

A071528


Number of 1's among the elements of the simple continued fraction for e(n)=sum(k=1,n,1/k!).


0



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
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OFFSET

1,3


COMMENTS

It seems that lim n >infinity a(n)/A069880(n) = C = 0.5... which is different from (log(4)log(3))/log(2)=0.415... the expected density of 1's (cf. measure theory of continued fractions).


LINKS

Table of n, a(n) for n=1..71.


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.


PROG

(PARI) for(n=1, 150, if(prod(i=1, length(contfrac((1+1/n)^n)), ncomponent(contfrac((1+1/n)^n), i)) == 0, print1(n, ", ")))


CROSSREFS

Sequence in context: A260295 A024683 A244017 * A056902 A089676 A230059
Adjacent sequences: A071525 A071526 A071527 * A071529 A071530 A071531


KEYWORD

easy,nonn


AUTHOR

Benoit Cloitre, Jun 02 2002


STATUS

approved



