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Search: id:A006466
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| A006466 |
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Continued fraction expansion of C = 2*sum( 1/2^(2^n), n=0 to infinity ).
(Formerly M0049)
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+0
5
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| 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2
(list; graph; listen)
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OFFSET
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0,5
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COMMENT
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C arises when looking for a sequence b(n) such that : b(1)=0, b(n+1) is the smallest integer > b(n) such that the continued fraction for 1/2^b(1)+1/2^b(2)+...+1/2^b(n+1) contains only 1's or 2's. Because b(n)=2^n-1 and C = sum(k>=0,1/2^b(k)). - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 03 2002
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REFERENCES
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J. O. Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.
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LINKS
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J. O. Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.
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FORMULA
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Recurrence: a(5n)=a(5n+1)=a(2)=a(5n+3)=a(20n+14)=a(40n+9)=1, a(20n+4)=a(40n+29)=2, a(5n+2)=3-a(5n-1), a(20n+19)=a(10n+9). - Ralf Stephan (ralf(AT)ark.in-berlin.de), May 17 2005
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CROSSREFS
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Sequence in context: A061704 A050361 A053150 this_sequence A086597 A031214 A056059
Adjacent sequences: A006463 A006464 A006465 this_sequence A006467 A006468 A006469
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KEYWORD
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nonn,cofr
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AUTHOR
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njas
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EXTENSIONS
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Better description and more terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 19 2001
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