
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, ...  JeanFrançois 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


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]]
