- This reference only shows that I independently proved Mertens' theorem (my formula 9).It still doesn't 'answer' 'why' we need to adjust the calculated probability with this constant to get the prime count. As mentioned, I am worried about this giant wave of prime multiples that is building up with this factor. But maybe it is all self-regulary by adjusting the dx.Also, so far, I haven't seen anyone using the probability product in an integration (my formula 6), which is just the key to the accurate prime counting results.

**From:**"primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>

**To:**primenumbers@yahoogroups.com

**Sent:**Monday, February 24, 2014 2:36 PM

**Subject:**[PrimeNumbers] Digest Number 3824

There is 1 message in this issue.

Topics in this digest:

1a. Re: Probabilistic approach to prime counting

From: djbroadhurst

Message

________________________________________________________________________

1a. Re: Probabilistic approach to prime counting

Posted by: david.broadhurst@... djbroadhurst

Date: Sun Feb 23, 2014 4:34 am ((PST))

Chris de Corte wrote:

> The question of why we had to correct our probabilities

Answer:

> with a factor alpha [i.e. exp(Euler)] from 2 onward remains

> open. And to be honest, we can’t give a good mathematical answer.

http://mathworld.wolfram.com/MertensTheorem.html

F. Mertens,

"Ein Beitrag zur analytischen Zahlentheorie",

J. reine angew. Math. 78 (1874) 46--62.

http://gdz.sub.uni-goettingen.de/en/dms/load/toc/?PPN=PPN243919689_0078

David

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------------------------------------------------------------------------ - Chris De Corte wrote:
> I independently proved Mertens' theorem

You proved nothing. Rather you made a faulty heuristic

and compared it with some data on small primes.

Your heuristic contained one very obvious mistake:

you sieved out primes less than x, while we

know that it is sufficient to sieve out primes less

than or equal to sqrt(x), to ensure that x is prime.

By this means you lost an obvious factor of

2 = log(x)/log(sqrt(x)).

Had you included this, your "fugde factor"

would have been the famous factor

2/exp(Euler) =~ 1.123

by which the sieve of Eratosthenes outperforms

a random sieve modelled by a Mertens product.

To prove this, simply combine Mertens' theorem (1874)

and the prime number theorem of Hadamard and

de la Vallée-Poussin (1896), neither of which

were proven in your "document".

The factor 2/exp(Euler) has been discussed several times on this list:

https://groups.yahoo.com/neo/groups/primenumbers/conversations/messages/21565

https://groups.yahoo.com/neo/groups/primenumbers/conversations/topics/20936

Hans Riesel discusses it on pp 66-67 of his book, following Theorem 3.1:

http://tinyurl.com/y9whej4

> the sieve of Eratosthenes sieves out numbers

David

> more efficiently than does a "random" sieve