668 Square root smooth numbers.sf

#!/usr/bin/ruby

# Daniel "Trizen" Șuteu
# Date: 07 June 2019
# https://github.com/trizen

# https://projecteuler.net/problem=668

# Runtime: 1.535s (previously: 2.402s)

# Formula:
#   R(n) = Sum_{sqrt(n) < p <= n} floor(n/p)
#
#   a(n) = n - R(n) - Sum_{p <= sqrt(n)} (p-1) - pi(sqrt(n))
#        = n - R(n) - Sum_{p <= sqrt(n)} p
#
# where p runs over the prime numbers.

# The interesting part is computing R(n) efficiently.

# See also:
#   https://oeis.org/A064775 -- Card{ k<=n, k such that all prime divisors of k are <= sqrt(k) }.

func R(n) {

    var p = next_prime(n.isqrt)
    var t = idiv(n,p)

    var sum = 0

    while (p <= n) {

        var u = idiv(n,p)

        if (u == t) {
            var q = next_prime(idiv(n,u))
            sum += u*prime_count(p, q-1)
            u = idiv(n,q)
            p = q
        }

        sum += u

        t = u
        p = p.next_prime
    }

    sum
}

func square_root_smooth_count (n) {
    n - n.isqrt.primes_sum - R(n)
}

say square_root_smooth_count(10_000_000_000)