A256102 Numbers m such that gcd(A001008(m), m) > 1, in increasing order.

20, 42, 77, 110, 156, 272, 342, 506, 812, 930, 1247, 1332, 1640, 1806, 2162, 2756, 3422, 3660, 4422, 4970, 5256, 6162, 6806, 7832, 9312, 9328, 10100, 10506, 11342, 11772, 12656, 16002, 17030, 18632, 19182, 22052, 22650, 24492, 26406, 27722, 29756, 31862, 32580, 36290, 37056, 38612, 39402

Offset: 1

Author

Wolfdieter Lang, Apr 16 2015

Keywords

• nonn
• easy

Comments

For the corresponding values of GCD(A001008(a(n)), a(n)) see A256103(n).

A001008(a(n))/A175441(a(n)) = A256103(n), n >= 1.

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.

Of course, HS(n)*HM(n) = n, n >= 1, where HS(n) = A001008(n)/A002805(n) and HM(n) = A102928(n)/A175441(n).

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

Examples

			n = 1: gcd(A001008(20), 20) = gcd(55835135, 20) = 5 = A256103(1) > 1.
A001008(20)/A175441(20) = 55835135/11167027 = 5 = A256103(1).
Because 19 is not in this sequence 1 = gcd(A001008(19), 19) = gcd(275295799, 19).
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A256103, A001008, A002805, A102928, A175441.

Programs

Mathematica

Select[Range[10^4], !CoprimeQ[#, Numerator @ HarmonicNumber[#]] &] (* Amiram Eldar, Feb 24 2020 *)

Formula

a(n) is the n-th smallest element of the set M:= {m positive inter | gcd(A001008(m), m) > 1}, n >= 1.