Problems & Puzzles: Puzzles

Puzzle 884. Almost consecutive integers type k-almost primes A natural number is called k-almost prime if it has exactly k prime factors, counted with multiplicity. Then, a semiprime integer is just a 2-almost prime integer. For k=2, all integers (2^2)n>2^2 are k-almost prime integers such that k>2. Accordingly the maximal quantity of consecutive integers type 2-almost prime is (2^2)-1=3. The first such trio is {33, 34, 35} = {3*11, 2*17, 5*7} But what about an almost-consecutive set of L integers type 2-almost prime, that jumps over the integers type (2^2)n? How large can be L for such sets? This is my best set: k=2, L=8: {143095, 143096, 143097, 143098, 143099, 143100, 143101, 143102, 143103, 143104, 143105} = {5*28619, 2^3*31*577, 3*47699, 2*71549, 11*13009, 2^2*3^3*5^2*53, 7*20443, 2*71551, 3*47701, 2^8*13*43, 5*28621} Q1. For k=2, can it be L larger than 8? Q2. Redo Q1 for k=3, [jumping over the integers (2^3)n]

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Contributions came from Emmanuel Vantieghem and Fred Schneider

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Emmanuel wrote:

Q1.The maximum is eight.  That can be reached (as in the example) between two odd mupltiples of  6.  No odd multiple of  6  (greater than 6) can be a semiprime.

 

Q2.The first set of seven consecutive numbers with three prime factor is :

   {211673,211674,211675,211676,211677,211678,211679}

 

Disregarding multiples of eigth, I think the maximum reachable is  20  (between two odd multiples of  12)  but I could only find a set of  15  almost consecutive numbers with 3 prime factors : 

{9449720273, 9449720274, 9449720275, 9449720276, 9449720277, 9449720278, 9449720279, -> (9449720280) ->, 9449720281, 9449720282, 9449720283,

 9449720284, 9449720285, 9449720286, 9449720287, -> (9449720288) ->, 9449720289}

No doubt a maximal set will consist of bigger numbers ...

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Fred wrote:

For Q1, every 9th number will be a multiple of 3^2.
For Q2... See Puzzle 885

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