A126016 Numbers whose aliquot sequence does not terminate in 1.

6, 25, 28, 95, 119, 143, 220

Offset

1,1

Comments

Sequence continues 276?, 284, 306?, 396?, 417, 445, 496, .... Because 276, 306 and 396 are all in the same family, either all 3 are present or none are. It is not known whether any aliquot sequence grows without bound; 276 is the smallest number for which this is unknown.

Additional tentative terms: 552, 562, 564, 565, 608, 650, 652, 660, 675, 685, 696, 780, 783, 790, 828, 840, 888, 909, 913, 966, 996, 1064, 1074, 1086, 1098, ... - Jean-François Alcover, Nov 14 2013

For additional terms, if the Goldbach Conjecture is assumed, take any odd term, subtract 1, and find two distinct primes that sum to it. For some numbers there will not be any pair of distinct primes. Multiply the two primes and the product is an element of the sequence. Note that this process does not work if the term - 1 is power of a prime. - Nathaniel J. Strout, Nov 25 2018

Links

Table of n, a(n) for n=1..7.

Eric Weisstein's World of Mathematics, Aliquot Sequence

P. Zimmermann, Latest information

Mathematica

maxAliquot = 10^45; A131884 = {}; s[1] = 1; s[n_] := DivisorSigma[1, n] - n; selQ[n_ /; n <= 5] = True; selQ[n_] := NestWhile[s, n, If[{##}[[-1]] > maxAliquot, Print["A131884: ", n]; AppendTo[A131884, n]; False, Length[{##}] < 4 || {##}[[-4 ;; -3]] != {##}[[-2 ;; -1]]] & , All] == 1; Reap[For[k = 1, k < 1100, k++, If[!selQ[k], Print[k]; Sow[k]]]][[2, 1]]

Crossrefs

Complement of A080907. Includes A000396, A063990 and other sociable numbers, A063769, numbers whose aliquot sequence reaches a sociable number and numbers whose aliquot sequence grows without bound.

Sequence in context: A184388 A136606 A074096 * A237286 A046416 A042879

Adjacent sequences: A126013 A126014 A126015 * A126017 A126018 A126019

Keyword

hard,nonn

Author

Franklin T. Adams-Watters, Dec 14 2006

Status

approved