## Re: Lucas super-pseudoprime puzzle

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• ... That s a beautiful function tn(). I suspected Mod()s would be involved, but could never have come up with that polcoeff() construct. It can be speeded up
Message 1 of 33 , May 24, 2010
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>
> Using Pari-GP, I found a candidate solution as follows:
>
> tn(n,a)=2*polcoeff(lift(Mod((a+x)/Mod(2,n),x^2+4-a^2)^n),0)==Mod(a,n);
> ts(n)=tn(n,3)&&tn(n,4)&&tn(n,5)&&tn(n,6)&&tn(n,7)&&tn(n,8);
> {forstep(n=3,10^8,2,if(n%3&&!isprime(n)&&ts(n),
> print(n);print(round(gettime/10^3)" seconds");break()));}
>
> 7056721
> 84 seconds

That's a beautiful function tn().
I suspected Mod()s would be involved, but could never have come up with that polcoeff() construct.

It can be speeded up by a factor of 6/5 by omitting the tn(n,7) test, at the expense of a very few spurions, which can easily be eliminated later by hand. Just 4 up to n=10^8, in fact:
8646121, 17639999, 39528721, 90336961.

Your code scales amazingly well with n: 90 secs for first 10^7, increasing by a mere 27 secs for the slab of 10^7 starting at n=10^8, on my 3.6Ghz box.

> 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.

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

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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