A335071 Numbers m such that the delta(m) = abs(sigma(m+1)/(m+1) - sigma(m)/(m)) is smaller than delta(k) for all k < m.

1, 2, 14, 21, 62, 81, 117, 206, 897, 957, 1334, 1634, 2685, 2974, 4364, 14282, 14841, 18873, 19358, 24957, 33998, 36566, 42818, 56564, 64665, 74918, 79826, 79833, 92685, 109214, 111506, 116937, 122073, 138237, 145215, 15511898, 16207345, 17714486, 17983593, 18077605

Offset: 1

Author

Amiram Eldar, May 22 2020

Keywords

• nonn

Comments

Can two consecutive numbers have the same abundancy: sigma(m)/m = sigma(m+1)/(m+1)? If yes, then this sequence is finite.

There is no disproof of existence, but this would require both of the consecutive numbers to be k-perfect with the same k >= 2, and it is conjectured that such numbers are multiples of k!. It is very unlikely that an odd k-perfect number will ever be found, and even much less probable that it will be just next to an even k-perfect number. - M. F. Hasler, Jun 06 2020

Examples

			The values of delta(k) for the first terms are 0.5, 0.166..., 0.114..., 0.112..., 0.102..., ...
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..260
• SeqFan thread, A335071 question, SeqFan Mailing List, May 2020.

Crossrefs

Cf. A000203, A002961, A017665/A017666.

Programs

Mathematica

ab[n_] := DivisorSigma[1, n]/n; dm = 2; ab1 = ab[1]; s = {}; Do[ab2 = ab[n]; d = Abs[ab2 - ab1]; If[d < dm, dm = d; AppendTo[s, n]]; ab1 = ab2, {n, 2, 10^5}]; s

PARI

lista(nn) = {my(d=oo, newd, lastm=1, ab=1); for (m=2, nn, nab = sigma(m)/m; if ((newd=abs(nab-ab)) < d, print1(m-1, ", "); d = newd;); ab = nab;);} \\ Michel Marcus, May 24 2020