

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
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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), 11261128.


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++ ]; k1, {n0, 3, 80}]] (* T. D. Noe *)


CROSSREFS

Cf. A059751, A059752.
Cf. A101083 (largest k such that the product (n+1)(n+2)...(n+k) has at least k distinct prime factors).
Sequence in context: A129456 A030412 A160371 * A101083 A097935 A188940
Adjacent sequences: A059683 A059684 A059685 * A059687 A059688 A059689


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



