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

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

Offset: 0

Author

Paul D. Hanna, Nov 28 2002

Keywords

• nonn

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A078531, A078532, A078534, A078535.

Programs

Mathematica

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

PARI

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

Formula

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)