# Average Exponents and the Divisor Function

Here’s something I’ve been thinking about lately: patterns in the prime factorizations of natural numbers. Every number can be written in exactly one way as a product of powers of primes:

$\displaystyle n = p_1^{a_1}p_2^{a_2}\dotsm p_k^{a_k}$

This factorization is also linked to the number of proper divisors of n. If we want to know how many divisors has, we are really asking how many numbers can be made from the same primes as n. If some can be written as:

$\displaystyle d = p_1^{b_1}p_2^{b_2}\dotsm p_k^{b_k}, \hspace{10 mm} 0 \le b_i \le a_i$

then clearly divides n, and every divisor of can be written this way. So how many such can there be? The only things to change are the exponents  $\displaystyle b_i$ , and these can each be any number from  $\displaystyle \{0, 1, 2, \ldots , a_i\}$  . This gives  $\displaystyle a_i + 1$  choices for each exponent, so the total number of divisors of is

$\displaystyle \prod_{i=1}^k \left(a_i + 1 \right)$

Before I go on, I should warn you that the following are just some observations on the behavior of factorizations and numbers of divisors. Perhaps someone reading this will have some insight, or has already investigated this. I have not managed to say anything meaningful about the divisor function, but I do find it linked to an expression that I think is interesting.

Because there are patterns in the values that these exponents tend to take. For instance, if I pick a number at random and ask you to guess the exponent of some prime in its factorization, what should you guess? It turns out that there is an elegant form for the expected value of these exponents.

Start with 2, and let’s say we’re picking from all natural numbers. Let  $\displaystyle \mu(p)$  be the expected value of the exponent of p. Even numbers are divisible by 2, so the exponent of 2 for half of the numbers is at least 1. That gives  $\displaystyle \mu(2) \ge 1/2$.

One fourth of numbers are divisible by 4, so they each contribute a 2 to the average. Think of them as contributing 1 along with the general even numbers, then an extra 1. Then one eighth of all numbers contribute again, and one sixteenth contribute a fourth time, etc. In the end we have

$\displaystyle \mu(2) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dotsb = 1$

This can be done just the same with the other primes

$\displaystyle \mu(p) = \sum_{k=1}^\infty \frac{1}{p^k}$

This can be simplified in a cute way if you’ll trust me that it’s convergent. Multiply the sum by to get

$\displaystyle p\mu(p) = 1 + \frac{1}{p} + \frac{1}{p^2} + \dotsb = 1 + \mu(p)$

$\displaystyle (p - 1)\mu(p) = 1$

$\displaystyle \mu(p) = \frac{1}{p-1}$

So the average exponent for each prime is  $\displaystyle \frac{1}{p-1}$ . I would end that sentence with an exclamation point, but I have gotten in trouble before with the whole excitement vs. factorial issue.

Well what can you do with these exponents? Back to the divisor function. Since we’re picking over all numbers, we can treat the exponents of prime terms like independent variables (if we were picking under say 100, an exponent of 2 like 6 would certainly affect the possibilities for 3). That means that expected values are preserved over addition and multiplication. So we can plug our expected exponents into the divisor formula to get an expected value for the number of divisors:

$\displaystyle \prod_{p \in \mathbb{P}} \left(1 + \frac{1}{p-1} \right)$

Wouldn’t it be a kick if it were convergent? Well, it’s divergent. In fact, the expanded sum contains every term of the form  $\displaystyle \frac{1}{p-1}$  , which is certainly greater than the sum of all terms of the form  $\displaystyle \frac{1}{p}$  , which by no proof of my own is divergent. So this is most definitely divergent.

So what does that get us? The average value of the divisor function grows with , very slowly but without limit. And it is related to a very pretty product over primes. If you have some insight on the topic, leave a reply!

1. fgsfds said:

The final product over the primes diverges because it is equal to the Euler Product for the Riemann zeta function for s=1. But I assume you know this.

• Peter said:

Ah, good point! I actually did not realize this, but yes, it is. So then this expression is also the sum of the harmonic series. Very interesting.