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Search: id:A071529
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| A071529 |
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Number of 1's among the elements of the simple continued fraction for (1+1/n)^n. |
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+0
2
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| 0, 0, 1, 1, 1, 4, 7, 7, 12, 5, 8, 10, 23, 18, 25, 14, 18, 17, 14, 24, 22, 22, 35, 15, 21, 30, 29, 33, 37, 30, 27, 47, 49, 44, 54, 55, 53, 51, 46, 46, 43, 60, 64, 65, 79, 64, 64, 67, 73, 66, 79, 68, 60, 76, 86, 85, 85, 83, 86, 74, 90, 84, 93, 106, 90, 85, 98, 107, 88, 104, 86
(list; graph; listen)
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OFFSET
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1,6
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COMMENT
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It seems that lim n ->infinity a(n)/A069887(n) = C = 0,41... which is closed to (ln(4)-ln(3))/ln(2)=0,415... the expected density of 1's (cf. measure theory of continued fractions).
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EXAMPLE
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(1+1/14)^14 has for continued fraction [2, 1, 1, 1, 2, 6, 1, 7, 1, 6, 2, 1, 4, 21, 1, 1, 7, 1, 1, 1, 3, 2, 7, 2, 7, 1, 2, 4, 1, 3, 2, 1, 1, 1, 5, 1, 2, 5, 1, 2] which contains 18 "1's" hence a(14)=18.
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PROGRAM
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(PARI) for(n=1, 100, s=(1+1/n)^n; print1(sum(i=1, length(contfrac(s)), if(1-component(contfrac(s), i), 0, 1)), ", "))
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CROSSREFS
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Sequence in context: A088744 A061891 A063194 this_sequence A154922 A060409 A151968
Adjacent sequences: A071526 A071527 A071528 this_sequence A071530 A071531 A071532
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KEYWORD
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easy,nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 02 2002
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