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

%I A359056 #32 Dec 30 2024 17:06:11
%S A359056 3,8,15,23,39,59,159,179,383,503,543,719,879,1203,1319,1383,1439,1623,
%T A359056 1823,2019,2559,2579,2859,2903,3063,3119,3779,4283,4359,4443,4679,
%U A359056 4703,5079,5099,5583,5639,5703,5939,6339,6639,6663,6719,6999,7419,8223,8783,8819,9183,9663,9903,10079,10839
%N A359056 Numbers k >= 3 such that 1/d(k - 2) + 1/d(k - 1) + 1/d(k) is an integer, d(i) = A000005(i).
%C A359056 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?
%C A359056 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
%H A359056 Seiichi Manyama, <a href="/A359056/b359056.txt">Table of n, a(n) for n = 1..10000</a>
%H A359056 Thomas F. Bloom, <a href="https://arxiv.org/abs/2112.03726">On a density conjecture about unit fractions</a>, arXiv:2112.03726 [math.NT], 2021.
%H A359056 Ernest S. Croot, III, <a href="https://annals.math.princeton.edu/wp-content/uploads/annals-v157-n2-p04.pdf">On a coloring conjecture about unit fractions</a>, Annals of Mathematics, Volume 157 (2003), 545-556.
%e A359056 k = 3:
%e A359056 1/d(1) + 1/d(2) + 1/d(3) = 1/1 + 1/2 + 1/2 = 2. Thus k = 3 is a term.
%e A359056 k = 8:
%e A359056 1/d(6) + 1/d(7) + 1/d(8) = 1/4 + 1/2 + 1/4 = 1. Thus k = 8 is a term.
%t A359056 Select[Range[11000], IntegerQ[Total[1/DivisorSigma[0, # - {0, 1, 2}]]] &] (* _Amiram Eldar_, Dec 14 2022 *)
%Y A359056 Cf. A000005, A317670, A350675.
%K A359056 nonn
%O A359056 1,1
%A A359056 _Ctibor O. Zizka_, Dec 14 2022