- In a message dated 04/02/2005 01:54:40 GMT Standard Time, decio@...

writes:

So, if we're looking for an accurate

numerical value of the probability that a random integer near n is prime,

shouldn't we use the heuristic 1.123/log n instead?

No, we shouldn't.

>by Mertens theorem, the probability that an integer [n] isn't

This is true for integers n >> x.

>divisible by any primes up to x is exp(-gamma)/log x

But fails when x gets as large as sqrt(n).

-Mike Oakes

[Non-text portions of this message have been removed] - As expected, my idea proposed below is not new. David Broadhurst and Phil

Carmody (both missed members of this list, unlike the Goldbach-proving

spammers) pointed me towards the twin-prime constant C2 = 0.660162, which is

the ratio between Mertens' product and `my' product below. One can use

Cohen's methods from http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi to

compute C2 to high-precision, as pointed out by David Broadhurst. Mike Oakes

also pointed out (off list) an interesting site regarding the twin-prime

constant, the page by Xavier Gourdon at

http://numbers.computation.free.fr/Constants/Primes/twin.html. If this isn't

enough, the owners of Crandall & Pomerance's book can have a look at chapter

1, who has a derivation of the twin prime conjecture.

Décio

On Thursday 03 February 2005 23:49, you wrote:

> For p prime, the integer p-1 is an important quantity in certain analyses

> (particularly if they involve the group (Z/pZ)*), but as I recall, this

> integer is usually considered as `a random integer in the vicinity of p' in

> such analyses (except perhaps by considering that it is even).

>

> However, a neat idea occurred to me: just as the probability that 2 divides

> this integer is skewed in comparison to a random integer (it's 1 in this

> case but 1/2 for a random integer), so is the probability that it is

> divisible by any other prime. Consider for instance the residue class of p

> mod 3; unless p itself is 3, then p == 1 or 2 mod 3, with equal probability

> for each residue class. So p-1 == 0 or 1 mod 3 also with equal probability.

> Thus, p-1 is divisible by 3 with probability 1/2, not 1/3 as expected.

> Using the same argument, p-1 is divisible by a given prime q with

> probability 1/(q-1), not 1/q.

>

> The first implication that occurred to me is that one shouldn't use

> Mertens' theorem in any analyses involving p-1 where p is prime. For

> instance, I computed the probability that primes from 3 to 31337 divide a

> given integer using Mertens' theorem; the result is 0.1085. Using PARI/GP

> to evaluate the product (1 - 1/(p-1)) for primes p from 3 to 31337, I

> obtained 0.07157. The difference is quite real.

>

> I'll try to work out an estimate, similar to Mertens' theorem, but for the

> product (1 - 1/(p-1)) instead. However, I'm pretty sure David Broadhurst or

> Mike Oakes will beat me to it (:

>

> I don't know how useful this is (probably not useful at all), but I thought

> it was neat and worth mentioning.

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