- --- In primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> Why, Bill, do you post such wild claims

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

> when you have done such little testing?

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 - --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>

You did very much what I did.

> 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!

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,

Not very.

> given the variability of the overall factor with a?

Would you buy an appeal to Occam's razor, mon vieux?

Mike