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Old 2020-03-27, 15:51   #30
sweety439
 
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Nov 2016

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Quote:
Originally Posted by sweety439 View Post
9*1024^n-1 is unlikely to have a covering set, like the simple cases: 3*2^n+-1, 5*2^n+-1, 7*2^n+-1, 9*2^n+-1, 11*2^n+-1, etc. they are unlikely to have a covering set, but also no proof.
Well... since all prime factors of k*2^n+-1 are odd, thus if there exist an n such that k*2^n+-1 or its dual (|2^n+-k|) is power of 2 (including 1), then k*2^n+-1 cannot have a covering set.

Thus these k*2^n+-1 cannot have a covering set:

1*2^n+1 (for n=0, the value is 2)
3*2^n+1 (for n=0, the value is 4)
7*2^n+1 (for n=0, the value is 8)
15*2^n+1 (for n=0, the value is 16)
31*2^n+1 (for n=0, the values is 32)
1*2^n-1 (the dual form |2^n-1|, for n=1, the value is 1)
3*2^n-1 (for n=0, the value is 2, also the dual form |2^n-3|, for n=2, the value is 1)
5*2^n-1 (for n=0, the value is 4)
7*2^n-1 (the dual form |2^n-7|, for n=3, the value is 1)
9*2^n-1 (for n=0, the value is 8)
15*2^n-1 (the dual form |2^n-15|, for n=4, the value is 1)
17*2^n-1 (for n=0, the value is 16)
31*2^n-1 (the dual form |2^n-31|, for n=5, the value is 1)
33*2^n-1 (for n=0, the value is 32)
etc.

However, there is no proof that 5*2^n+1, 9*2^n+1, 11*2^n+-1, 13*2^n+-1, 17*2^n+1, ... have no covering set (i.e. the sequence of the smallest prime factor of k*2^n+-1 is unbounded above).

Last fiddled with by sweety439 on 2020-03-27 at 15:54
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