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## Re: Prime Conjecture

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• ... Yes, yes. Maybe I ll share that proof later, hehe. Along similar lines: Given a prime p, is there always an r
Message 1 of 7 , Mar 4, 2008
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--- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
wrote:
>
> Hello All:
>
> Can anyone prove this conjecture?
>
> For all n>2 ;
>
> for all i: 1<= i<= n exist j: n+1 <= j <= n^(3/2)
>
> (1+pj)/(-1+pi) is integer.
>
>
> pi is the ith prime number.
>
> Sincerely:
>
> Sebastian Martín Ruiz

Yes, yes. Maybe I'll share that proof later, hehe.

Along similar lines:
Given a prime p, is there always an r < (log(2*p+1))^2
such that p*r - 1 is prime?
(Or such that p*r + 1 is prime?)

Example: let p = 740191.
The first r such that p*r - 1 is prime is r = 192.
But (log(2*740191+1))^2 =~201.9 > 192.

Mark
• Sebastian asked the following. 1a. Prime Conjecture Posted by: Sebastian Martin sebi_sebi@yahoo.com sebi_sebi Date: Sun Mar 2, 2008 8:25 am ((PST)) Hello
Message 2 of 7 , Mar 4, 2008
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Sebastian asked the following.

1a. Prime Conjecture
Posted by: "Sebastian Martin" sebi_sebi@... sebi_sebi
Date: Sun Mar 2, 2008 8:25 am ((PST))

Hello All:

Can anyone prove this conjecture?

For all n>2 ;

for all i: 1<= i<= n exist j: n+1 <= j <= n^(3/2)

(1+pj)/(-1+pi) is integer.

pi is the ith prime number.

Sincerely:

Sebastian Martín Ruiz

Sebastian, this should be quite easy to prove for just about any subsequence of the positive integers.

Prove that it's true for the entire set of positive integers, then it would be trivial to show that it's
also true for the prime number subsequence.

Kermit Rose < kermit@... >
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