- 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

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[Non-text portions of this message have been removed] - --- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>

wrote:>

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

> 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

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@... 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@... >