Hi!
To do a congurence analysis is fairly simple by means of some
math program.
For k-values of interest, I generated k*2^n-1 for n=0 to 30.
Then I factored the values.
In this table, the cyclic behavior becomes very obvious.
A number of small factors will occur.
The table for 885 was interesting.
Out of 30 number, 14 were primes!
Torbjörn Alm
----- Original Message -----
From: "Torbjörn Alm" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, June 27, 2001 9:11 PM
Subject: Re: Mersenne: Proth observations
> 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.
>
> Torbjörn 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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