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A380013 Continued fraction expansion of Sum_{i>=0} (-1)^i/(q(i)*q(i+1)) where q(0)=q(1)=1, q(2n+2)=q(2n+1)+q(2n), and q(2n+3)=q(2n+1)*(q(2n+2)+1).

%I A380013 #69 Feb 12 2025 16:54:19
%S A380013 0,1,1,1,1,3,1,18,1,432,1,196992,1,38895676416,1,
%T A380013 1512881323731695591424,1,2288809899755012359448064967916189926490112,
%U A380013 1
%N A380013 Continued fraction expansion of Sum_{i>=0} (-1)^i/(q(i)*q(i+1)) where q(0)=q(1)=1, q(2n+2)=q(2n+1)+q(2n), and q(2n+3)=q(2n+1)*(q(2n+2)+1).
%C A380013 a(19) has 85 decimal digits and a(21) has 170 decimal digits.
%C A380013 This number is transcendental.
%C A380013 q(n) is the denominator of the convergent resulting from terms a(0..n).
%C A380013 The continued fraction is constructed by successively appending a pair of terms 1 and its own q(n) so far, so a(2*n) = 1 and a(2*n+1) = q(2*n-1) for n>=1
%C A380013 The series and the recurrence for q follows from that construction.
%C A380013 The series can also be written Sum_{i>=0} (-1)^i/x(i) where x(i) = q(i)*q(i+1) and in that case x(0)=1, x(2n+1) divides x(2n+2), and x(2n+3) = ((x(2n+2)/x(2n+1))*(x(2n)/x(2n-1))*...*(x(2)/x(1)))^2 + x(2n+2).
%H A380013 Andrew Howroyd, <a href="/A380013/b380013.txt">Table of n, a(n) for n = 0..25</a>
%H A380013 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 A380013 0 + 1/(1 + 1/(1 + 1/(1 + ... ))) = 0.6087912199223083952132365...
%o A380013 (PARI)
%o A380013 Q(n) = {my(v=vector(n+1)); v[1]=v[2]=1; for(i=2, n, v[i+1] = if(i%2==0, v[i]+v[i-1], v[i-1]*(v[i]+1))); v}
%o A380013 seq(n)=my(q=Q(max(2,n-2))); vector(n+1, n, if(n%2 || n<4, n>1, q[n-2])) \\ _Andrew Howroyd_, Jan 13 2025
%Y A380013 Cf. A006280, A019426.
%K A380013 nonn,cofr
%O A380013 0,6
%A A380013 _Khalil Ayadi_, Jan 09 2025