## Counting Primes

### February 9, 2016

We have studied functions for counting the prime numbers less than a given number in two previous exercises. All of them were based on Legendre’s phi function that counts the numbers less than x not stricken by sieving with the first a primes.

Today we look at a rather different method of counting the primes that is due to G. H. Hardy and Edward M. Wright. Their method is based on factorials, and their formula is $\pi(n) = -1 + \sum_{j=3}^{n} \left( (j-2)! - j\lfloor \frac{(j-2)!}{j} \rfloor \right)$

for n > 3, where ⌊n⌋ is the greatest integer less than n. The expression inside the big parentheses is 1 when n is prime and 0 when n is composite, by Wilson’s Theorem, which states that an integer n > 1 is prime if and only if (n − 1)! ≡ −1 (mod n); the theorem was first stated by Ibn al-Haytham (c. 1000AD), but is named for John Wilson, who first published it in 1770, and it was first proved by Lagrange in 1771.

Your task is to write a program that counts primes by the Hardy-Wright method. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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### 7 Responses to “Counting Primes”

1. Very interesting algo, albeit somewhat impractical. It was a good refresher though for the BigInt type (Julia language) that makes it usable for inputs over 23.

2. Jan Van lent said
(use-package :iterate)

(defun prime-count (n)
(+ -1 (iter (for j from 3 to n)
(for f first 1 then (* f (- j 2)))
(summing (- f (* j (floor f j)))))))

3. Jan Van lent said

My previous comment has a solution in Common Lisp.

4. Mike said
from itertools import accumulate, count, islice
from operator import mul

def pi(n):
factorials = accumulate(count(1), mul)
summands = (f - (f//j)*j for j,f in enumerate(factorials, 3))
return sum(islice(summands, n-2)) - 1

5. DanN said
import functools

def fact(n):

return functools.reduce(lambda x, y: x * y, list(range(1, n + 1)))

def primeCounting(n):

return (-1 + sum([(fact(j - 2) - j * (fact(j - 2) // j)) for j in range(3, n + 1)]))

6. Jussi Piitulainen said

Two Scheme implementations, or one in two ways. I felt quite stupid for quite a while before I understood that I want to update the factorial with j-1 instead of j-2 when I also update j. (Now I don’t understand why the model implementation also works :)

(define (pi n) ;for n > 3
(do ((S 0 (+ S (- f (* j (quotient f j)))))
(f 1 (* (- j 1) f)) ; sic!
(j 3 (+ j 1)))
((< n j) (- S 1))))

(define (pi1 n) ;for n > 3
(let loop ((S 0) (f 1) (j 3))
(if (< n j)
(- S 1)
(loop (+ S (- f (* j (quotient f j))))
(* (- j 1) f) ; sic!
(+ j 1)))))

7. […] this weekend I was going through the Programming Praxis website, and this problem caught my […]