## Re: faster than the APR algorithm

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• ... Nonsense. Your Lucas tests are very weak, since you require only that V(3,1,n) = 3 mod n V(4,1,n) = 4 mod n Here are the double-pseudoprimes less than
Message 1 of 33 , May 22, 2010
Bill Bouris <leavemsg1@...> wrote:

> 231 is the only composite that gives a mis-read

Nonsense. Your Lucas tests are very weak, since
you require only that
V(3,1,n) = 3 mod n
V(4,1,n) = 4 mod n
Here are the double-pseudoprimes less than 10^6:

231, 11395, 17119, 31535, 80189, 84419, 117215, 120581,
133399, 163081, 194833, 228241, 516559, 545279, 588455,
721801, 737471, 797819, 873181 ...

Why, Bill, do you post such wild claims
when you have done such little testing?

David
• ... You did very much what I did. I nearly fell off the chair when I averaged everything out and saw 0.57... :-) ... Not very. Would you buy an appeal to
Message 33 of 33 , May 27, 2010
>
> I tried 1/n^c:
>
> v=[1237.1, 328.7, 105.4, 28.01, 6.22 , 1.510, 0.439, 0.0939];
> print(vector(8,k,-log(v[k]/10^6)/(k+4)/log(10)));
>
> [0.5815, 0.5805, 0.5682, 0.5691, 0.5785, 0.5821, 0.5780, 0.5856]
>
> and then A/n^c, using the first datum to remove A:
>
> print(vector(7,k,-log(v[k+1]/v[1])/k/log(10)));
>
> [0.5756, 0.5348, 0.5484, 0.5747, 0.5827, 0.5750, 0.5885]
>
> In both cases c =~ Euler looks rather convincing,
> given the statistics. Well spotted, Sir!

You did very much what I did.
I nearly fell off the chair when I averaged everything out and saw 0.57... :-)

> How strongly are you committed to A = 1, for the average,
> given the variability of the overall factor with a?

Not very.
Would you buy an appeal to Occam's razor, mon vieux?

Mike
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