A335071 - OEIS

%I #30 May 15 2023 08:46:17

%S 1,2,14,21,62,81,117,206,897,957,1334,1634,2685,2974,4364,14282,14841,

%T 18873,19358,24957,33998,36566,42818,56564,64665,74918,79826,79833,

%U 92685,109214,111506,116937,122073,138237,145215,15511898,16207345,17714486,17983593,18077605

%N 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.

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

%C 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

%H Amiram Eldar, <a href="/A335071/b335071.txt">Table of n, a(n) for n = 1..260</a>

%H SeqFan thread, <a href="http://list.seqfan.eu/oldermail/seqfan/2020-May/020757.html">A335071 question</a>, SeqFan Mailing List, May 2020.

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

%t 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

%o (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

%Y Cf. A000203, A002961, A017665/A017666.

%K nonn

%O 1,2

%A _Amiram Eldar_, May 22 2020