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

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

Offset: 0

Author

Paul D. Hanna, Nov 28 2002

Keywords

• nonn

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A078531, A078532, A078533, A078534.

Programs

Mathematica

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

PARI

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

Formula

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)

Extensions

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