## Re: [PrimeNumbers] Re: Nash weights

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• 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
Message 1 of 6 , Mar 12, 2001
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
asking for this just last week. Great minds think alike. The logic is

Joe.

Date: 12/03/01 01:38

Jack Brennen wrote:

> I have done some extensive calculations looking for high
> Proth weights for both k*2^n+1 and k*2^n-1

I like low weights, too.

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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• ... 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
Message 2 of 6 , Mar 12, 2001
On Mon, 12 March 2001, joe.mclean@... wrote:
> 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.

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?
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
> help. How about that Phil. Spookily, I was actually thinking about
> asking for this just last week. Great minds think alike. The logic is

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.

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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