OFFSET

1,2

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

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..260

SeqFan thread, A335071 question, SeqFan Mailing List, May 2020.

EXAMPLE

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

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

PROG

(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

CROSSREFS

KEYWORD

nonn

AUTHOR

Amiram Eldar, May 22 2020

STATUS

approved