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 2020-07-17, 21:14 #23 pepi37     Dec 2011 After milion nines:) 2×3×227 Posts Mnash doesnot take in calculation is sequence has algebra factors
2020-07-17, 21:26   #24
sweety439

Nov 2016

3·5·132 Posts

Quote:
 Originally Posted by pepi37 Mnash doesnot take in calculation is sequence has algebra factors
so how to get the true Nash weight?

Also, I want to calculate the Nash weight for the general case (k*b^n+c)/gcd(k+c,b-1) (k>=1, b>=2, c != 0, gcd(k,c)=1, gcd(b,c)=1)

2020-07-18, 05:12   #25
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

2·3·751 Posts

Quote:
 Originally Posted by sweety439 so how to get the true Nash weight?
Read the definition of Nash weight. Apply it.

2020-07-19, 14:34   #26
Thomas11

Feb 2003

1,907 Posts

Quote:
 Originally Posted by sweety439 There is a bug: but 9*4^n-1 should have weight 0, since it is proven composite by full algebra factors (3*2^n-1) * (3*2^n+1)
The tool doesn't check for algebraic factors...

2020-08-04, 23:48   #27
sweety439

Nov 2016

3×5×132 Posts

Quote:
 Originally Posted by VBCurtis Read the definition of Nash weight. Apply it.
See page https://www.rieselprime.de/ziki/Riesel_k%3D1-300, 9*2^n-1 has Nash weight 1674, and 25*2^n-1 has Nash weight 1571

 2020-08-04, 23:52 #28 sweety439     Nov 2016 3·5·132 Posts Also see post https://mersenneforum.org/showpost.p...7&postcount=24, some forms like 25*12^n-1 and 144*19^n-1 have zero weight. 25*12^n-1: for even n let n = 2*q; factors to: (5*12^q - 1) * (5*12^q + 1) odd n: factor of 13 144*19^n-1: for even n let n = 2*q; factors to: (12*19^q - 1) * (12*19^q + 1) odd n: factor of 5 Last fiddled with by sweety439 on 2020-08-04 at 23:53

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