Problems & Puzzles: Puzzles

 Puzzle 885. Square-free almost consecutive integers type k-almost primes This is a follow up to Puzzle 884, after a suggestion given by Fred Schneider. Q. Send your largest square-free almost consecutive integers type k-almost primes. Do this for k=2. For k=3 and for k=4.

Contributions came from Emmanuel Vantieghem and Jan van Delden

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Emmanuel wrote:

The earliest set of  12  consecutive squarefree integers with exactly two prime factors is :
{374243, 374246, 374249, 374251, 374254, 374257, 374258, 374259, 374261, 374263, 374266, 374267}
there is no bigger set with numbers < 10^9.

The earliest set of  21  consecutive squarefree integers with exactly three prime factors is :
{650291611, 650291613, 650291614, 650291615, 650291617, 650291618, 650291619, 650291621, 650291622, 650291623, 650291626, 650291627, 650291629, 650291630, 650291633, 650291635, 650291637, 650291638, 650291639, 650291641, 650291645}
there is no bigger set with numbers < 10^9.

The earliest set of  12  consecutive squarefree integers with exactly four prime factors is :
{372536810, 372536813, 372536814, 372536815, 372536817, 372536818, 372536819, 372536821, 372536822, 372536823, 372536826, 372536827}
there is no bigger set with numbers < 10^10.

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Jan wrote:

k=2: Length 8   [143095, 143097, 143098, 143099, 143101, 143102, 143103, 143105]

Tested until 10^9
k=3: Length 12 [969833, 969834, 969835, 969837, 969838, 969839, 969841, 969842, 969843, 969845, 969846, 969847]
Tested until 5.10^8

k=4: Length 9 [118932833, 118932834, 118932835, 118932837, 118932838, 118932839, 118932841, 118932842, 118932843]

Tested until 5.10^8

I “jumped” only 1 number which is not squarefree

For almost consecutive k-almost primes the sequence breaks down if there is more than 1 number in the sequence of the wrong type (not square free or too few/ many factors). If we just focus on a sequence of two or more numbers which are not square free one can find an upperbound on the maximum length.

k=2: Maximum length<=17, starting number should then be 10 mod 36 for length 17 if present

k=3: Maximum length<=18, starting number should be 477 mod 900 for length 18 if present

k=4: Maximum length<=18, starting numbers: 51 different residues mod 44100 for length 18 if present

It seems that it is easier to find solutions to the k=3 case. Maybe someone could give an heuristic to the distribution of these lengths?

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