. Hasler, Nov 12 2024

			G.f.: 1 + 3*x + 18*x^2 + 126*x^3 + 945*x^4 + 7371*x^5 + 58968*x^6 + 480168*x^7 + ...

Gheorghe Coserea, Table of n, a(n) for n = 0..200
A. Bostan, S. Boukraa, J.-M. Maillard, and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
A. Straub, V. H. Moll, and T. Amdeberhan, The p-adic valuation of k-central binomial coefficients, Acta Arith. 140 (2009) 31-41, eq (1.10).

Cf. A004117, A007559, A047657, A004988, A054861, A034689, A053101, A072888, A216316.
Related to diagonal of rational functions: A268545-A268555.
Cf. A008931, A078532, A245114, A247029, A376282, A376636.

List([0..25], n-> 3^n*Product([0..n-1], k-> 3*k+1)/Factorial(n) ); # G. C. Greubel, Aug 22 2019

[1] cat [3^n*&*[3*k+1: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019

a:= n-> (3^n/n!)*mul(3*k+1, k=0..n-1); seq(a(n), n=0..25); # G. C. Greubel, Aug 22 2019

Table[(-9)^n Binomial[-1/3, n], {n, 0, 25}] (* Jean-François Alcover, Sep 28 2016, after Peter Luschny *)

a(n) = prod(k=0, n-1, 3*k + 1)*3^n/n! \\ Michel Marcus, Jun 30 2013

my(x='x, y='y);
R = (1 - 9*x*y) / (1 - 2*x - 3*y + 3*y^2 + 9*x^2*y);
diag(n, expr, var) = {
  my(a = vector(n));
  for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
  for (k = 1, n, a[k] = expr;
       for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
  return(a);
};
diag(20, R, [x,y])  \\ Gheorghe Coserea, Jul 01 2016

Vec((1-9*x+O(x^25))^(-1/3)) \\ yields the same as:
apply( {A004987(n)=prod(k=0, n-1, 9*k+3)\n!}, [0..24]) \\ M. F. Hasler, Nov 12 2024

[9^n*rising_factorial(1/3, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019

G.f.: (1 - 9*x)^(-1/3).
a(n) = (3^n/n!)*A007559(n), n >= 1, a(0) := 1.
a(n) ~ Gamma(1/3)^-1*n^(-2/3)*3^(2*n)*{1 - 1/9*n^-1 + ...}.
Representation as n-th moment of a positive function on (0, 9): a(n) = Integral_{x=0..9} ( x^n*(1/(Pi*sqrt(3)*6*(x/9)^(2/3)*(1-x/9)^(1/3))) ), n >= 0. This function is the solution of the Hausdorff moment problem on (0, 9) with moments equal to a(n). As a consequence this representation is unique. - Karol A. Penson, Jan 30 2003
D-finite with recurrence: n*a(n) + 3*(2-3*n)*a(n-1)=0. - R. J. Mathar, Jun 07 2013
0 = a(n) * (81*a(n+1) - 15*a(n+2)) + a(n+1) * (-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jan 27 2014
G.f. A(x)=:y satisfies 0 = y'' * y - 4 * y' * y'. - Michael Somos, Jan 27 2014
a(n) = (-9)^n*binomial(-1/3, n). - Peter Luschny, Mar 23 2014
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([1/3], [1], 9*x). - Karol A. Penson, Dec 19 2015
Sum_{n>=0} 1/a(n) = (sqrt(3)*Pi + 3*(12 + log(3)))/32 = 1.3980385924595932... - Ilya Gutkovskiy, Jul 01 2016
Binomial transform of A216316. - Peter Bala, Jul 02 2023
From Peter Bala, Mar 31 2024: (Start)
a(n) = (9^n)*Sum_{k = 0..2*n} (-1)^k*binomial(-1/3, k)* binomial(-1/3, 2*n - k).
(9^n)*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
Sum_{k = 0..n} a(k)*a(n-k) = A004988(n).
Sum_{k = 0..2*n} a(k)*a(2*n-k) = 18^n/(2*n)! * Product_{k = 1..n} (6*k - 1)*(3*k - 2). (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^3). - Seiichi Manyama, Jun 20 2025

More terms from Ralf Stephan, Mar 13 2004

More terms from Benoit Cloitre, Jun 05 2004

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

Original entry on oeis.org

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

A376636 G.f. A(x) satisfies A(x) = (1 + 9xA(x)^2)^(1/3).

Original entry on oeis.org

1, 3, 9, 18, 0, -162, -567, 0, 8019, 31590, 0, -520506, -2160756, 0, 38480265, 164549880, 0, -3072083274, -13390246485, 0, 258054995016, 1139882486490, 0, -22474826957232, -100257845970825, 0, 2011064804461548, 9039247392729582, 0, -183769714890451800

Offset: 0

Seiichi Manyama, Oct 23 2024

sign

Paolo Xausa, Table of n, a(n) for n = 0..1000

Cf. A004987, A008931, A078532, A245114, A247029, A376282.

Cf. A385117.

A376636[n_] := 9^n*Binomial[(2*n + 1)/3, n]/(2*n + 1);
Array[A376636, 35, 0] (* Paolo Xausa, Aug 04 2025 *)

a(n) = 9^n*binomial(2*n/3+1/3, n)/(2*n+1);

a(n) = 9^n * binomial(2*n/3 + 1/3,n)/(2*n+1).
From Seiichi Manyama, Jun 20 2025: (Start)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)).
a(3*n+1) = 0 for n > 0. (End)
D-finite with recurrence n*(n-2)*a(n) +54*(2*n-5)*(n-4)*a(n-3)=0. - R. J. Mathar, Jul 30 2025

A078534 Coefficients of power series that satisfies A(x)^5 - 25xA(x)^6 = 1, A(0)=1.

Original entry on oeis.org

1, 5, 100, 2625, 78125, 2502500, 84150000, 2929265625, 104646953125, 3814697265625, 141323284375000, 5305403695312500, 201382633183593750, 7715985752343750000, 298023223876953125000, 11591412585295166015625, 453601640704152832031250

Offset: 0

Paul D. Hanna, Nov 28 2002

nonn

If A(x) = Sum_{k>=1} a(k)*x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(n-1) = n^(2*n-3) and a(2*n-1) = n^(4*n-2) (conjecture).
From Emeric Deutsch, Dec 10 2002: (Start)
If A(x) = Sum_{k>=1} a(k)*x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(k) = n^(2*k)*binomial(k/n+1/n+k-1,k)/(k+1) and, consequently, a(n-1) = n^(2*n-3) and a(2*n-1) = n^(4*n-2).
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 + x*A(x)^2. (End)

			A(x)^5 - 25*x*A(x)^6 = 1 since A(x)^5 = 1 + 25*x + 750*x^2 + 24375*x^3 + 831250*x^4 + ... and A(x)^6 = 1 + 30*x + 975*x^2 + 33250*x^3 + ... also a(4) = 5^7, a(9) = 5^18 = 3814697265625.

G. C. Greubel, Table of n, a(n) for n = 0..610

Cf. A078531, A078532, A078533, A078535.

Table[5^(2n) Binomial[(6n-4)/5,n]/(n+1),{n,0,25}]  (* Harvey P. Dale, Mar 27 2011 *)

for(n=0,50, print1(5^(2*n)*binomial((6*n-4)/5, n)/(n+1), ", ")) \\ G. C. Greubel, Jan 30 2017

a(n) = 5^(2*n)*binomial(6*n/5 - 4/5, n)/(n+1). - Emeric Deutsch, Dec 10 2002
a(n) ~ sqrt(3) * 6^(6*n/5 - 4/5) * 5^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
From Seiichi Manyama, Jun 21 2025: (Start)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7).
G.f.: ( (1/x) * Series_Reversion(x/(1+25*x)^(6/5)) )^(1/6). (End)

More terms from Harvey P. Dale, Mar 27 2011

A245114 G.f. A(x) satisfies A(x)^3 = 1 + 9xA(x)^5.

Original entry on oeis.org

1, 3, 36, 585, 10935, 221697, 4740120, 105225318, 2402040420, 56029889025, 1329627118248, 31998624800220, 779102941714461, 19157195459506230, 475034438632316400, 11865382635213387504, 298265217964573747095, 7539795161286074350785, 191548870595159091038640, 4888023169106780049244275

Offset: 0

Paul D. Hanna, Jul 31 2014

			G.f.: A(x) = 1 + 3*x + 36*x^2 + 585*x^3 + 10935*x^4 + 221697*x^5 +...
where A(x)^3 = 1 + 9*x*A(x)^5:
A(x)^3 = 1 + 9*x + 135*x^2 + 2430*x^3 + 48195*x^4 + 1015740*x^5 +...
A(x)^5 = 1 + 15*x + 270*x^2 + 5355*x^3 + 112860*x^4 + 2480058*x^5 +...

Robert Israel, Table of n, a(n) for n = 0..696

Cf. A004987, A008931, A078532, A247029, A376282, A376636.

Cf. A245112, A385119.

rec:= 2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13):
f:= gfun:-rectoproc({rec,a(0)=1,a(1)=3,a(2)=36},a(n),remember):
map(f, [$0..30]); # Robert Israel, Jan 30 2018

nmax = 19; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - (1 + 9 x A[x]^5) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. HoldPattern[a[n_] -> k_] :> Set[a[n], k];
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)

/* From A(x)^3 = 1 + 9*x*A(x)^5 : */
{a(n) = local(A=1+x);for(i=1,n,A=(1 + 9*x*A^5 +x*O(x^n))^(1/3));polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

{a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n+1)}
for(n=0,20,print1(a(n),", "))

a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n + 1).
2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13). - Robert Israel, Jan 30 2018
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7). - Seiichi Manyama, Jun 20 2025
a(n) ~ 3^n * 5^(5*n/3-1/6) / (sqrt(Pi) * 2^(2*(n+2)/3) * n^(3/2)). - Amiram Eldar, Sep 02 2025

A008931 Expansion of (2/(1+sqrt(1-36*x)))^(1/3).

Original entry on oeis.org

1, 3, 45, 936, 22572, 592515, 16434495, 473825700, 14058408519, 426438391743, 13164565835421, 412255067017248, 13064028812911440, 418149414542496168, 13498863325944967656, 439006511643775469856, 14369623854340007790108, 473027210589699351461700

Offset: 0

N. J. A. Sloane

nonn

G. C. Greubel, Table of n, a(n) for n = 0..640

Cf. A004987, A078532, A245114, A247029, A376282, A376636.

a:=[1];; for n in [2..20] do a[n]:=6*(5-21*(n-1)+18*(n-1)^2)*a[n-1]/((n-1)*(3*n-2)); od; a; # G. C. Greubel, Sep 13 2019

I:=[1]; [n le 1 select I[n] else 6*(5-21*(n-1)+18*(n-1)^2)*Self(n-1)/((n-1)*(3*n-2)): n in [1..20]]; // G. C. Greubel, Sep 13 2019

seq(9^n*binomial(2*n +1/3, n)/(6*n+1), n=0..20); # G. C. Greubel, Sep 13 2019

CoefficientList[Series[Surd[2/(1+Sqrt[1-36x]),3],{x,0,20}],x] (* Harvey P. Dale, Aug 12 2016 *)
Table[9^n Binomial[2 n + 1/3, n]/(6 n + 1), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 12 2016 *)

my(x='x+O('x^20)); Vec((2/(1+sqrt(1-36*x)))^(1/3)) \\ G. C. Greubel, Apr 11 2017

[9^n*binomial(2*n +1/3, n)/(6*n+1) for n in (0..20)] # G. C. Greubel, Sep 13 2019

From Vladimir Reshetnikov, Oct 12 2016: (Start)
a(n) = 9^n*binomial(2*n + 1/3, n)/(6*n + 1).
D-finite with recurrence: n*(3*n+1)*a(n) = 6*(18*n^2-21*n+5)*a(n-1). (End)
a(n) ~ 2^(2*n-2/3)*3^(2*n-1)/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Oct 13 2016
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^9). - Seiichi Manyama, Jun 20 2025
G.f.: 2F1(1/6, 2/3 ; 4/3 ; 36*x). - R. J. Mathar, Jul 30 2025

A376282 G.f. A(x) satisfies A(x) = (1 + 9xA(x)^7)^(1/3).

Original entry on oeis.org

1, 3, 54, 1368, 40365, 1299078, 44223732, 1565864784, 57079952046, 2127818007315, 80742077597610, 3108398557803480, 121107814518484872, 4766365291226837508, 189209375036491438800, 7567095678024459993120, 304603864960375133224533, 12331716699093681951702810

Offset: 0

Seiichi Manyama, Oct 23 2024

nonn

Paolo Xausa, Table of n, a(n) for n = 0..600

Cf. A004987, A008931, A078532, A245114, A376636.

A376282[n_] := 9^n*Binomial[(7*n + 1)/3, n]/(7*n + 1);
Array[A376282, 20, 0] (* Paolo Xausa, Aug 04 2025 *)

a(n) = 9^n*binomial(7*n/3+1/3, n)/(7*n+1);

a(n) = 9^n * binomial(7*n/3 + 1/3,n)/(7*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^11). - Seiichi Manyama, Jun 20 2025
D-finite with recurrence 8*n*(n-1)*(n-2)*(4*n-5)*(2*n-1)*(4*n+1)*a(n) -189*(7*n-11)*(7*n-17)*(7*n-2)*(7*n-20)*(7*n-5)*(7*n-8)*a(n-3)=0. - R. J. Mathar, Jul 30 2025

A078533 Coefficients of power series that satisfies A(x)^4 - 16x*A(x)^5 = 1, A(0)=1.

Original entry on oeis.org

1, 4, 56, 1024, 21216, 473088, 11075328, 268435456, 6677665280, 169514369024, 4373549027328, 114349209288704, 3023068543631360, 80675644291153920, 2170389180446539776, 58798996734949195776, 1602737048880933109760, 43924199383151211970560

Offset: 0

Paul D. Hanna, Nov 28 2002

nonn

If A(x) = Sum_{k>=1} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2) (conjecture).
If A(x) = Sum_{k>=1} 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

			A(x)^4 - 16x*A(x)^5 = 1 since A(x)^4 = 1 + 16x + 320x^2 + 7040x^3 + 163840x^4 + ... and A(x)^5 = 1 + 20x + 440x^2 + 10240x^3 + ... also a(3) = 4^5, a(7) = 4^14 = 268435456.

G. C. Greubel, Table of n, a(n) for n = 0..675

Cf. A078531, A078532, A078534, A078535.

Table[4^(2*n)*Binomial[5*n/4-3/4, n]/(n+1),{n,0,20}] (* Vaclav Kotesovec, Dec 03 2014 *)

for(n=0,50, print1(2^(4*n)*binomial((5*n-3)/4,n)/(n+1), ", ")) \\ G. C. Greubel, Jan 30 2017

a(n) = 4^(2n)*binomial(5n/4 - 3/4, n)/(n+1). - Emeric Deutsch, Dec 10 2002
a(n) ~ 5^(5*n/4 - 1/4) * 2^(2*n - 1/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
From Seiichi Manyama, Jun 21 2025: (Start)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^6).
G.f.: ( (1/x) * Series_Reversion(x/(1+16*x)^(5/4)) )^(1/5). (End)

A078535 Coefficients of power series that satisfies A(x)^6 - 36x*A(x)^7 = 1, A(0)=1.

Original entry on oeis.org

1, 6, 162, 5760, 232254, 10077696, 458960580, 21634449408, 1046465787510, 51644846702592, 2590092194793948, 131621703842267136, 6762649550214036780, 350714987252652441600, 18334388441036020419720, 965148007553698721955840, 51116742846877582931249574

Offset: 0

Paul D. Hanna, Nov 28 2002

nonn

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(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2) (conjecture).
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

			A(x)^6 - 36x*A(x)^7 = 1 since A(x)^6 = 1 +36x +1512x^2 +68040x^3 +3193344x^4 +... and A(x)^7 = 1 +42x +1890x^2 +88704x^3 +... also a(5)=6^9, a(11)=6^22 = 131621703842267136.

Andrew Howroyd, Table of n, a(n) for n = 0..200

Cf. A078531, A078532, A078533, A078534.

Table[6^(2*n)*Binomial[7*n/6-5/6, n]/(n+1),{n,0,20}] (* Vaclav Kotesovec, Dec 03 2014 *)

a(n) = {6^(2*n)*binomial((7*n-5)/6, n)/(n+1)} \\ Andrew Howroyd, Nov 05 2019

a(n) = 6^(2n)*binomial(7n/6-5/6, n)/(n+1). - Emeric Deutsch, Dec 10 2002
a(n) ~ 7^(7*n/6-1/3) * 6^n / (sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
From Seiichi Manyama, Jun 21 2025: (Start)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^8).
G.f.: ( (1/x) * Series_Reversion(x/(1+36*x)^(7/6)) )^(1/7). (End)

Terms a(13) and beyond from Andrew Howroyd, Nov 05 2019

A247029 G.f. A(x) satisfies A(x) = A(x)^4 - 9*x.

Original entry on oeis.org

1, 3, -18, 180, -2187, 29484, -424116, 6377292, -99034650, 1576075644, -25569752274, 421325812440, -7031733125508, 118620405322020, -2019349799669160, 34647126360607440, -598525520999144643, 10401492640172342940, -181721630178565389900, 3189811189331825319492

Offset: 0

Paul D. Hanna, Sep 09 2014

sign

			G.f.: A(x) = 1 + 3*x - 18*x^2 + 180*x^3 - 2187*x^4 + 29484*x^5 - 424116*x^6 +...
where
A(x)^4 = 1 + 12*x - 18*x^2 + 180*x^3 - 2187*x^4 + 29484*x^5 - 424116*x^6 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..500

Cf. A004987, A008931, A078532, A245114, A376282, A376636.

Cf. A224884.

FullSimplify[Table[-(-1)^n * 3^(2*n-1) * 4^(n-1) * Gamma[n/3 + 1/6] * Gamma[2*n/3 - 1/6] / (Pi * Gamma[n + 1]), {n, 0, 20}]] (* Vaclav Kotesovec, Nov 18 2017 *)

{a(n)=polcoeff(x/serreverse(x*(1+9*x +x^2*O(x^n))^(1/3)), n)}
for(n=0, 25, print1(a(n), ", "))

G.f.: x / Series_Reversion( x*(1 + 9*x)^(1/3) ).
Recurrence: (n-2)*(n-1)*n*a(n) = -216*(2*n - 5)*(4*n - 13)*(4*n - 7)*a(n-3). - Vaclav Kotesovec, Nov 18 2017
a(n) ~ -(-1)^n * 2^(8*n/3 - 13/6) * 3^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Nov 18 2017
G.f. A(x) satisfies A(x) = 1/A(-x/A(x)^5). - Seiichi Manyama, Jun 20 2025

cp's OEIS Frontend

A004987 a(n) = (3^n/n!)*Product_{k=0..n-1} (3*k + 1). 3-central binomial coefficients.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

PARI

PARI

Sage

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

PARI

PARI

Formula

A376636 G.f. A(x) satisfies A(x) = (1 + 9*x*A(x)^2)^(1/3).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A078534 Coefficients of power series that satisfies A(x)^5 - 25*x*A(x)^6 = 1, A(0)=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A245114 G.f. A(x) satisfies A(x)^3 = 1 + 9*x*A(x)^5.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

A008931 Expansion of (2/(1+sqrt(1-36*x)))^(1/3).

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

A004987 a(n) = (3^n/n!)Product_{k=0..n-1} (3k + 1). 3-central binomial coefficients.

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

A376636 G.f. A(x) satisfies A(x) = (1 + 9xA(x)^2)^(1/3).

A078534 Coefficients of power series that satisfies A(x)^5 - 25xA(x)^6 = 1, A(0)=1.

A245114 G.f. A(x) satisfies A(x)^3 = 1 + 9xA(x)^5.

A376282 G.f. A(x) satisfies A(x) = (1 + 9xA(x)^7)^(1/3).