A078534 -id:A078534 - 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

A078532 Coefficients of power series that satisfies A(x)^3 - 9xA(x)^4 = 1, A(0)=1.

Original entry on oeis.org

1, 3, 27, 315, 4158, 59049, 880308, 13586859, 215233605, 3479417370, 57168561996, 951892141473, 16026585711660, 272383068872700, 4666865660812044, 80521573261807755, 1397858693681272230, 24398716826612190447, 427921056863230599900, 7537621933880388620010

Offset: 0

Paul D. Hanna, Nov 28 2002

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
Radius of convergence of g.f. A(x) is r = 1/(3*4^(4/3)) where A(r) = 4^(1/3). - Paul D. Hanna, Jul 24 2012
Self-convolution cube yields A214668.

			A(x)^3 - 9x*A(x)^4 = 1 since A(x)^3 = 1 +9x +108x^2 +1458x^3 +21060x^4 +... and A(x)^4 = 1 +12x +162x^2 +2340x^3 +... also a(2)=3^3, a(5)=3^10.

G. C. Greubel, Table of n, a(n) for n = 0..750

Cf. A214668, A078531, A078533, A078534, A078535.
Cf. A004987, A008931, A245114, A247029, A376282, A376636.
Cf. A004990.

Table[3^(2n) Binomial[(4n-2)/3,n]/(n+1),{n,0,20}] (* Harvey P. Dale, Nov 03 2011 *)

for(n=0,25, print1(9^n * binomial((4*n-2)/3, n)/(n+1), ", ")) \\ G. C. Greubel, Jan 26 2017

a(n) = 3^(2n)*binomial(4n/3-2/3, n)/(n+1). - Emeric Deutsch, Dec 10 2002
Sequence with offset 1 is expansion of reversion of g.f. x*(1-9*x)^(1/3), which equals x times the g.f. of A004990.
a(n) ~ 2^(8*n/3-5/6) * 3^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
D-finite with recurrence n*(n-1)*(n+1)*a(n) -216*(4*n-5)*(2*n-1)*(4*n-11)*a(n-3)=0. - R. J. Mathar, Mar 24 2023
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^5). - Seiichi Manyama, Jun 20 2025

More terms from Harvey P. Dale, Nov 03 2011

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

A385205 G.f. A(x) satisfies A(x) = ( 1 + 25xA(x)^4 )^(1/5).

Original entry on oeis.org

1, 5, 50, 500, 4375, 27500, 0, -3562500, -70078125, -876562500, -6926562500, 0, 1189169921875, 25690820312500, 346441406250000, 2911880859375000, 0, -550017993164062500, -12339622131347656250, -171953389892578125000, -1487552714691162109375, 0

Offset: 0

Seiichi Manyama, Jun 21 2025

sign

Cf. A078534, A299958, A385203, A385204.

Cf. A182122, A376636, A385207.

a(n) = 25^n*binomial(4*n/5+1/5, n)/(4*n+1);

a(n) = 25^n * binomial(4*n/5+1/5,n)/(4*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^3).
G.f.: ( (1/x) * Series_Reversion(x/(1+25*x)^(4/5)) )^(1/4).
a(5*n+1) = 0 for n > 0.
G.f.: 1/B(x), where B(x) is the g.f. of A299958.

A385203 G.f. A(x) satisfies A(x) = ( 1 + 25xA(x)^2 )^(1/5).

Original entry on oeis.org

1, 5, 0, -125, 625, 5625, -87500, 0, 9140625, -60156250, -653125000, 11654296875, 0, -1470068359375, 10353515625000, 118916992187500, -2225148925781250, 0, 302784667968750000, -2199076690673828125, -25952287445068359375, 497460246276855468750, 0

Offset: 0

Seiichi Manyama, Jun 21 2025

sign

Cf. A078534, A385204, A385205.

a(n) = 25^n*binomial(2*n/5+1/5, n)/(2*n+1);

a(n) = 25^n * binomial(2*n/5+1/5,n)/(2*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x/A(x)).
G.f.: ( (1/x) * Series_Reversion(x/(1+25*x)^(2/5)) )^(1/2).
a(5*n+2) = 0 for n >= 0.

A385204 G.f. A(x) satisfies A(x) = ( 1 + 25xA(x)^3 )^(1/5).

Original entry on oeis.org

1, 5, 25, 0, -1250, -6875, 65625, 1062500, 0, -116796875, -782031250, 8609375000, 155390625000, 0, -19950927734375, -141498046875000, 1635108642578125, 30759411621093750, 0, -4223049316406250000, -30787073669433593750, 364567847442626953125

Offset: 0

Seiichi Manyama, Jun 21 2025

sign

Cf. A078534, A385203, A385205.

a(n) = 25^n*binomial(3*n/5+1/5, n)/(3*n+1);

a(n) = 25^n * binomial(3*n/5+1/5,n)/(3*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)).
G.f.: ( (1/x) * Series_Reversion(x/(1+25*x)^(3/5)) )^(1/3).
a(5*n+3) = 0 for n >= 0.
G.f.: 1/B(-x), where B(x) is the g.f. of A385203.

A380465 G.f. A(x) satisfies A(x) = 1/( 1 - 25xA(x)^2 )^(1/5).

Original entry on oeis.org

1, 5, 125, 4250, 166250, 7052500, 315459375, 14648437500, 699404062500, 34120414453125, 1693355782421875, 85222795492187500, 4339218139648437500, 223115431527734375000, 11568972340119140625000, 604249120575386718750000, 31761084429202554931640625, 1678825356066226959228515625

Offset: 0

Seiichi Manyama, Jun 23 2025

nonn

Cf. A034688, A078534, A385203, A385204, A385205, A380466, A380471.

a(n) = 25^n*binomial(7*n/5+1/5, n)/(7*n+1);

G.f. A(x) satisfies A(x) = ( 1 + 25*x*A(x)^7 )^(1/5).
a(n) = 25^n * binomial(7*n/5+1/5,n)/(7*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^9).
G.f.: ( (1/x) * Series_Reversion(x/(1+25*x)^(7/5)) )^(1/7).

A380466 G.f. A(x) satisfies A(x) = 1/( 1 - 25xA(x)^3 )^(1/5).

Original entry on oeis.org

1, 5, 150, 6250, 301875, 15868125, 881237500, 50865750000, 3021240234375, 183454158593750, 11336659803906250, 710625236343750000, 45075347315400390625, 2887845039367675781250, 186601230428607421875000, 12146710229056792968750000, 795792421294273872070312500

Offset: 0

Seiichi Manyama, Jun 23 2025

nonn

Cf. A034688, A078534, A385203, A385204, A385205, A380465, A380471.

a(n) = 25^n*binomial(8*n/5+1/5, n)/(8*n+1);

G.f. A(x) satisfies A(x) = ( 1 + 25*x*A(x)^8 )^(1/5).
a(n) = 25^n * binomial(8*n/5+1/5,n)/(8*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^11).
G.f.: ( (1/x) * Series_Reversion(x/(1+25*x)^(8/5)) )^(1/8).

A380471 G.f. A(x) satisfies A(x) = 1/( 1 - 25xA(x)^4 )^(1/5).

Original entry on oeis.org

1, 5, 175, 8625, 495000, 30980625, 2050781250, 141187921875, 10006590468750, 725240531640625, 53503504196484375, 4004478454589843750, 303320955472031250000, 23207794539155419921875, 1791025435519151367187500, 139250846557940616210937500, 10897102765738964080810546875

Offset: 0

Seiichi Manyama, Jun 23 2025

nonn

Cf. A034688, A078534, A385203, A385204, A385205, A380465, A380466.

a(n) = 25^n*binomial(9*n/5+1/5, n)/(9*n+1);

G.f. A(x) satisfies A(x) = ( 1 + 25*x*A(x)^9 )^(1/5).
a(n) = 25^n * binomial(9*n/5+1/5,n)/(9*n+1).
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^13).
G.f.: ( (1/x) * Series_Reversion(x/(1+25*x)^(9/5)) )^(1/9).

cp's OEIS Frontend

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

A078532 Coefficients of power series that satisfies A(x)^3 - 9*x*A(x)^4 = 1, A(0)=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A385205 G.f. A(x) satisfies A(x) = ( 1 + 25*x*A(x)^4 )^(1/5).

Views

Author

Keywords

Crossrefs

Programs

PARI

Formula

A385203 G.f. A(x) satisfies A(x) = ( 1 + 25*x*A(x)^2 )^(1/5).

Views

Author

Keywords

Crossrefs

Programs

PARI

Formula

A385204 G.f. A(x) satisfies A(x) = ( 1 + 25*x*A(x)^3 )^(1/5).

Views

Author

Keywords

Crossrefs

Programs

PARI

Formula

A380465 G.f. A(x) satisfies A(x) = 1/( 1 - 25*x*A(x)^2 )^(1/5).

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

A078532 Coefficients of power series that satisfies A(x)^3 - 9xA(x)^4 = 1, A(0)=1.

A385205 G.f. A(x) satisfies A(x) = ( 1 + 25xA(x)^4 )^(1/5).

A385203 G.f. A(x) satisfies A(x) = ( 1 + 25xA(x)^2 )^(1/5).

A385204 G.f. A(x) satisfies A(x) = ( 1 + 25xA(x)^3 )^(1/5).

A380465 G.f. A(x) satisfies A(x) = 1/( 1 - 25xA(x)^2 )^(1/5).

A380466 G.f. A(x) satisfies A(x) = 1/( 1 - 25xA(x)^3 )^(1/5).

A380471 G.f. A(x) satisfies A(x) = 1/( 1 - 25xA(x)^4 )^(1/5).