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Search: id:A007400
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| A007400 |
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Continued fraction for Sum[ 1/2^(2^n),{n,0,Infinity} ] = 0.8164215090218931... |
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+0
15
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| 0, 1, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 6, 4
(list; graph; listen)
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OFFSET
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0,3
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REFERENCES
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Shallit, Jeffrey; 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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H. Cohn, Symmetry and specializability in continued fractions
J. O. Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.
A. J. van der Poorten, An introduction to continued fractions
G. Xiao, Contfrac
Index entries for continued fractions for constants
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FORMULA
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Recurrence: a(0)=0, a(1)=1, a(2)=4, a(8n)=a(8n+3)=2, a(8n+4)=a(8n+7)=a(16n+5)=a(16n+14)=4, a(16n+6)=a(16n+13)=6, a(8n+1)=a(4n+1), a(8n+2)=a(4n+2). - Ralf Stephan (ralf(AT)ark.in-berlin.de), May 17 2005
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PROGRAM
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(PARI) a(n)=if(n<3, [0, 1, 4][n+1], if(n%8==1, a((n+1)/2), if(n%8==2, a((n+2)/2), [2, 0, 0, 2, 4, 4, 6, 4, 2, 0, 0, 2, 4, 6, 4, 4][(n%16)+1]))) /* from R. Stephan */
(PARI) a(n)=contfrac(suminf(n=0, 1/2^(2^n)))[n+1]
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CROSSREFS
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Cf. A007404.
Cf. A089267.
Adjacent sequences: A007397 A007398 A007399 this_sequence A007401 A007402 A007403
Sequence in context: A095382 A135282 A103859 this_sequence A019921 A061469 A021706
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KEYWORD
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nonn,cofr
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AUTHOR
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Simon Plouffe (plouffe(AT)math.uqam.ca), Jeff Shallit (shallit(AT)graceland.uwaterloo.ca)
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