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• ## Re: Lucas super-pseudoprime puzzle

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• ... David, I have at last started to digest that paper. In particular, I believe that the relation between Chebyshev and Lucas polynomials is: T_n(x) =
Message 1 of 33 , May 26, 2010
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>
> This was the puzzle:
>
> > Puzzle: Find a composite positive odd integer n that passes
> > the Lucas test V(a,1,n) = a mod n, for every integer a.
>
>[snip]
>
> Then by googling
> > pseudoprime 7056721
> I arrived at
> which proves that 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]
> This paper claims that 7056721 is the unique solution with n < 10^10.

David,
I have at last started to digest that paper.
In particular, I believe that the relation between Chebyshev and Lucas polynomials is:
T_n(x) = (1/2)*V(2*x,1,n)
Yes?

Their definition of a "Chebyshev number" is a composite number n such that
T_n(a) = a mod n
for all integers a.

This translates into:
V(2*a,1,n) = 2*a mod n
for all integers a.

That's a weaker requirement than yours (above), which needs the modular equality to hold for _odd_ first argument of V() as well.

I wonder whether or not it is really a weaker condition.
Any views on this?

Mike
• ... 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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