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 A059686 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. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS R. K. Guy, Unsolved Problems in Theory of Numbers, 2nd ed., Section B32, discusses some conjectures of Grimm that could produce related sequences. REFERENCES D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XII.15, p. 438. LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 C. A. Grimm, A conjecture on consecutive composite numbers, Amer. Math. Monthly, 76 (1969), 1126-1128. EXAMPLE 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. MATHEMATICA 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

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Last modified November 18 21:04 EST 2018. Contains 317331 sequences. (Running on oeis4.)