OFFSET

1,1

COMMENTS

The sets {2, 4, 4}, {2, 3, 6} and {3, 3, 3} including permutations of elements of the set are the solutions of this unit fraction. There is no k for which {d(k - 2), d(k - 1), d(k)} equals {3, 3, 3}. May the set {2, 3, 6} exist for some k?

LINKS

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Thomas F. Bloom, On a density conjecture about unit fractions, arXiv:2112.03726 [math.NT], 2021.

Ernest S. Croot, III, On a coloring conjecture about unit fractions, Annals of Mathematics, Volume 157 (2003), 545-556.

EXAMPLE

k = 3:

1/d(1) + 1/d(2) + 1/d(3) = 1/1 + 1/2 + 1/2 = 2. Thus k = 3 is a term.

k = 8:

1/d(6) + 1/d(7) + 1/d(8) = 1/4 + 1/2 + 1/4 = 1. Thus k = 8 is a term.

MATHEMATICA

Select[Range[11000], IntegerQ[Total[1/DivisorSigma[0, # - {0, 1, 2}]]] &] (* Amiram Eldar, Dec 14 2022 *)

CROSSREFS

KEYWORD

nonn

AUTHOR

Ctibor O. Zizka, Dec 14 2022

STATUS

approved