Re: [PrimeNumbers] Re: Number of factors
>Interestingly enough, if p>>1, sum=B_1+log(log(p)) and we get
> Provided that p << N, I guess that the average value
> of omega(n/p) for the numbers n up to N with least
> prime divisor p is
> log(log(N)) + B_1 - sum(prime q < p, 1/q) + o(1)
> with B_1 =
The chance of N being prime is 1/log(N). Invoking Mertens' 3rd theorem, the
chance of a number sieved up to p being prime should be
log(p)/(log(N)*exp(-gamma)), where gamma is the Euler-Mascheroni constant.
If the log(N) in the number of divisors is distantly related to the chance
of N being prime, one may squeeze this result into our formula obtaining
log(log(N)*exp(-gamma)/log(p))=log(log(N))-log(log(p))-gamma. Not sure if
this last passage makes sense but at this point I guess
log(log(N))-log(log(p)) is a very reasonable answer to the problem. By the
way, N>p^2 means log(log(N))-log(log(p))>log(2) which looks reasonable
It's even very elegant if one writes it as log(lg(N)), where lg is the
logarithm to base p...
[Non-text portions of this message have been removed]
- --- In firstname.lastname@example.org,
Bernardo Boncompagni <RedGolpe@...> wrote:
> > log(log(N)) + B_1 - sum(prime q < p, 1/q) + o(1)For log(p) << log(N), the average of omega(n/p),
> Interestingly enough, if p>>1, sum=B_1+log(log(p))
> and we get log(log(N))-log(log(p)).
with n running from p to N and p the least
prime divisor of n, is (I claim) asymptotic to the
sum of 1/q, with prime q running from q = p up to
the greatest prime q <= N/p.
All I did was to use Mertens at the upper limit.
For log(2) << log(p) << log(N), one may also use
Mertens at the lower limit to get log(log(N)/log(p)),
as Bernardo has observed. However one should not push
this up to p = O(sqrt(N)), since there we may run into
a problem that worried Chebyshev:
David (per proxy Pafnuty Lvovich)