login
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
OFFSET
1,1
COMMENTS
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
REFERENCES
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.
LINKS
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<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 *)
CROSSREFS
Cf. A101083 (largest k such that the product (n+1)(n+2)...(n+k) has at least k distinct prime factors).
Sequence in context: A030412 A329526 A160371 * A101083 A373557 A097935
KEYWORD
nonn,easy,nice,look
AUTHOR
N. J. A. Sloane, Feb 06 2001
EXTENSIONS
More terms from Fabian Rothelius, Feb 08 2001
Corrected and extended by Naohiro Nomoto, Feb 28 2001
STATUS
approved