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A380194 Continued fraction expansion of Sum_{i>=0} (-1)^i/(q(i)*q(i+1)) where q(0)=q(1)=1, q(3n+2)=q(3n+1)+q(3n), q(3n+3)=q(3n+2)+q(3n+1), and q(3n+4)=q(3n+2)*(q(3n+2)*q(3n+3)+1).

%I A380194 #45 Feb 11 2025 00:01:54
%S A380194 0,1,1,1,4,1,1,289,1,1,81126049,1,1,2128359349797626142548649,1,1,
%T A380194 38565134716822109850786884343127955049217538196275147632486387905655060249,
%U A380194 1,1
%N A380194 Continued fraction expansion of Sum_{i>=0} (-1)^i/(q(i)*q(i+1)) where q(0)=q(1)=1, q(3n+2)=q(3n+1)+q(3n), q(3n+3)=q(3n+2)+q(3n+1), and q(3n+4)=q(3n+2)*(q(3n+2)*q(3n+3)+1).
%C A380194 This is a transcendental number.
%C A380194 The n-th convergent of a(0..n) has q(n) as denominator.
%C A380194 Thus a(3*n+2) = a(3*n+3)=1 and a(3*n+4) = q(3*n+2)^2 for n>=1 are the results of repeatedly appending a triple of terms 1,1,Q^2 where Q is the convergent denominator after the first new 1.
%C A380194 The recurrence for q follows from this construction, and the alternating series is the continued fraction value for any sequence of convergent denominators.
%C A380194 This structure leads to the series and the recurrence for q.
%C A380194 Sum_{i>=0} (-1)^i/x(i) is another way to write the series, where x(i) = q(i)*q(i+1). When x(0)=1 , x(3n+2) divides x(3n+3), x(3n+2)-x(3n+1)=((x(3n+1))/x(3n))*(x(3n-1)/x(3n-2))*(x(3n-3)/x(3n-4))...(x(2)/x(1)))^2,x(3n+4)-x(3n+3)=(x(3n+3)/x(3n+2))^2*(x(3n+2)-x(3n+1)).
%H A380194 Khalil Ayadi, Chiheb Ben Bechir, and Maher Saadaoui, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL28/Ayadi/ayadi12.html">Continued Fractions with Predictable Patterns and Transcendental Numbers</a>, Journal of Integer Sequences, Vol. 28 (2025), Article 25.1.4.
%e A380194 0 + 1/(1 + 1/(1 + 1/(1 + ... ))) = 0.645164877940276...
%o A380194 (PARI) q(n) = if (n<=1, 1, if (n%3==1, q(n-2)*(q(n-2)*q(n-1)+1), q(n-1)+q(n-2)));
%o A380194 a(n) = if (n==0, 0, if ((n%3)==1, q(n-2)^2, 1)); \\ _Michel Marcus_, Jan 17 2025
%Y A380194 Cf. A006280, A019426, A003417.
%K A380194 nonn,cofr
%O A380194 0,5
%A A380194 _Khalil Ayadi_, Jan 15 2025