Re: GF density
There were more my messages about GF density some weeks ago.
I try to show all ideas in this one.
Let consider GF(n,b). Let N=2^n, q=2*k*N+1 is prime
(k is natural), Euler=0.577215...
It's known that, given n, exactly N/q GFs are divisible by q. Namely,
exactly N GFs are divisible by q for every set of q consecutive even b's.
It's known that the density of primes around x tends to 1/log(x) when
It's also known that
lim(x->+infinity, product(p<x, 1-1/p)*log(x)) = 2/exp(Euler),
where p runs over ODD primes, Euler = 0.577215...
So, after few minutes of work with pencil, I conjectured that
C(n) = lim(Q->+infinity, product(q<Q, 1-N/q)*log(Q)*exp(Euler)/2),
where q runs over primes 2*k*N+1.
For large n first such k is enough big, so we have
But if first such k is small, e.g., k=1 for n=15 (there are no more known
k=1 cases for n>15 because 2^16+1 is the largest known Fermat
number), we have extremely small C(n): a very large amount of GF(b,n)
is divisible by these small q's.
I suggested to use A(n)=C(n)/log(N), so the number of GF primes for
fixed n and b<B approximately equals to A(n)*li(B)*ln(N)/N.
Taking about m=10^6 first q's gives good result, the relative error seems
to be about O[sqrt(N/Q)].
For example, C(1000)=621.9, A(1000)=0.8963 obtained quickly with