A359056 Numbers k >= 3 such that 1/d(k - 2) + 1/d(k - 1) + 1/d(k) is an integer, d(i) = A000005(i).

3, 8, 15, 23, 39, 59, 159, 179, 383, 503, 543, 719, 879, 1203, 1319, 1383, 1439, 1623, 1823, 2019, 2559, 2579, 2859, 2903, 3063, 3119, 3779, 4283, 4359, 4443, 4679, 4703, 5079, 5099, 5583, 5639, 5703, 5939, 6339, 6639, 6663, 6719, 6999, 7419, 8223, 8783, 8819, 9183, 9663, 9903, 10079, 10839

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?

Because no numbers exist such that {p, p+1 = q^2, k}, {p, k, p+2 = q^2}, {p-2 = q^2, k, p}, {p-1 = q^2,p, k}, {k, p, p+1 = q^2}, {k, p-1 = q^2, p}, p, q prime numbers and k some number with 6 divisors, the answer is no. - Ctibor O. Zizka, Dec 30 2024

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

Cf. A000005, A317670, A350675.

Sequence in context: A331943 A064356 A157606 * A191414 A086959 A335753

Adjacent sequences: A359053 A359054 A359055 * A359057 A359058 A359059

Keyword

nonn,changed

Author

Ctibor O. Zizka, Dec 14 2022

Status

approved