Mersenne prime exponent not randomly distributed?

[R. Gerbicz]

Originally Posted by R. Gerbicz

So the probability that
for p prime number 2^p-1 is also prime is about
prod(k=1,k<2^p/p,isprime(2*k*p+1)*(1-1/(2*k))) and this will be about
c/p, where c is a constant. This proof isn't hard. Use that about every p-th
prime number is in the 2*k*p+1 sequence and 1-x is about exp(-x) if abs(x)
is small and also Mertens' theorem.

Something is wrong here. Because the original expected probability c*log(p)/p
so I've lost log(p) factor, this is important, otherwise there would be only
c*log(log(log(n))) Mersenne primes up to n (this can't be, looking the Mersenne
primes, the expected Mersennes is c*log(log(n)) up to n). So there is two possibilities:
prod(k=1,k<2^p/p,isprime(2*k*p+1)*(1-1/(2*k))) is c*log(p)/p or somewhere
else there is an error in my argument.


It is possible to extend Wagstaff's (wrong) conjecture to other modulo's. To give
c1 and c2 for that: the expected probability that 2^p-1 is prime is e^gamma*log(c1*p)/(p*log(2))
if p==1 mod 3 and e^gamma*log(c2*p)/(p*log(2)) if p==2 mod 3. Looking the data c1>c2.
But if for example we see the remainders modulo 15 then we get the highest
probability for p==4 mod 15. So it's interesting that not p==1 mod 15 gives the highest
probability. ( for that (p-1) is divisible by 3 and 5).

[ewmayer]

OK, now that the latest discovery has been confirmed, we have a brand-spanking-new data point to read more into than it deserves ;)

32582657 = 2¹⁰*47*677 + 1 .

[alpertron]

The second largest prime factor 47 is slightly greater than the fifth root of the exponent, about 31.8. So there is no significant statistic departure here from the expected value.

[ewmayer]

Originally Posted by alpertron

The second largest prime factor 47 is slightly greater than the fifth root of the exponent, about 31.8. So there is no significant statistic departure here from the expected value.

Next thing you're going to tell us that the 2¹⁰ was "fully within the expected average range," right?

"This number is not smooth - it only pretends to be smooth." :D

[R. Gerbicz]

Originally Posted by ewmayer

Next thing you're going to tell us that the 2¹⁰ was "fully within the expected average range," right?

No, it was luck. The exponent of M15 is 1279 and this is the first Mersenne prime, whose exponent>2^10. We know exactly 30 Mersenne primes larger than M14. So the probability that none of the p-1 is divisible by 1024 is (1-1/512)^30=0.9430 so it has got about 5.7% chance that at least one of them is divisible by 1024. Note that it isn't very exactly because for example p==1 mod 4 has got a slightly higher chance than p==3 mod 4 for Mersenne exponents.

We can continue this. p-2=3^6*5*7*1277 so p-2 is divisible by 3^6=729. Not bad, p-1 is divisible by 2^10, and it's neighbour divisible by 3^7. What's this chance?

[alpertron]

Originally Posted by ewmayer

Next thing you're going to tell us that the 2¹⁰ was "fully within the expected average range," right?

"This number is not smooth - it only pretends to be smooth." :D

Please see post #4 on this thread for the expected values of the second largest factor of p-1.

[wpolly]

32582657=2*3*5430443-1

[akruppa]

The average exponent of 2 in a prime minus 1 is 2. For the Mersenne primes in Dario's list we have an average exponent of 67/29 = 2.31... without M44, and 77/30 = 2.566... with M44. It's a bit above the expected value, but not an awful lot...

Alex

[Flatlander]

Originally Posted by alpertron

I don't know if this was noticed before.

This is the factorization of the exponent - 1 of Mersenne primes.

...

Look at the second largest factor, in general it is much smaller than a second largest factor of a random number, so it appears that the exponents are not randomly distributed.

Is there any reason 'exponent - 1' was chosen?
If you look at 'exponent + 1', the first few factor very uninterestingly.

[R.D. Silverman]

Originally Posted by Flatlander

Is anything any reason 'exponent - 1' was chosen?
If you look at 'exponent + 1', the first few factor very uninterestingly.

I don't want to burst everyone's bubble, but this entire discussion
is *meaningless*, because noone has defined 'randomly distributed'.

If one means 'uniformly random', that is clearly impossible.

One must specify a *density function* before one can even begin to
talk about 'random distribution'. So far, I have not seen such a
definition or discussion.

The gaps between Mersenne exponents can not follow *any* stationary
distribution. The gaps must increase as p --> oo.

All this talk about 'random' is meaningless without some rigorous definitions.

[alpertron]

Originally Posted by R.D. Silverman

I don't want to burst everyone's bubble, but this entire discussion
is *meaningless*, because noone has defined 'randomly distributed'.

If one means 'uniformly random', that is clearly impossible.

One must specify a *density function* before one can even begin to
talk about 'random distribution'. So far, I have not seen such a
definition or discussion.

The gaps between Mersenne exponents can not follow *any* stationary
distribution. The gaps must increase as p --> oo.

All this talk about 'random' is meaningless without some rigorous definitions.

If f(2,p) is the second largest factor of p-1 (when p is prime), I'm looking for the average of log(f(2,p))/log p in the range (n, 10n) when n -> inf. I think that it should be a constant near 0.2. Please see post #4 again.