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Numbers n such that n through n+4 are divisible by the same number of distinct primes.
+10
6
54, 91, 92, 115, 141, 142, 143, 144, 158, 205, 212, 213, 214, 215, 295, 301, 323, 324, 325, 391, 535, 685, 721, 799, 1135, 1345, 1465, 1535, 1711, 1941, 1981, 2101, 2215, 2302, 2303, 2304, 2425, 2641, 2664, 2714, 3865, 3912, 4411, 5450, 5461, 6354, 6505
MATHEMATICA
SequencePosition[PrimeNu[Range[7000]], {x_, x_, x_, x_, x_}][[All, 1]] (* Harvey P. Dale, Jun 13 2022 *)
Numbers n such that n through n+6 are divisible by the same number of distinct primes.
+10
4
141, 142, 212, 213, 323, 2302, 10280, 16682, 19052, 20212, 21195, 25779, 33332, 35118, 35164, 35202, 39693, 39694, 40269, 41390, 41780, 42342, 42410, 44360, 44361, 44362, 48919, 48920, 48921, 48922, 53734, 54349, 54350, 56014, 56015
Numbers n such that each of the 6 consecutive numbers n through n+5 has exactly two distinct prime factors.
+10
2
91, 141, 142, 143, 212, 213, 214, 323, 324, 2302, 2303
COMMENTS
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
MATHEMATICA
Select[Range[3000], AllTrue[# + Range[0, 5], Length@FactorInteger[#] == 2 &] &] (* Giovanni Resta, May 09 2017 *)
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