A078535 - OEIS

%I #17 Oct 11 2023 04:40:47

%S 1,6,162,5760,232254,10077696,458960580,21634449408,1046465787510,

%T 51644846702592,2590092194793948,131621703842267136,

%U 6762649550214036780,350714987252652441600,18334388441036020419720,965148007553698721955840,51116742846877582931249574

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

%C 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).

%C 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

%C 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

%H Andrew Howroyd, <a href="/A078535/b078535.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = 6^(2n)*binomial(7n/6-5/6, n)/(n+1). - _Emeric Deutsch_, Dec 10 2002

%F a(n) ~ 7^(7*n/6-1/3) * 6^n / (sqrt(2*Pi) * n^(3/2)). - _Vaclav Kotesovec_, Dec 03 2014

%e 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.

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

%o (PARI) a(n) = {6^(2*n)*binomial((7*n-5)/6, n)/(n+1)} \\ _Andrew Howroyd_, Nov 05 2019

%Y Cf. A078531, A078532, A078533, A078534.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Nov 28 2002

%E Terms a(13) and beyond from _Andrew Howroyd_, Nov 05 2019