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%I #32 Jun 28 2022 10:59:34
%S 1,2,3,4,5,7,8,9,10,11,12,13,14,16,17,19,20,22,23,24,25,26,27,28,29,
%T 31,32,34,36,37,38,39,40,41,43,46,47,49,50,52,53,55,56,57,58,59,61,62,
%U 63,64,65,67,68,71,72,73,74,75,76,77,79,80,81,82,83,85,86,88,89,91,92,93
%N Numbers m such that the denominator of Sum_{k=1..n} 1/gcd(m,k) equals m.
%C Does lim_{n -> infinity} a(n)/n = 3/2?
%C Sum_{k=1..n} 1/gcd(n,k) = (1/n)*Sum_{d|n} phi(d)*d = (1/n)*Sum_{k=1..n} gcd(n,k)*phi(gcd(n,k))/phi(n/gcd(n,k)), where phi = A000010. - _Richard L. Ollerton_, May 10 2021
%C The numbers of terms not exceeding 10^k, for k = 1, 2, ..., are 9, 78, 709, 6713, 65135, 637603, 6275585, 61972835, 613362869, 6080312594, ... . Apparently, the asymptotic density of this sequence is 0 and the limit in the question above is infinite. - _Amiram Eldar_, Jun 28 2022
%H Amiram Eldar, <a href="/A071000/b071000.txt">Table of n, a(n) for n = 1..10000</a>
%e Sum_{k=1..12} 1/gcd(12,k) = 77/12 hence 12 is in the sequence.
%t Select[Range[100],Denominator[Sum[1/GCD[#,k],{k,#}]]==#&] (* _Harvey P. Dale_, Dec 13 2011 *)
%o (PARI) for(n=1,300,if(denominator(sum(i=1,n,1/gcd(n,i))) == n,print1(n,",")))
%Y Cf. A000010, A018804.
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, May 18 2002
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