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A338633 G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 2^3*x/(A(x) - 3^3*x/(A(x) - 4^3*x/(A(x) - 5^3*x/(A(x) - 6^3*x/(A(x) - ...)))))), a continued fraction relation.

%I A338633 #9 Nov 12 2020 11:48:14
%S A338633 1,1,7,250,21867,3725702,1096355494,513875333940,361121449989171,
%T A338633 362961084011245198,502496711191618404882,929337000359116522329132,
%U A338633 2238572532534241145084855934,6875030222633195280825967544508,26436454884630260855874989243890732
%N A338633 G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 2^3*x/(A(x) - 3^3*x/(A(x) - 4^3*x/(A(x) - 5^3*x/(A(x) - 6^3*x/(A(x) - ...)))))), a continued fraction relation.
%C A338633 Compare to the continued fraction relation for the g.f. of A158119 and A338634.
%H A338633 Paul D. Hanna, <a href="/A338633/b338633.txt">Table of n, a(n) for n = 0..150</a>
%F A338633 For n > 0, a(n) is odd iff n is a power of 2 (conjecture).
%F A338633 From _Vaclav Kotesovec_, Nov 12 2020: (Start)
%F A338633 a(n) ~ sqrt(3/(2*Pi)) * (6*Gamma(2/3)/Gamma(1/3)^2)^(3*n + 3/2) * (n!)^3 / sqrt(n).
%F A338633 a(n) ~ 2^(6*n + 4) * 3^(3*n/2 + 5/4) * Pi^(3*n + 5/2) * n^(3*n + 1) / Gamma(1/3)^(9*(n + 1/2)) / exp(3*n). (End)
%e A338633 G.f.: A(x) = 1 + x + 7*x^2 + 250*x^3 + 21867*x^4 + 3725702*x^5 + 1096355494*x^6 + 513875333940*x^7 + 361121449989171*x^8 + 362961084011245198*x^9 + ...
%e A338633 where
%e A338633 1 = A(x) - x/(A(x) - 2^3*x/(A(x) - 3^3*x/(A(x) - 4^3*x/(A(x) - 5^3*x/(A(x) - 6^3*x/(A(x) - 7^3*x/(A(x) - 8^3*x/(A(x) - 9^3*x/(A(x) - ...))))))))), a continued fraction relation.
%o A338633 (PARI) {a(n) = my(A=[1],CF=1); for(i=1,n, A=concat(A,0); for(i=1,#A, CF = Ser(A) - (#A-i+1)^3*x/CF ); A[#A] = -polcoeff(CF,#A-1) );A[n+1] }
%o A338633 for(n=0,20,print1(a(n),", "))
%Y A338633 Cf. A000699, A158119, A216966, A338634.
%K A338633 nonn
%O A338633 0,3
%A A338633 _Paul D. Hanna_, Nov 04 2020