See

written in C
my task was to make a script that reads [m;n]
and then does
m,m+1,m+2,...,m+n;
and prints only the numbers that have a residue equal to 3 or 6 after diving the m`s by 7


#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main ()
{
int array[100],array3[100];
int n,i,y,m;
y=1;
printf("Type m\t");
    scanf("%d",&m);
printf("Type n\t");
    scanf("%d",&n);
for (i=0;i<=n;i++)
 {
    array[i]=m+i;
 }
for (i=0;i<=n;i++)
    {
        if(3==array[i]%7)
        {
            array[i]=array3[y];
            y++;
            }
            else
            if (6==array[i]%7)
            {
            array[i]=array3[y];
            y++;
            }
        }
 for (y=1;y<=n;y++)      
 printf("%d\n",array3[y]);

    return 0;
}

#include <sdtio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
    int n, m;
   
    if(argc != 3)
        exit(1);

    n = atoi(argv[1]);
    m = n + atoi(argv[2]);

    n -= (n%7 <= 3)? n%7 : printf("%d", n+=6-n%7), -1;
    while(n+6 <= m) {
        printf("%d\n%d\n", n+3, n+6);
        n += 7;
    }
    if(n+3 <= m)
        printf("%d\n", n+3);
}
[spoiler]HAX[spoiler]

>>7
Replace
n -= (n%7 <= 3)? n%7 : printf("%d", n+=6-n%7), -1;
with
n -= (n%7 <= 3)? n%7 : (printf("%d\n", n+=6-n%7), -1);