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

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

Offset: 0

Author

Paul D. Hanna, Nov 28 2002

Keywords

• nonn
• easy

Comments

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A214668, A078531, A078533, A078534, A078535.

Cf. A004987, A008931, A245114, A247029, A376282, A376636.

Cf. A004990.

Programs

Mathematica

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

PARI

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

Formula

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

Extensions

More terms from Harvey P. Dale, Nov 03 2011