A335291 Numbers m such that the delta(m) = abs(h(m+1) - h(m)) is smaller than delta(k) for all k < m, where h(m) is the harmonic mean of the divisors of m.

1, 2, 4, 91, 272, 20118, 20712, 33998, 42818, 61695, 25274946, 27194929, 34883654, 40406622, 43176318, 47350866, 52680050, 149736013, 154957034, 162929406, 171560153, 187012577, 208015843, 267361097, 300087726, 325189758, 355153181, 443360633, 584803578, 605883413

Offset: 1

Author

Amiram Eldar, May 30 2020

Keywords

• nonn

Comments

Apparently, most of the terms m have h(m+1) > h(m) and numerator(delta(m)) = 1.

Can two consecutive numbers have the same harmonic mean of divisors? If yes, then this sequence is finite.

Examples

			The values of delta(k) for the first terms are 0.333..., 0.166..., 0.047..., 0.0357..., ...
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..86

Crossrefs

Cf. A000005, A000203, A099377, A099378.

Cf. A238380, A335071.

Programs

Mathematica

h[n_] := n * DivisorSigma[0, n]/DivisorSigma[1, n]; dm = 1; h1 = h[1]; s = {}; Do[h2 = h[n]; d = Abs[h2 - h1]; If[d < dm, dm = d; AppendTo[s, n-1]]; h1 = h2, {n, 2, 10^5}]; s