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A071528
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Number of 1's among the elements of the simple continued fraction for e(n)=sum(k=1,n,1/k!).
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0
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1, 1, 2, 2, 4, 5, 6, 8, 7, 12, 11, 11, 15, 13, 13, 16, 19, 17, 18, 19, 23, 25, 25, 27, 29, 32, 32, 27, 40, 40, 46, 35, 44, 38, 41, 43, 40, 46, 45, 55, 54, 57, 62, 53, 57, 52, 59, 67, 61, 67, 66, 69, 74, 80, 79, 85, 77, 78, 76, 83, 85, 88, 96, 78, 101, 93, 89, 101, 88, 106, 95
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| It seems that lim n ->infinity a(n)/A069880(n) = C = 0,5... which is different from (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
| e(10) has for continued fraction [1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 11, 1, 1, 29, 1, 1, 2] which contains 12 "1's" hence a(10)=12.
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PROG
| (PARI) for(n=1, 150, if(prod(i=1, length(contfrac((1+1/n)^n)), n-component(contfrac((1+1/n)^n), i)) == 0, print1(n, ", ")))
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CROSSREFS
| Sequence in context: A064574 A059015 A024683 * A056902 A089676 A062436
Adjacent sequences: A071525 A071526 A071527 * A071529 A071530 A071531
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KEYWORD
| easy,nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 02 2002
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