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A101083
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Largest k such that the product (n+1)(n+2)...(n+k) has at least k distinct prime factors.
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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, 13, 12, 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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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
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REFERENCES
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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Table[k=2; While[Length[FactorInteger[Times@@Range[n0+1, n0+k]]]>=k, k++ ]; k-1, {n0, 100}]
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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