%I #25 Jul 21 2021 15:42:37
%S 1,2,2,172,342,26556,67220,4875160,14125030,973837420,3087573628,
%T 204536051176,692771715836,44412235657176,158358513025896,
%U 9874709152875568,36706645561910150,2234840966950941260,8601116786415880940,512801585354912006600,2032977466125710169236
%N Expansion of ((1 + 16*x)/(1 - 16*x))^(1/16).
%C Let ((1 + k*x)/(1 - k*x))^(1/k) = a(0) + a(1)*x + a(2)*x^2 + ...
%C Then n*a(n) = 2*a(n-1) + k^2*(n-2)*a(n-2) for n > 1.
%H Seiichi Manyama, <a href="/A304915/b304915.txt">Table of n, a(n) for n = 0..500</a>
%F n*a(n) = 2*a(n-1) + 16^2*(n-2)*a(n-2) for n > 1.
%F a(n) ~ 2^(4*n + 1/16) / (Gamma(1/16) * n^(15/16)) * (1 - (-1)^n * sqrt(2 - sqrt(2 + sqrt(2))) * 2^(7/8) * Gamma(1/16)^2 / (64*Pi*n^(1/8))). - _Vaclav Kotesovec_, May 21 2018
%t CoefficientList[Series[((1+16x)/(1-16x))^(1/16),{x,0,30}],x] (* _Harvey P. Dale_, Jul 21 2021 *)
%o (PARI) N=66; x='x+O('x^N); Vec(((1+16*x)/(1-16*x))^(1/16))
%Y Cf. A063886, A303537, A303538.
%K nonn
%O 0,2
%A _Seiichi Manyama_, May 21 2018