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A235063 Continued fraction expansion of Sum(i=1..inf, 1/2^(2^i+1) ).

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%I A235063 #13 Jun 27 2023 15:07:58
%S A235063 2,2,4,2,8,3,8,1,8,3,4,2,12,2,8,1,8,3,4,2,8,3,8,1,12,2,4,2,12,2,8,1,8,
%T A235063 3,4,2,8,3,8,1,8,3,4,2,12,2,8,1,12,2,4,2,8,3,8,1,12,2,4,2,12,2,8,1,8,
%U A235063 3,4,2,8,3,8,1,8,3,4,2,12,2,8,1,8,3,4,2,8,3,8,1,12,2,4,2,12,2,8,1,12,2,4
%N A235063 Continued fraction expansion of Sum(i=1..inf, 1/2^(2^i+1) ).
%H A235063 H. Cohn, <a href="http://arXiv.org/abs/math.NT/0008221">Symmetry and specializability in continued fractions</a>. arXiv preprint math/0008221 (2000), published in Acta Arithmetica 75 (1996): 4.
%H A235063 Jeffrey Shallit, <a href="http://www.cs.uwaterloo.ca/~shallit/Papers/scf.pdf">Simple continued fractions for some irrational numbers</a>, J. Number Theory 11 (1979), no. 2, 209-217.
%H A235063 <a href="/index/Con#confC">Index entries for continued fractions for constants</a>
%F A235063 0.40821075451094657... = (1/2) A007400.
%F A235063 Recurrence: a(8n)=1, a(8n+4)=a(16n+14)=a(32n+26)=2, a(16n+6)=a(32n+10)=3, a(8n+3)=4, a(8n+7)=a(16n+5)=a(32n+9)=8, a(16n+13)=a(32n+25)=12, a(8n+1)=a(4n+1), a(8n+2)=a(4n+2), starting 2,2,4 (conjectured).
%e A235063 0.40821075451094657... = 2/(2+1/(2+1/(4+1/(2+1/(8+1/(3+1/8...
%o A235063 (PARI) a(n)=contfrac(suminf(i=0, 1/2^(2^i+1)))[n+1]
%Y A235063 Cf. A007400.
%K A235063 cofr,nonn
%O A235063 0,1
%A A235063 _Ralf Stephan_, Jan 03 2014