## Re: Lucas super-pseudoprime puzzle

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• ... Yes. So far I have run the faster test up to n = 9*10^9. Confirmation of the claim that the next solution has n 10^10 should not take much longer. I did
Message 1 of 33 , May 24, 2010
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"mikeoakes2" <mikeoakes2@...> wrote:

>> n is a solution if and only if it is an odd
>> square-free composite integer such that for each prime p|n
>> n = +/- 1 mod p-1 ... [1]
>> n = +/- 1 mod p+1 ... [2]

> It might be interesting to program this test as it might be
> significantly faster for large n, not so?

Yes. So far I have run the faster test up to n = 9*10^9.
Confirmation of the claim that the next solution has
n > 10^10 should not take much longer.

I did not see Jacobs, Rayes and Trevisan remark
on the fact that n cannot be divisible 3, so
please pat yourself on the back, Mike.

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
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>
> 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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