Problems & Puzzles: Puzzles

Puzzle 1072 Exactly k consecutive odd numbers with ... JM Bergot, sent the following nice puzzle: For exactly one consecutive odd number having just ONE distinct prime factor:  37 (because 35=5*7, 37, 39=3*13) Exactly two consecutive odd numbers with one distinct prime factor each: 17 and 19 (because 15=3*5, 17, 19, 21=3*7) Exactly three consecutive odd numbers with one distinct prime factor each: 79,81,83 (because 77=7*11, 79, 81=3^4, 83, 85=5*17) Exactly five consecutive odd numbers with one distinct prime factor each: 23,25,27,29,31 (because 21=3*7, 23, 25=5^2, 27=3^3, 29, 31, 33=3*11) Q. Can you find exactly  k consecutive odd numbers with each having just one prime factor, for k=4, 6, 7, 8, 9, 10, ...?

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During the week 15-21 Jan, 2022, contributions came from Giorgos Kalogeropoulos, Emmanuel Vantieghem

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Giorgos wrote:

If we have 6 consecutive odd numbers, then two of them must be divisible by 3 and differ by 6.

This means that 3^a + 6 = 3^b. This is only true for a=1 and b=2.
The only 6 consecutive odd numbers are {3, 5, 7, 9=3^2, 11, 13}.

So there are no other solutions for k=6 and also for k>6.

For the case of n=4 we have that at least one number k must be divisible by 3.

a) For  {3^x, s1, s2 , 3^y} we only have {3, 5, 7, 9 = 3^2} which is invalid because 11 is prime

b) For the cases {s1, n, s2, s3} and {s1, s2, n, s3} (where s1,s2,s3 have one prime factor and n is divisible by 3),
    searching for powers of 3 didn't return any results. Also, in the case where s1,s2,s3 are primes, we can confirm that there is no such power of 3 up to 3^(10^6) from OEIS.

    It seems that n has always at least 2 prime factors (one being 3) but I don't have a proof.
 

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Emmanuel wrote:

The best I could find is this :
6  values : (1 has 0 prime factors), 3, 5, 7, 9, 11, 13, (15 = 3*5).

In my opinion, the chance to find more solutions is very small.

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