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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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16
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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
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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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Harry J. Smith, Table of n, a(n) for n = 0..20000
H. Cohn, Symmetry and specializability in continued fractions
__Jeffrey 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, May 17 2005
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EXAMPLE
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0.816421509021893143708079737... = 0 + 1/(1 + 1/(4 + 1/(2 + 1/(4 + ...))))
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PROG
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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]
(PARI) { allocatemem(932245000); default(realprecision, 26000); x=suminf(n=0, 1/2^(2^n)); x=contfrac(x); for (n=1, 20001, write("b007400.txt", n-1, " ", x[n])); } [From Harry J. Smith, May 07 2009]
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CROSSREFS
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Cf. A007404, A089267.
Sequence in context: A135282 A179411 A103859 * A057696 A057697 A019921
Adjacent sequences: A007397 A007398 A007399 * A007401 A007402 A007403
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KEYWORD
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nonn,cofr
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AUTHOR
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Simon Plouffe, _Jeffrey Shallit_
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EXTENSIONS
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Deleted old PARI program Harry J. Smith, May 20 2009
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STATUS
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approved
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