- Hello all.

k = 138847 has a standard (negative) Nash weight of 29, which is very

respectable for a reasonably small number. The lowest positive Nash

weight I found was 9 (yes, nine), so I'm sure that I could find

something for the negative side in the low teens if I put my mind to

it, though this would take some time with my current computers.

However, if someone could write a very fast program for me, that would

help. How about that Phil. Spookily, I was actually thinking about

asking for this just last week. Great minds think alike. The logic is

dead easy, honest.

Joe.

______________________________ Reply Separator _________________________________

Subject: [PrimeNumbers] Re: Nash weights

Author: d.broadhurst@... at Internet

Date: 12/03/01 01:38

Jack Brennen wrote:

> I have done some extensive calculations looking for high

I like low weights, too.

> Proth weights for both k*2^n+1 and k*2^n-1

They have been studied for k*2^n+1.

But in the Riesel case, k*2^n-1,

I do not know of any lower non-zero

weight than that for k=138847.

138847*2^n-1 is prime for n=33 and for no other n<380,000.

Is there any lower non-zero Nash weight with k<10^6,

for k*2^n +/- 1 ?

David

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---------------------------------------------------------------------------- - On Mon, 12 March 2001, joe.mclean@... wrote:
> k = 138847 has a standard (negative) Nash weight of 29, which is very

I find the measurements to be bizarre, in my mind I want a fractional value, and all these integers are out of range, so to speak. I guess that they're scaled, but why should they be?

> respectable for a reasonably small number. The lowest positive Nash

> weight I found was 9 (yes, nine), so I'm sure that I could find

> something for the negative side in the low teens if I put my mind to

> it, though this would take some time with my current computers.

What is the "average" Nash weight? (not specifying which average, probably geometric). And what if this is normalised to 1?

> However, if someone could write a very fast program for me, that would

Jack sent me a mail with a third modifier in it, independent of my first two modifiers. This was to do with (Aurefeulian?) factorisations of prime-power factors of k.

> help. How about that Phil. Spookily, I was actually thinking about

> asking for this just last week. Great minds think alike. The logic is

> dead easy, honest.

I believe that if I have the definitive expression for a single number, then I can turn that into a (windowed) sieve technique. Who knows, I may even be able to write some "very fast code" for it. It appears to be a O(sqrt(N)) space, O(N) time algorithm, where N is the number itself, not the number of digits.

Code like this must already have been written. I'm sure it;s easier for me to tweak than it is to start from scratch?

Phil

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