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

%I #17 Jan 31 2017 02:46:32

%S 1,5,100,2625,78125,2502500,84150000,2929265625,104646953125,

%T 3814697265625,141323284375000,5305403695312500,201382633183593750,

%U 7715985752343750000,298023223876953125000,11591412585295166015625,453601640704152832031250

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

%C 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).

%C 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

%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 G. C. Greubel, <a href="/A078534/b078534.txt">Table of n, a(n) for n = 0..610</a>

%F a(n) = 5^(2n)*binomial(6n/5 - 4/5, n)/(n+1). - _Emeric Deutsch_, Dec 10 2002

%F a(n) ~ sqrt(3) * 6^(6*n/5 - 4/5) * 5^n / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Dec 03 2014

%e A(x)^5 - 25x*A(x)^6 = 1 since A(x)^5 = 1 + 25x + 750x^2 + 24375x^3 + 831250x^4 + ... and A(x)^6 = 1 + 30x + 975x^2 + 33250x^3 + ... also a(4) = 5^7, a(9) = 5^18 = 3814697265625.

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

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

%Y Cf. A078531, A078532, A078533, A078535.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Nov 28 2002

%E More terms from _Harvey P. Dale_, Mar 27 2011