Find Mersenne Primes twice as fast?

[a1call]

Originally Posted by CRGreathouse

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.

[CRGreathouse]

You could just as easily take 15^n - 1 or 6^n - 36.

[science_man_88]

Originally Posted by CRGreathouse

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.