- Let a and b be two relatively prime numbers. f(a,b) = a^2 - b^2 - a*b The

conjecture is that: f(a,b) = 1mod10 or (-1)mod 10 or 0mod5. and that:

f(a,b) will generate the set of every prime number of the form (+/-1)mod10

and the prime number 5 itself. There are an infinity of a and b which will

generate each member of this set and that: f(a,b) also generates are all

the possible composites of the prime numbers contained in the above set.

I have a web site skywebsite.com/mistermac/mistermac/ which contains the

above conjecture.

_________________________________________________________________________

Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com >It is easy to show that the repunits in a given base (i.e., the

Very true - in fact see:

>sequence enumerated above) cannot have a finite covering set

>of primes.

http://www.users.globalnet.co.uk/~perry/maths/uniqueantidivisor.htm#infinite

primes

I was hoping this would allow us to see the true differential between

Mersenne's (in which the sums do sum to 2^k-1), and other similar sequences

(which don't (sum to 2^k-1 or p^k-1)).

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths

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-----Original Message-----

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

Sent: 22 May 2002 19:55

To: primenumbers@yahoogroups.com

Subject: [PrimeNumbers] Re: New conjecture

--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:

>

> [A conjecture...]

> 1

> 1+p

> 1+p+p^2

> 1+p+p^2+p^3

> 1+p+p^2+p^3+p^4

> etc...

>

> contains an infinite number of prime terms.

Probably quite difficult to prove of course; proving this would

include the lesser task of proving the infinitude of Mersenne

primes, and we know that isn't easy to prove.

It is easy to show that the repunits in a given base (i.e., the

sequence enumerated above) cannot have a finite covering set

of primes.

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