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A059686
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Grimm numbers (1): a(n) = largest k so that for each m in {n+1, n+2, ..., n+k} there corresponds a different prime factor p_m.
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4
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2, 3, 4, 4, 3, 5, 4, 6, 6, 7, 6, 7, 6, 5, 8, 8, 7, 8, 7, 7, 8, 7, 6, 7, 9, 8, 8, 11, 10, 11, 10, 11, 11, 10, 12, 12, 11, 10, 9, 9, 8, 11, 10, 9, 10, 9, 8, 11, 13, 13, 12, 11, 10, 11, 14, 15, 14, 13, 12, 14, 13, 12, 13, 13, 14, 14, 13, 12, 11, 10, 9, 15, 14, 13, 14, 13, 13, 17, 16, 17
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OFFSET
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1,1
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COMMENTS
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Guy (2004) discusses some conjectures of Grimm that could produce related sequences.
The name "Grimm numbers" refers to the American mathematician Carl Albert Grimm (1926-2018). - Amiram Eldar, Apr 23 2024
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REFERENCES
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Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B32, pp. 133-134.
József Sándor, Dragoslav S. Mitrinovic, and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, Chapter XII, p. 438, Section XII.15.
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LINKS
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EXAMPLE
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For n=4 we look at the sequence {5,6,7,8,9,...} and we must pick a different prime factor for as many as we can. We can choose 5 for 5, 3 for 6, 7 for 7, 2 for 8, but now we are stuck, so k=4, a(4) = 4.
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MATHEMATICA
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Needs["DiscreteMath`Combinatorica`"]; factors[n_Integer] := First[Transpose[FactorInteger[n]]]; Join[{2, 3}, Table[k=2; While[s=Table[{}, {n0+k}]; prms=0; Do[If[PrimeQ[n], prms++, t=factors[n]; s[[n]]=t; Do[i=t[[j]]; If[i<n, AppendTo[s[[i]], n]], {j, Length[t]}]], {n, n0+1, n0+k}]; Length[BipartiteMatching[FromAdjacencyLists[s]]]+prms == k, k++ ]; k-1, {n0, 3, 80}]] (* T. D. Noe *)
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CROSSREFS
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Cf. A101083 (largest k such that the product (n+1)(n+2)...(n+k) has at least k distinct prime factors).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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