The length of a phone number varies by country, but US a phone number is a 10 digit number, and 10-digit numbers are often divisible by three different prime numbers, give or take a couple. Assuming phone numbers are scattered among possible 10-digit numbers in a way that doesn’t bias their number of prime factors, these numbers will often have between 1 and 5 prime factors. If a country has 9- or 11-digit phone numbers, the result is essentially the same.

Let ω(*n*) be the number of distinct prime factors of *n*. Then the Erdős–Kac theorem says roughly that ω(*n*) is distributed like a normal random variable with mean and variance log log *n*. More precisely, fix two numbers *a* and *b*. For a given value of *x*, count the proportion of positive integers less than *x* where

(ω(*n*) – log log *n*) / sqrt( log log *n*)

is between *a* and *b*. Then in the limit as *x* goes to infinity, this proportion approaches the probability that a standard normal random variable is between *a* and *b*.

So, by that heuristic, the number of distinct prime factors of a 10-digit number is approximately normally distributed with mean and variance log log 10^11 = 3.232, and such a distribution is between 1 and 5 around 73% of the time.

My business phone number, for example, is 8324228646. Obviously this is divisible by 2. In fact it equals 2 × 3^{2} × 462457147, and so it has exactly three distinct prime factors: 2, 3, and 462457147.

Here’s how you could play with this using Python.

```
from sympy.ntheory import factorint
def omega(n):
return len(factorint(n))
```

I looked in SymPy and didn’t see an implementation of ω(*n*) directly, but it does have a function factorint that returns the prime factors of a number, along with their multiplicities, in a dictionary. So ω(*n*) is just the size of that dictionary.

I took the first 20 phone numbers in my contacts and ran them through `omega`

and got results consistent with what you’d expect from the theory above. One was prime, and none had more than five factors.

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