Answer L7

#include <stdio.h>
#include <stdlib.h>

/* Puzzle L07 -- print the first N odd integers. Ask the user for N. */
int main()
{
  int j, N;
  printf("Enter n: ");
  scanf("%d", &N );
 
  for (j=0; j < N; j++ )
  {
    printf("%3d\n", 2*j+1 );
  }

  return 0;
}

Comment: The loop is a counting loop. It counts from zero to N-1. The counts, j, are transformed into odd numbers by using 2*j+1. The value 2*j is always even, of course. Adding 1 to that gives an odd number.

To check that this works, look at what happens when j==0. This gives 2*0+1 == 1, the first odd number. Now you need to worry that successive odd numbers are calculated, and that the last one calculated is the N'th odd number. Practice worrying about this.

It is good to practice worrying. As a programmer, you will worry about fussy details a lot.