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A101913 G.f. satisfies: A(x) = 1/(1 + x*A(x^3)) and also the continued fraction: 1+x*A(x^4) = [1;1/x,1/x^3,1/x^9,1/x^27,...,1/x^(3^(n-1)),...].

%I A101913 #12 Mar 01 2022 11:01:19
%S A101913 1,-1,1,-1,2,-3,4,-6,9,-13,19,-28,41,-61,90,-132,195,-288,424,-625,
%T A101913 922,-1359,2004,-2955,4356,-6423,9471,-13963,20587,-30355,44755,
%U A101913 -65987,97293,-143449,211503,-311844,459785,-677912,999524,-1473709,2172854,-3203685,4723551,-6964461,10268490,-15139986
%N A101913 G.f. satisfies: A(x) = 1/(1 + x*A(x^3)) and also the continued fraction: 1+x*A(x^4) = [1;1/x,1/x^3,1/x^9,1/x^27,...,1/x^(3^(n-1)),...].
%F A101913 From _Joerg Arndt_, Oct 15 2011: (Start)
%F A101913 For the sequence abs(a(n)) we have
%F A101913 g.f. B(x) 1/(1-x/(1-x^3/(1-x^9/(1-x^27(1- ... ))))) and
%F A101913 B(x) satisfies B(x) = 1 + x*B(x)*B(x^3)  (cf. A000621)
%F A101913 (End)
%F A101913 G.f.: T(0), where T(k) = 1 - (-x)^(3^k)/((-x)^(3^k) - 1/T(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Oct 18 2013
%F A101913 a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/3)} a(k) * a(n-3*k-1). - _Ilya Gutkovskiy_, Mar 01 2022
%o A101913 (PARI) {a(n)=local(A);A=1-x;for(i=1,n\3+1, A=1/(1+x*subst(A,x,x^3)+x*O(x^n)));polcoeff(A,n,x)}
%o A101913 (PARI) {a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(3))+1,n,1/x^(3^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(4*n+1)),4*n+1)}
%Y A101913 Cf. A101912, A101914-A101918.
%K A101913 sign
%O A101913 0,5
%A A101913 _Paul D. Hanna_, Dec 20 2004