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Old 2016-09-08, 04:02   #23
a1call
 
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Quote:
Originally Posted by CRGreathouse View Post
Yes, well done sm88.

My example of 4^n - 64 was constructed to take advantage of this property in a way that guarantees that I get the divisibility promised: since it's 0 at n = 3 , any prime not dividing 4 will loop back through that modular 0.
Corrections are welcome:

4^n-64= 64(4^(n-3)-1)
Dropping multiple of power of 2:
>>4^m-1= 2^(2q)-1
The function has 1/2 the number of infinite instances of 2^n-1, which is still infinite.
Is that enough to guarantee divisibility by all primes?
Yes, I think so since 2^(2q)-1 will always be divisible by 2^q-1.

The 2 functions are equivalent in this regard.
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Old 2016-09-08, 07:11   #24
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You could just as easily take 15^n - 1 or 6^n - 36.
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Old 2016-09-08, 11:45   #25
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Quote:
Originally Posted by CRGreathouse View Post
You could just as easily take 15^n - 1 or 6^n - 36.
where the latter always divides by 5. and for non zero n, 2 and 3. for all even n values it also divides by 7.
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