Calculate the sequence where each term an is the smallest natural number greater than the previous term, that has exactly n divisors.



Task
Show here, on this page, at least the first 15 terms of the sequence.



See also
 OEIS:A069654


Related tasks
 Sequence: smallest number with exactly n divisors
 Sequence: nth number with exactly n divisors‎‎


#include <stdio.h>
 
#define MAX 15
 
int count_divisors(int n) {
    int i, count = 0;
    for (i = 1; i * i <= n; ++i) {
        if (!(n % i)) {
            if (i == n / i)
                count++;
            else
                count += 2;
        }
    }
    return count;
}
 
int main() {
    int i, next = 1;
    printf("The first %d terms of the sequence are:\n", MAX);
    for (i = 1; next <= MAX; ++i) {
        if (next == count_divisors(i)) {           
            printf("%d ", i);
            next++;
        }
    }
    printf("\n");
    return 0;
}