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A359056
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Numbers k >= 3 such that 1/d(k - 2) + 1/d(k - 1) + 1/d(k) is an integer, d(i) = A000005(i).
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2
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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
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OFFSET
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1,1
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COMMENTS
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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?
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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Select[Range[11000], IntegerQ[Total[1/DivisorSigma[0, # - {0, 1, 2}]]] &] (* Amiram Eldar, Dec 14 2022 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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