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A101912 G.f. satisfies: A(x) = 1/(1 + x*A(x^2)) and also the continued fraction: 1 + x*A(x^3) = [1; 1/x, 1/x^2, 1/x^4, 1/x^8, ..., 1/x^(2^(n-1)), ...].

%I A101912 #26 Mar 01 2022 10:57:06
%S A101912 1,-1,1,0,-1,1,0,-2,3,-1,-3,6,-4,-4,12,-10,-5,23,-25,-2,43,-57,12,74,
%T A101912 -124,56,120,-258,172,170,-516,454,187,-989,1095,40,-1811,2487,-604,
%U A101912 -3128,5375,-2567,-4991,11140,-7704,-6976,22164,-20062,-7220,42288,-48020,-36,76928,-108334,29476,131898,-233020,117166
%N A101912 G.f. satisfies: A(x) = 1/(1 + x*A(x^2)) and also the continued fraction: 1 + x*A(x^3) = [1; 1/x, 1/x^2, 1/x^4, 1/x^8, ..., 1/x^(2^(n-1)), ...].
%C A101912 Sequence appears to have a rational g.f. - _Ralf Stephan_, May 17 2007
%C A101912 The conjecture is wrong. The g.f. is dependent on the number of terms. - _Johannes W. Meijer_, Aug 08 2011
%F A101912 G.f.: 1/(1 + x/(1 + x^2/(1 + x^4/(1 + x^8/(1 + ...))))) (continued fraction). - _Joerg Arndt_, Oct 19 2012
%F A101912 G.f. A(x) = 1/B(x) where B(x) is the g.f. of A218031. - _Joerg Arndt_, Oct 19 2012
%F A101912 a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/2)} a(k) * a(n-2*k-1). - _Ilya Gutkovskiy_, Mar 01 2022
%p A101912 nmax:=57: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101912(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^2)),x,k+1); for n from 0 to k do x(n):=coeff(f(x),x,n) od: A101912(k):=x(k): od: seq(A101912(n), n=0..nmax); # _Johannes W. Meijer_, Aug 08 2011
%t A101912 m = 58; A[_] = 0; Do[A[x_] = 1/(1 + x A[x^2]) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* _Jean-François Alcover_, Nov 03 2019 *)
%o A101912 (PARI) {a(n)=local(A);A=1-x;for(i=1,n\2+1, A=1/(1+x*subst(A,x,x^2)+x*O(x^n)));polcoeff(A,n,x)}
%o A101912 (PARI) {a(n)=local(M=contfracpnqn(concat(1, vector(#binary(n)+1,n,1/x^(2^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(3*n+1)),3*n+1)}
%Y A101912 Cf. A101913, A101914, A101915, A101916, A101917, A101918.
%K A101912 sign
%O A101912 0,8
%A A101912 _Paul D. Hanna_, Dec 20 2004