%I #13 Feb 24 2020 08:41:13
%S 20,42,77,110,156,272,342,506,812,930,1247,1332,1640,1806,2162,2756,
%T 3422,3660,4422,4970,5256,6162,6806,7832,9312,9328,10100,10506,11342,
%U 11772,12656,16002,17030,18632,19182,22052,22650,24492,26406,27722,29756,31862,32580,36290,37056,38612,39402
%N Numbers m such that gcd(A001008(m), m) > 1, in increasing order.
%C For the corresponding values of GCD(A001008(a(n)), a(n)) see A256103(n).
%C A001008(a(n))/A175441(a(n)) = A256103(n), n >= 1.
%C This means that for all values n not in the present sequence the numerator of the harmonic sum (HS) of the first n positive integers coincides with the denominator of the harmonic mean (HM) of the first n positive integers. That is, n divides the HM(n) numerator A102928(n) for n not in the present sequence.
%C Of course, HS(n)*HM(n) = n, n >= 1, where HS(n) = A001008(n)/A002805(n) and HM(n) = A102928(n)/A175441(n).
%C All terms are composite. Sequences contains all numbers of the form p*(p - 1), where p is a prime >= 5. This is because p^2 divides numerator(Sum_{i=1..p-1) 1/(k*p + i)), and p divides numerator(Sum_{i=1..p-1} 1/(i*p)), so p divides numerator(Sum_{i=1..p*(p-1)} 1/i). - _Jianing Song_, Dec 24 2018
%H Amiram Eldar, <a href="/A256102/b256102.txt">Table of n, a(n) for n = 1..500</a>
%F a(n) is the n-th smallest element of the set M:= {m positive inter | gcd(A001008(m), m) > 1}, n >= 1.
%e n = 1: gcd(A001008(20), 20) = gcd(55835135, 20) = 5 = A256103(1) > 1.
%e A001008(20)/A175441(20) = 55835135/11167027 = 5 = A256103(1).
%e Because 19 is not in this sequence 1 = gcd(A001008(19), 19) = gcd(275295799, 19).
%t Select[Range[10^4], !CoprimeQ[#, Numerator @ HarmonicNumber[#]] &] (* _Amiram Eldar_, Feb 24 2020 *)
%Y Cf. A256103, A001008, A002805, A102928, A175441.
%K nonn,easy
%O 1,1
%A _Wolfdieter Lang_, Apr 16 2015