Given a positive integer, count the number of its positive divisors.



Task
Show the result for the first   100   positive integers.



Related task
  Tau number


import std.stdio;
 
// See https://en.wikipedia.org/wiki/Divisor_function
uint divisor_count(uint n) {
    uint total = 1;
    // Deal with powers of 2 first
    for (; (n & 1) == 0; n >>= 1) {
        ++total;
    }
    // Odd prime factors up to the square root
    for (uint p = 3; p * p <= n; p += 2) {
        uint count = 1;
        for (; n % p == 0; n /= p) {
            ++count;
        }
        total *= count;
    }
    // If n > 1 then it's prime
    if (n > 1) {
        total *= 2;
    }
    return total;
}
 
void main() {
    immutable limit = 100;
    writeln("Count of divisors for the first ", limit, " positive integers:");
    for (uint n = 1; n <= limit; ++n) {
        writef("%3d", divisor_count(n));
        if (n % 20 == 0) {
            writeln;
        }
    }
}