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%I #13 Sep 24 2022 08:18:12
%S 1,10,110,1870,8602,249458,1247290,51138890,218502530,2316126818,
%T 136651482262,136651482262,570720896506,6277929861566,521068178509978,
%U 46375067887388042,2016307299451654,203647037244617054
%N a(n) = denominator((1/2)*(1 + Product_{k=0..n-1} 2*(1 + 3*k)/(5 + 6*k))).
%C Previous name was: a(n)=denominator of k_n such that Integrate[(1+x^(3n))/Sqrt[1-x^3],{x,0,1}]= k_n*(Gamma[1/3]^3)/(2^(1/3)Sqrt[3]Pi) where n >= 0.
%C General formula: Integral_{x=0..1} ((1+x^(3n))/sqrt(1-x^3)) dx = G_3 * k_n = G_3*A146752(n)/A146753(n) = A118292*A146752(n)/A146753(n) where G_3 = (Gamma(1/3)^3)/(2^(1/3)*sqrt(3)*Pi).
%F a(n) = denominator((1/2)*(1 + Product_{k=0..n-1} 2*(1 + 3*k)/(5 + 6*k))).
%t Table[Denominator[(1/2) (1 + Product[(2 (1 + 3 k))/(5 + 6 k), {k, 0, n - 1}])], {n, 0, 30}]
%Y Cf. A146752 (numerator), A118292 (G_3).
%K nonn,frac
%O 0,2
%A _Artur Jasinski_, Nov 01 2008
%E New name (using given formula) from _Joerg Arndt_, Sep 24 2022
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