This is in line the the existence of Sirpinski numbers
(no primes exists) and Riesel numbers for k*2^n+1.
They are proved by means of congurences.
The following values of k have given an exceptional harvest:
753 (9 primes up to 48000),
755 (8 primes up to 48000),
765 (9 primes up to 48000).
Other good k-values are 783, 789, and 885.
I guess that a congurence analysis will find much fewer
eliminating congurences.
Torbj�rn Alm
----- Original Message -----
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, June 26, 2001 9:08 PM
Subject: RE: Mersenne: Proth observations
>
> Andy Hedges <[EMAIL PROTECTED]> asks:
>
> > Anyone have any idea why for k = 659 there are very little primes? In
fact
> > for k up to 200000 there are none (I haven't found any in this range
yet!).
>
> Let k = 659.
>
> If n == 1 (mod 2) then k*2^n == 1 (mod 3)
> If n == 2 (mod 4) then k*2^n == 1 (mod 5)
> If n == 0 (mod 3) then k*2^n == 1 (mod 7)
> If n == 4 (mod 12) then k*2^n == 1 (mod 13)
> If n == 8 (mod 9) then k*2^n == 1 (mod 73)
>
> Therefore, if k*2^n - 1 is prime, then n == 20 or 32 (mod 36).
> Other useful congruences include
>
> If n == 2 (mod 5) then k*2^n == 1 (mod 31)
> if n == 0 (mod 23) then k*2^n == 1 (mod 47)
>
> This doesm't explain the total lack of primes, but
> shows that many potential n can be eliminated early.
>
>
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