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A088983
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Numbers n such that each of the 6 consecutive numbers n through n+5 has exactly two distinct prime factors.
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2
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91, 141, 142, 143, 212, 213, 214, 323, 324, 2302, 2303
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OFFSET
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1,1
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COMMENTS
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Initial segment of A045934 is identical to this sequence but in A045934 the 12th term is divisible by 3 prime factors. Is the present sequence complete?
We're looking for at least 6 consecutive positive integers that each have exactly two distinct prime divisors. I.e. 6 consecutive positive integers m with omega(m) = 2. Now of exactly 6 consecutive integers, exactly one of them is divisible by 6, i.e. m is of the form 2*3*k. However m has exactly 2 distinct prime divisors, so k can only have prime divisors 2 or 3. Now, suppose m ends in 6 or higher. Then one of the consecutive integers is divisible by 10 = 2*5. I.e. it's of the form 2*5*t. Then t can only have prime divisors 2 and 5. (End)
This sequence has no run of four consecutive integers, since Eggleton and MacDougall prove that there are no more than 9 consecutive integers with A001221(k) = 2. They conjecture that A007774 contains no runs of 9 consecutive integers, and has only two runs of size 8 (at 141 and 212) and two maximal runs of size 7 (at 323 and 2302); they add that the maximal run of size 6 at 91 might be the only such run, so A088983 might be complete. - Roger Eggleton via Jason Kimberley, Jul 12 2017
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LINKS
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MATHEMATICA
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Select[Range[3000], AllTrue[# + Range[0, 5], Length@FactorInteger[#] == 2 &] &] (* Giovanni Resta, May 09 2017 *)
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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