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Old 2006-06-23, 20:07   #1
alpertron
 
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Aug 2002
Buenos Aires, Argentina

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Default Distribution of Mersenne prime factors mod 6

Let M be a Mersenne number with prime exponent q.

Then 2q = 1 (mod M)

2aq = 1 (mod M)

2aq+1 = 2 (mod M)

Since q is not multiple of 3, we can set a=-q mod 3, so the exponent is also multiple of 3, say 3r.

23r = 2 (mod M)

(2r)3 = 2 (mod M)

Let x = 2r:

x3 = 2 (mod M)

This means that for every prime factor p of M we have:

x3 = 2 (mod p)

This congruence is always satisfied when p mod 6 = 5, but it is interesting to notice that this is not the case for p mod 6 = 1.

Primes p up to 100000:
p = 5 (mod 6): 4806
p = 1 (mod 6): 1559

Primes p up to 1000000:
p = 5 (mod 6): 39231
p = 1 (mod 6): 13032

So it appears that it is more common to have prime factors of the form 6k+5 than from the form 6k+1.

It should be interesting to computing the real distribution from the known prime factors of Mersenne numbers (say from Will Edgington database).

Last fiddled with by alpertron on 2006-06-23 at 20:08
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