 mersenneforum.org > Math What's the average number of factors of an integer of size n?
 Register FAQ Search Today's Posts Mark Forums Read 2021-04-13, 09:48 #1 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 3×149 Posts What's the average number of factors of an integer of size n? Are there any estimates/theories/conjectures of the number of factors a random integer of size n has? Clearly any integer could be prime, but could also have factors of multiple factor of 2. So 64 has 6 factors, and 67 has no factors. I rather suspect that someone has worked out the average number of factors an integer of size n would have. I tried a quick test of this in Mathematica, but its factoring algorithm is very dumb - it uses only one core of my computer, and common sense says it could test multiple factors at the same time, not just one. I have 52 cores, but the license is for 4-cores. Dave Last fiddled with by drkirkby on 2021-04-13 at 09:49 Reason: Remove the number of unnecessary spaces this forum inserts between paragraphs   2021-04-13, 09:55   #2
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

61·97 Posts Quote:
 Originally Posted by drkirkby Are there any estimates/theories/conjectures of the number of factors a random integer of size n has? Clearly any integer could be prime, but could also have factors of multiple factor of 2. So 64 has 6 factors, and 67 has no factors. I rather suspect that someone has worked out the average number of factors an integer of size n would have. I tried a quick test of this in Mathematica, but its factoring algorithm is very dumb - it uses only one core of my computer, and common sense says it could test multiple factors at the same time, not just one. I have 52 cores, but the license is for 4-cores. Dave
https://en.m.wikipedia.org/wiki/Erd%...three%20primes.
The average number of factors is Loglog(n). This was a quick Google.   2021-04-13, 09:58   #3
Nick

Dec 2012
The Netherlands

174410 Posts Quote:
 Originally Posted by drkirkby So 64 has 6 factors, and 67 has no factors.
That is inconsistent counting.   2021-04-13, 11:19 #4 charybdis   Apr 2020 7628 Posts A fun exercise is to deduce from Mertens' theorem $\sum_{p\leq x}\frac{1}{p} = \log\log x + M + o(1)$ (where M is a constant) that $\sum_{n\leq x}|\omega(n)-\log\log n|^2 = O(x\log\log x)$ where ω(n) is the number of distinct prime factors of n. This implies that ω(n) cannot stray too far from loglog(n) very often. Of course this is much weaker than Erdos-Kac but the proof, including that of Mertens' theorem, is a lot simpler. Last fiddled with by charybdis on 2021-04-13 at 11:20   2021-04-13, 12:15 #5 Dr Sardonicus   Feb 2017 Nowhere 2·47·53 Posts The "normal order" and "average order" are not necessarily the same. Number of distinct prime factors, sum of prime-power exponents, and the divisor function (number of positive integer divisors) are discussed in the following paper: The normal number of prime factors of a number n Curiously, the "normal order" of the divisor function is significantly smaller than the "average order."   2021-04-13, 12:50   #6
charybdis

Apr 2020

7628 Posts Quote:
 Originally Posted by Dr Sardonicus The "normal order" and "average order" are not necessarily the same.
Indeed, and it's trivial to construct a counterexample. But they are the same for ω(n), and in fact to prove the result I gave above one would want to deal with $$\sum_{n\leq x}\omega(n)$$ first, from which we derive the average order.

Quote:
 Curiously, the "normal order" of the divisor function is significantly smaller than the "average order."
Technically speaking, it doesn't have a normal order in the strict sense, although its logarithm does. I suppose if you're Hardy and Ramanujan you're allowed to be a bit flexible with your terminology    2021-04-13, 13:10   #7
drkirkby

"David Kirkby"
Jan 2021
Althorne, Essex, UK

1BF16 Posts Quote:
 Originally Posted by Nick That is inconsistent counting.

I take your point. The only factor of 64 is 2 - I was considering 2x2x2x2x2x2=64 to be 6 factors.

Dave   2021-04-13, 14:06   #8
Dr Sardonicus

Feb 2017
Nowhere

115668 Posts Quote:
 Originally Posted by charybdis Indeed, and it's trivial to construct a counterexample. But they are the same for ω(n), and in fact to prove the result I gave above one would want to deal with $$\sum_{n\leq x}\omega(n)$$ first, from which we derive the average order.
Check. It's in the paper I linked to.
Quote:
 Technically speaking, it doesn't have a normal order in the strict sense, although its logarithm does. I suppose if you're Hardy and Ramanujan you're allowed to be a bit flexible with your terminology Yes, but I was being even more "flexible" than they were. As a non-trivial example where normal and average orders both exist but are different:

For positive integers n, let r2(n) be the number of representations of n as the sum of the squares of two integers. For example, r2(1) = 4, using the pairs (-1,0), (1,0), (0,-1), and (0,1).

Considering the points with integer coordinates inside the circle x^2 + y^2 = n, we see that

so the "average" order of r2(n) is π.

OTOH, it is well known that r2(n) = 0 unless every prime-power factor pe where p == 3 (mod 4) has an even exponent e. It then follows (though not easily) that the proportion of n with r2(n) > 0 tends to 0 as n increases without bound, so the "normal" order of r2(n) is 0.

IIRC r2(n) = 4(d1(n) - d3(n)), where d1(n) and d3(n) are the number of positive integer divisors of n which are congruent to 1 and 3 (mod 4) respectively.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Detheris Math 10 2021-04-15 14:55 MPT2019 Factoring 32 2019-11-06 23:42 aketilander Operazione Doppi Mersennes 1 2012-11-09 21:16 Robertcop Math 10 2006-04-30 22:23 wblipp ElevenSmooth 1 2003-11-25 15:47

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