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A328899
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Numbers k such that the k-th repunit (10^k-1)/9 sets a new record for the number of distinct prime factors.
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2
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1, 2, 3, 6, 12, 18, 24, 30, 42, 60, 84, 96, 120, 168, 180, 210, 240, 252, 300, 360, 420, 630, 660
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OFFSET
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1,2
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COMMENTS
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The corresponding numbers of distinct prime factors are in A095371.
a(20) > 322.
Since A095371(19) = 40, to show that a(20) > 323 we use the fact that (10^323-1)/9) is a product of 4 primes and a 271-digit composite number C. We then use a computer search to show that C has no prime factor <= floor(C^(1/(41-4))) = 19858291. This implies that (10^323-1)/9) has less than 41 distinct prime factors. Applying this same approach to 337 and 353 (the only numbers between 323 and 359 for which the complete factorization of the corresponding repunit is not known) and using the factorization of (10^360-1)/9 with 44 distinct prime factors show that a(20) = 360 and A095371(20) = 44. This approach also shows that a(21) = 420 and A095371(21) = 55.
(End)
a(24) <= 840. Conjecture: a(24) = 840, a(25) = 1260, a(26) = 1680, a(27) = 1980, a(28) = 2520, a(29) = 3360, a(30) = 3780, a(31) = 3960, a(32) = 4620, a(33) = 5040, a(34) = 6300, a(35) = 7560, a(36) = 9240, a(37) = 10080. - Chai Wah Wu, Nov 01 2019
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LINKS
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MATHEMATICA
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r[n_] := (10^n - 1)/9; L = {}; bst = -1; Do[v = PrimeNu[r[n]]; If[v > bst, bst = v; AppendTo[L, n]], {n, 60}]; L
(* or, based on the b-file of A095370: *)
w = Last /@ Cases[Import["https://oeis.org/A095370/b095370.txt", "Table"], {_Integer, _Integer}]; L={}; bst=-1; Do[If[w[[j]] > bst, AppendTo[L, j]; bst = w[[j]]], {j, Length@w}]; L
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CROSSREFS
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KEYWORD
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nonn,base,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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