## Re: Number of factors

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• In primenumbers@yahoogroups.com, ... Provided that p
Message 1 of 7 , Oct 1, 2011
Bernardo Boncompagni <RedGolpe@...> wrote:

> Let's say we have a random number N:
> we know that we can expect
> log(log(N)) distinct prime factors.
> Great, we found the least of them, say p.
>
> In the end, my question is:
> how many factors should one expect in N/p?

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 =
0.261497212847642783755426838608695859051566648261199206192...
but this is only a guess. (It is true for p = 2
and looks sensible for N >> p >> 2.)

David
• ... I have proven this for primes p up to 19, but the general proof eludes me. My proof for p = 19 was already laborious and is equivalent to showing that
Message 2 of 7 , Oct 2, 2011

> 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 =
> 0.261497212847642783755426838608695859051566648261199206192...

I have proven this for primes p up to 19,
but the general proof eludes me.
My proof for p = 19 was already laborious and is
equivalent to showing that test(19) is true, here:

{T(n)=local(F=factor(n)[,1]);
sum(k=1,n,sum(j=1,#F,gcd(k,F[j])==1))/n^2;}

{test(p)=local(P=1,D=1,A=1-1/p);
forprime(q=2,p-1,P*=q;D*=1-1/q;A-=1/q);
sumdiv(P,n,moebius(n)*T(n*p))*p/D == A;}

{gettime;forprime(p=2,19,if(test(p),
print("p = "p" proven in "gettime" ms")));}

p = 2 proven in 0 ms
p = 3 proven in 0 ms
p = 5 proven in 0 ms
p = 7 proven in 0 ms
p = 11 proven in 8 ms
p = 13 proven in 141 ms
p = 17 proven in 2926 ms
p = 19 proven in 66089 ms

Challenge: prove that test(p) is true for all prime p.

David (unable to this, at present)
• ... Interestingly enough, if p 1, sum=B_1+log(log(p)) and we get log(log(N))-log(log(p)). The chance of N being prime is 1/log(N). Invoking Mertens 3rd
Message 3 of 7 , Oct 3, 2011
>
> 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 =
> 0.261497212847642783755426838608695859051566648261199206192...
>
Interestingly enough, if p>>1, sum=B_1+log(log(p)) and we get
log(log(N))-log(log(p)).

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
enough.

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]
• ... For log(p)
Message 4 of 7 , Oct 3, 2011
Bernardo Boncompagni <RedGolpe@...> wrote:

> > log(log(N)) + B_1 - sum(prime q < p, 1/q) + o(1)
>
> Interestingly enough, if p>>1, sum=B_1+log(log(p))
> and we get log(log(N))-log(log(p)).

For log(p) << log(N), the average of omega(n/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: