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

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

Offset

0,2

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^(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)

Links

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

Formula

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)

Example

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.

Mathematica

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

Prog

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

Crossrefs

Cf. A078531, A078532, A078533, A078535.

Sequence in context: A208348 A128784 A369535 * A380242 A141120 A384017

Adjacent sequences: A078531 A078532 A078533 * A078535 A078536 A078537

Keyword

nonn

Author

Paul D. Hanna, Nov 28 2002

Extensions

More terms from Harvey P. Dale, Mar 27 2011

Status

approved