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Numbers k such that the simple continued fraction for (1+1/k)^k contains k.
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%I #17 Sep 05 2020 07:45:53

%S 2,3,7,8,15,17,18,24,32,45,54,59,90,100,112,114,128,135,144,249,270,

%T 419,452,517,557,674,701,822,916,1074,1274,1643,2137,2180,2272,2519,

%U 2539,3049,3294,3666,3907,4405,4463,4478,4848,5226,5446,7058,7189,7857,8966,9430

%N Numbers k such that the simple continued fraction for (1+1/k)^k contains k.

%H Amiram Eldar, <a href="/A071527/b071527.txt">Table of n, a(n) for n = 1..95</a>

%e (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.

%t For[n=1, n<=3000, n++, {r=(1+1/n)^n;If[MemberQ[ContinuedFraction[r], n], Print[n]]}]

%t Select[Range[6000],MemberQ[ContinuedFraction[(1+1/#)^#],#]&] (* _Harvey P. Dale_, May 03 2011 *)

%o (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,",")))

%K nonn

%O 1,1

%A _Benoit Cloitre_, Jun 02 2002

%E More terms from _John W. Layman_, Apr 26 2004

%E More terms from _Harvey P. Dale_, May 03 2011.

%E More terms from _Amiram Eldar_, Sep 05 2020