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Numbers m such that floor((1+1/m)^k)=k for an integer k>1.
1

%I #34 Feb 16 2019 06:45:23

%S 2,3,4,5,7,13,20,27,31,34,36,38,48,65,70,73,76,79,95,104,112,117,125,

%T 126,144,145,153,154,159,164,172,179,182,185,190,195,202,204,216,218,

%U 225,235,253,269,279,285,291,292,300,301,313,314,315,340,341,342,355,356,365

%N Numbers m such that floor((1+1/m)^k)=k for an integer k>1.

%H Vaclav Kotesovec, <a href="/A066661/b066661.txt">Table of n, a(n) for n = 1..68896</a> (terms 1..22681 from David A. Corneth)

%H David A. Corneth, <a href="/A066661/a066661.gp.txt">Identities floor((1+1/a(n)) ^ k) = k corresponding to a(n)</a>

%H David A. Corneth, <a href="/A066661/a066661_1.gp.txt">PARI program</a>

%H Vaclav Kotesovec, <a href="/A066661/a066661_1.jpg">Plot of a(n)/(n*log(n)) for n = 2..68896</a>

%F Conjecture: a(n)=C*n*log(n) asymptotically with C=1.5...

%e Floor((1+1/4)^11) = 11 hence 4 is in the sequence.

%t t = Table[Floor[Re[-LambertW[-1, -Log[1 + 1/n]]/Log[1 + 1/n]] + 1], {n, 1, 1000}]; Select[Range[Length[t]], Floor[(1 + 1/#1)^t[[#1]]] == t[[#1]]&] (* _Vaclav Kotesovec_, Feb 16 2019 *)

%o (PARI) See Corneth link \\ _David A. Corneth_, Feb 15 2019

%K nonn,easy

%O 1,1

%A _Benoit Cloitre_, Jan 14 2002

%E Terms 172 and 301 added by _Vaclav Kotesovec_, Feb 15 2019