It's likely that they are asymptotically equal.

In this moment I'm running a routine (I'm over 8.5 x 10^6) that counts

separately the number of primes ending with the digit 1, 3, 7, 9.

The fraction of the total number of primes found (that is: p1/PN, p3/PN,

p7/PN, p9/PN, where pk is a prime with k as last digit) is < 0.2500 for

the primes ending with 1 and 9, is > 0.2500 for the primes ending with 3

and 7.

They are all approaching 0.25: always from below p1 and p9 (0.2499),

always from above p3 and p7 (0.2501).

Gianni

>-----Original Message-----

>From: jbrennen [mailto:jack@...]

>Sent: Monday, December 06, 2004 3:19 PM

>To: primenumbers@yahoogroups.com

>Subject: [PrimeNumbers] Re: another way to calculate primes

>

>

>

>--- Mike Oakes wrote:

>>

>> In a message dated 06/12/2004 11:57:15 GMT Standard Time,

>> g.mazzarello@t... writes:

>>

>> >I was wondering this: for a fixed M, are there more primes

>< M ending

>> >with 3 and 7 than primes < M ending with 1 and 9?

>>

>> Everyone believes that asymptotically (as M -> oo) the numbers are

>> equal, but no-one can prove this.

>

>I thought that they were proven to be asymptotically equal by

>an extension of Dirichlet's Theorem.

>

> http://www.utm.edu/research/primes/notes/Dirichlet.html

>

>See the last equation on that page, which gives bounds for an

>error term.

>

>

>