Question
Consider the fraction, \frac{n}{d}, where n and d are positive integers. If n \lt d and \mathrm{HCF}(n,d)=1, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d \leq 8 in ascending order of size, we get:
\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}
It can be seen that there are 21 elements in this set.
How many elements would be contained in the set of reduced proper fractions for d \leq 1,000,000?
Haskell
farey :: Int -> Int
farey = (map f [0..] !!) where
f n = (n*(n + 3)) `quot` 2 - sum [farey (n `quot` k) | k <- [2..n]]
main :: IO ()
main = print $ farey 1000000 - farey 1$ ghc -O2 -o counting-fractions counting-fractions.hs
$ time ./counting-fractions
real 0m0.372s
user 0m0.364s
sys 0m0.008sPython
#!/usr/bin/env python
import math
from operator import mul
from functools import reduce
def prime_factors(n):
res = set()
# iterate over all even numbers first.
while n % 2 == 0:
res.add(2)
n //= 2
# try odd numbers up to sqrt(n)
limit = math.sqrt(n+1)
i = 3
while i <= limit:
if n % i == 0:
res.add(i)
n //= i
limit = math.sqrt(n+i)
else:
i += 2
if n != 1:
res.add(n)
return res
def totient(n):
if n == 1: return 1
return int(round(n * reduce(mul, [1 - 1.0 / p for p in prime_factors(n)])))
def farey_length(n):
return sum(totient(m) for m in range(1, n+1)) - 1
def main():
print(farey_length(1000000))
if __name__ == "__main__": main()$ time python3 farey.py
real 0m13.880s
user 0m13.879s
sys 0m0.000s