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 A101083 Largest k such that the product (n+1)(n+2)...(n+k) has at least k distinct prime factors. 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, 13, 12, 15, 14, 13, 14, 13, 13, 17, 16, 17 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS This sequence is based on a slightly weaker, but still unproved, version of Grimm's conjecture: If there is no prime in the interval [n+1, n+k], then the product (n+1)(n+2)...(n+k) has at least k distinct prime divisors. We have a(n) >= A059686(n), with the two sequences first differing at n=70. Computing a(n) is much faster than computing A059686. It seems that Grimm's conjecture could have another (but not a weak) form: let p(1)...p(i) be a subset of prime numbers such that while n is integer, 0 < n < i, for any n, p(n) < p(n+1). Then there exists such sequence c(1)...c(i) where each term is a composite number, c(n+1) = c(n) + 1, and c(n) == 0 (mod p(n)). - Sergey Pavlov, Mar 21 2017 REFERENCES See A059686 LINKS T. D. Noe, Table of n, a(n) for n=1..1000 M. Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004, pages 6-7. EXAMPLE a(6) = 5 because 7*8*9*10*11 has 5 prime factors and 7*8*9*10*11*12 does not have 6 prime factors. MATHEMATICA Table[k=2; While[Length[FactorInteger[Times@@Range[n0+1, n0+k]]]>=k, k++ ]; k-1, {n0, 100}] CROSSREFS Cf. A059686 (Grimm numbers). Sequence in context: A329526 A160371 A059686 * A097935 A188940 A286549 Adjacent sequences:  A101080 A101081 A101082 * A101084 A101085 A101086 KEYWORD easy,nonn AUTHOR T. D. Noe, Nov 30 2004 STATUS approved

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Last modified January 22 07:31 EST 2022. Contains 350481 sequences. (Running on oeis4.)