login
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.
2
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
1,2
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
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