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%I #17 Aug 29 2022 09:28:28
%S 1,3,1,3,4,4095,1,3,3,1,3,4722366482869645213695,1,2,1,3,3,1,4095,4,3,
%T 1,3,
%U 3121748550315992231381597229793166305748598142664971150859156959625371738819765620120306103063491971159826931121406622895447975679288285306290175
%N Continued fraction expansion for c=sum_{k>=0} 1/2^(k!).
%C Observation: if b(k) denotes the sequence of all elements of the continued fraction for c, b(k) = 4095 if k==6 or 19 (mod 24); b(k) = 4722366482869645213695 if k==12 or 37 (mod 48); .... If b(k) is not congruent to 5 (mod 10), it seems that b(k) = 1,2,3 or 4 only.
%C Conjecture: a(3*2^n) = -1 + 2^[(n+1)((n+2)!) ]. - _Ralf Stephan_, May 17 2005
%C The conjecture follows from the theorem in Shallit's paper. The continued fraction has a "folded" overall structure. - _Georg Fischer_, Aug 29 2022
%H Rick L. Shepherd, <a href="/A076157/b076157.txt">Table of n, a(n) for n = 1..47</a>
%H Jeffrey O. Shallit, <a href="https://doi.org/10.1016/0022-314X(82)90047-6">Simple Continued Fractions for Some Irrational Numbers II</a>, J. Number Theory 14 (1982), 228-231.
%H Rick L. Shepherd, <a href="/A076157/a076157.txt">Table of n, a(n) for n = 1..383 (has larger terms than b-files support)</a>
%F c=1.2656250596046447753906250000000000007... = A076187.
%o (PARI)
%o {allocatemem(220000000);
%o default(realprecision, 1000000);
%o contfrac(suminf(k=0, 1/(2^(k!))))}
%Y Cf. A076152, A076154, A076187.
%Y Cf. A007400, A004200, A006466.
%K cofr,nonn
%O 1,2
%A _Benoit Cloitre_, Nov 02 2002
%E More terms from _Ralf Stephan_, May 17 2005
%E b-file, a-file, PARI program, and corrected conjecture by _Rick L. Shepherd_, Jun 07 2013
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