## Re: faster than the APR algorithm

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• ... I remark that it took 114 milliseconds to prove Bill wrong: t3(n)=2*polcoeff(lift(Mod((3+x)/Mod(2,n),x^2-5)^n),0)==Mod(3,n);
Message 1 of 33 , May 22, 2010

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

I remark that it took 114 milliseconds to prove Bill wrong:

t3(n)=2*polcoeff(lift(Mod((3+x)/Mod(2,n),x^2-5)^n),0)==Mod(3,n);
t4(n)=2*polcoeff(lift(Mod((2+x)/Mod(1,n),x^2-3)^n),0)==Mod(4,n);
{forstep(k=3,10^6,2,if(k!=231&&!isprime(k)&&t3(k)&&t4(k),
print(k" found after "gettime" milliseconds");break()));}

11395 found after 114 milliseconds

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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