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A126016
Numbers whose aliquot sequence does not terminate in 1.
6
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
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
KEYWORD
hard,nonn
AUTHOR
STATUS
approved