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%I A078533 #15 Jan 31 2017 02:46:25
%S A078533 1,4,56,1024,21216,473088,11075328,268435456,6677665280,169514369024,
%T A078533 4373549027328,114349209288704,3023068543631360,80675644291153920,
%U A078533 2170389180446539776,58798996734949195776,1602737048880933109760,43924199383151211970560
%N A078533 Coefficients of power series that satisfies A(x)^4 - 16x*A(x)^5 = 1, A(0)=1.
%C A078533 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 A078533 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 A078533 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 A078533 G. C. Greubel, <a href="/A078533/b078533.txt">Table of n, a(n) for n = 0..675</a>
%F A078533 a(n) = 4^(2n)*binomial(5n/4 - 3/4, n)/(n+1). - _Emeric Deutsch_, Dec 10 2002
%F A078533 a(n) ~ 5^(5*n/4 - 1/4) * 2^(2*n - 1/2) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Dec 03 2014
%e A078533 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.
%t A078533 Table[4^(2*n)*Binomial[5*n/4-3/4, n]/(n+1),{n,0,20}] (* _Vaclav Kotesovec_, Dec 03 2014 *)
%o A078533 (PARI) for(n=0,50, print1(2^(4*n)*binomial((5*n-3)/4,n)/(n+1), ", ")) \\ _G. C. Greubel_, Jan 30 2017
%Y A078533 Cf. A078531, A078532, A078534, A078535.
%K A078533 nonn
%O A078533 0,2
%A A078533 _Paul D. Hanna_, Nov 28 2002

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