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A071527
Numbers k such that the simple continued fraction for (1+1/k)^k contains k.
1
2, 3, 7, 8, 15, 17, 18, 24, 32, 45, 54, 59, 90, 100, 112, 114, 128, 135, 144, 249, 270, 419, 452, 517, 557, 674, 701, 822, 916, 1074, 1274, 1643, 2137, 2180, 2272, 2519, 2539, 3049, 3294, 3666, 3907, 4405, 4463, 4478, 4848, 5226, 5446, 7058, 7189, 7857, 8966, 9430
OFFSET
1,1
LINKS
EXAMPLE
(1+1/17)^17 has for continued fraction [2, 1, 1, 1, 3, 1, 10, 1, 2, 1, 4, 5, 4, 3, 2, 1, 2, 37, 3, 5, 3, 2, 1, 14, 1, 17, 1, 22, 1, 2, 1, 1, 1, 1, 2, 2, 16, 2, 2, 1, 1, 1, 10, 2, 5, 2] which contains 17 hence 17 is in the sequence.
MATHEMATICA
For[n=1, n<=3000, n++, {r=(1+1/n)^n; If[MemberQ[ContinuedFraction[r], n], Print[n]]}]
Select[Range[6000], MemberQ[ContinuedFraction[(1+1/#)^#], #]&] (* Harvey P. Dale, May 03 2011 *)
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, ", ")))
CROSSREFS
Sequence in context: A033082 A084406 A075648 * A307328 A097095 A101755
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jun 02 2002
EXTENSIONS
More terms from John W. Layman, Apr 26 2004
More terms from Harvey P. Dale, May 03 2011.
More terms from Amiram Eldar, Sep 05 2020
STATUS
approved