Pattern 17

#include <stdio.h>

int main()
{
  int n = 5;
  int i,j;

  for(i = 1; i <= n; i++)
  {
    for (j = 1; j <= n; j++)
    {
      printf("%2d ", (2 * (i + j)) - 3);
    }
    printf("\n");
  }
  return 0;
}

#include <iostream.h>

int main()
{
  int n = 5;
  

  for(int i = 1; i <= n; i++)
  {
    for(int j = 1; j <= n; j++)
    {
     cout<<(2 * (i + j)) - 3 <<" ";
    }
    cout<<endl;
  }
  return 0;
}

  class PatternProg
  {
  public static void main(String args[])
  {
    int n = 5;


    for (int i = 1; i <= n; i++)
    {
      for (int j = 1; j <= n; j++)
      {
        System.out.printf("%2d ",(2 * (i + j)) - 3);
      }
      System.out.println();
    }

  }
}

using System;

class PatternProg
{
  public static void Main()
  {
    int n = 5;


    for (int i = 1; i <= n; i++)
    {
      for (int j = 1; j <= n; j++)
      {
        Console.Write("{0,2:D} ", (2 * (i + j)) - 3);
      }
      Console.WriteLine();
    }

    Console.ReadKey(true);
  }
}

n = 5

for x in range(1, n + 1):
  for y in range(1, n + 1):
      print("{:2d} ".format((2 * (x + y)) - 3), end="")
  print()