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A076157
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Continued fraction expansion for c=sum_{k>=0} 1/2^(k!).
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4
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1, 3, 1, 3, 4, 4095, 1, 3, 3, 1, 3, 4722366482869645213695, 1, 2, 1, 3, 3, 1, 4095, 4, 3, 1, 3, 3121748550315992231381597229793166305748598142664971150859156959625371738819765620120306103063491971159826931121406622895447975679288285306290175
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OFFSET
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1,2
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COMMENTS
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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.
Conjecture: a(3*2^n) = -1 + 2^[(n+1)((n+2)!) ]. - Ralf Stephan, May 17 2005
The conjecture follows from the theorem in Shallit's paper. The continued fraction has a "folded" overall structure. - Georg Fischer, Aug 29 2022
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LINKS
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FORMULA
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c=1.2656250596046447753906250000000000007... = A076187.
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PROG
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(PARI)
{allocatemem(220000000);
default(realprecision, 1000000);
contfrac(suminf(k=0, 1/(2^(k!))))}
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CROSSREFS
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KEYWORD
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cofr,nonn
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AUTHOR
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EXTENSIONS
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b-file, a-file, PARI program, and corrected conjecture by Rick L. Shepherd, Jun 07 2013
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STATUS
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approved
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