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## 15700RE: [PrimeNumbers] Re: another way to calculate primes

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• Dec 6, 2004
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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.
>
>
>
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