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• ## Re: [PrimeNumbers] Number of factors in average

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• ... Hash: SHA1 ... If no one else has a better approach, I suggest a statistical approach based on Dickman s theorem on smoothness of number, i.e. psi(n,
Message 1 of 6 , May 5, 2003
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On Monday 05 May 2003 06:24, Jose Ramón Brox wrote:
> Which is the best manner to calculate the average of factors of the
> numbers up to a fixed N? And between N and M?
>
> (for example, up to 10: 1,2...,9,10 have 1,1,1,2,1,2,1,3,2,2 factors, the
> average is 1,6)
>
> Is the average tending to infinite or to a specific ratio?
>

If no one else has a better approach, I suggest a statistical approach based
on Dickman's theorem on smoothness of number, i.e. psi(n, n^(1/x)) = x*u^(-u)
with u = log(x)/log(log(x)). Now compute the expected value E[ ] (you should
be very familiar with that given your telecommunications background!) and
luckily you'll be able to work out a formula.

Décio
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• In a message dated 05/05/03 10:25:14 GMT Daylight Time, ambroxius@terra.es ... It tends to infinity. The standard number theory function you are interested in
Message 2 of 6 , May 5, 2003
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In a message dated 05/05/03 10:25:14 GMT Daylight Time, ambroxius@...
writes:

> Which is the best manner to calculate the average of factors of the
> numbers up to a fixed N? And between N and M?
>
> (for example, up to 10: 1,2...,9,10 have 1,1,1,2,1,2,1,3,2,2 factors, the
> average is 1,6)
>
> Is the average tending to infinite or to a specific ratio?
>

It tends to infinity.

The standard number theory function you are interested in is d(n), defined to
be the number of divisors of n, including 1 and n, so that d(n) >= 2.

Then Hardy and Wright "An Introduction to the Theory of Numbers" (1979) (p.
264) have :-
"Theorem 319. The average order of d(n) is log n."

An alternative formulation of this result is:-
lim {n -> oo} [d(1) + d(2) + ... + d(n)] / log(n) = 1.

Mike Oakes

[Non-text portions of this message have been removed]
• Sent too quickly - I should have written: lim {n - oo} [d(1) + d(2) + ... + d(n)] / (n*log(n)) = 1. Mike [Non-text portions of this message have been removed]
Message 3 of 6 , May 5, 2003
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Sent too quickly - I should have written:
lim {n -> oo} [d(1) + d(2) + ... + d(n)] / (n*log(n)) = 1.

Mike

[Non-text portions of this message have been removed]
• Hi: I was thinking on prime factors rather than in divisors... the number of them is quite lesser than this of the divisors... what order has? Jose ... From:
Message 4 of 6 , May 5, 2003
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Hi:

I was thinking on prime factors rather than in divisors... the number of them is quite lesser than this of the divisors... what order has?

Jose
----- Original Message -----
From: mikeoakes2@...
Sent: Monday, May 05, 2003 1:39 PM
Subject: Re: [PrimeNumbers] Number of factors in average

In a message dated 05/05/03 10:25:14 GMT Daylight Time, ambroxius@...
writes:

> Which is the best manner to calculate the average of factors of the
> numbers up to a fixed N? And between N and M?
>
> (for example, up to 10: 1,2...,9,10 have 1,1,1,2,1,2,1,3,2,2 factors, the
> average is 1,6)
>
> Is the average tending to infinite or to a specific ratio?
>

It tends to infinity.

The standard number theory function you are interested in is d(n), defined to
be the number of divisors of n, including 1 and n, so that d(n) >= 2.

Then Hardy and Wright "An Introduction to the Theory of Numbers" (1979) (p.
264) have :-
"Theorem 319. The average order of d(n) is log n."

An alternative formulation of this result is:-
lim {n -> oo} [d(1) + d(2) + ... + d(n)] / log(n) = 1.

Mike Oakes

[Non-text portions of this message have been removed]

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://www.primepages.org/

[Non-text portions of this message have been removed]
• In a message dated 06/05/03 00:40:56 GMT Daylight Time, ambroxius@terra.es ... them is quite ... Sorry, my fault. That standard number theory function is
Message 5 of 6 , May 5, 2003
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In a message dated 06/05/03 00:40:56 GMT Daylight Time, ambroxius@...
writes:

> I was thinking on prime factors rather than in divisors... the number of
them is quite
> lesser than this of the divisors... what order has?

Sorry, my fault.
That standard number theory function is Omega(n), defined to be the total
number of prime factors of n; in other words, if there is the prime
factorisation
n = p_1^e_1 * ... * p_r^e_r,
then
Omega(n) = e_1 + ... + e_r.

So, in particular Omega(1) = 0. [As an aside: anyone who thinks 1 is a prime
would have a hard job defining Omega(); and 1 is certainly not composite...]

Omega(n) has average order log(log(n)).

Mike

[Non-text portions of this message have been removed]
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