# Riemann's Prime Power Counting Function

### We look at this mathematical equation and how to model it using the Python and Mathematica languages. Read on to learn more!

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Join For FreeThe prime number theorem says that π(*x*), the number of primes less than or equal to *x*, is asymptotically *x*/log *x*. So it's easy to estimate the number of primes below some number *N*. But what if we want to estimate the number of prime *powers* less than *N*? This is a question that comes up in finite fields, for example, since there is a finite field with elements if and only if *n* is a prime power. It's also important in finite simple groups because these groups are often indexed by prime powers.

**Riemann's prime-power counting function** Π(*x*) counts the number of prime powers less than or equal to *x*. Clearly Π(*x*) > π( *x*) for *x* ≥ 4 since every prime is a prime power, and 4 is a prime power but not a prime. Is Π(*x*) much bigger than π(*x*)? What is its limiting distribution, i.e. what is the analog of the prime number theorem for prime powers?

## Numerical Examples

It turns out Π(*x*) equals π(*x*) asymptotically. That is, even though Π(*x*) is always bigger than π(*x*), their ratio converges to 1 as *x* increases.

Why is that? Let's first look at *N* = 1,000,000. The number of primes less than one million happens to be 78,498. The number of prime powers less than *N* is 78,734. So the latter includes only 236 prime powers with exponent greater than 1.

If we increase *N* to 1,000,000,000, there are 50,847,534 primes less than *N* and 50,851,223 prime powers, a difference of 3,689. Said another way, 99.99% of the prime powers less than a billion have exponent 1.

## Equation for Π(*x*)

The number of prime powers less than *N* with exponent 2 equals the number of primes less than the square root of *N*. And the number of prime powers less than *N* with exponent 3 equals the number of primes less than the cube root of *N*. The number of prime powers with exponent 4 equals the number of primes less than the fourth root of *N*. Etcetera.

Even if *N* is large, these counts start getting small pretty soon. How soon? We're taking roots of order *r* until the *r*th root of *N *is less than 2, because then there are no more primes less than that root. That means we keep going until *r* > log _{2}*N*. And so we have the following equation:

## Mathematica and Python Code

I looked in Mathematica and SymPy for a function to compute Π(*x*) and didn't see one. Maybe I missed something. But in either case it's easy to implement our own using the equation above.

In Mathematica:

`pp[n_] := Sum[PrimePi[n^(1/r)], {r, 1, Log2[n]}]`

In Python:

```
from sympy import primepi
from math import log2
def pp(n):
top = int(log2(n))
return sum(
[primepi(n**(1/r)) for r in range(1, 1+top)]
)
```

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