Pattern 353

#include <stdio.h>

int main()
{
  int n = 5, x = 1;
  int i,j;
  for(i = 1; i <= n; i++)
  {
    for (j = 1; j <= i; j++)
    {
      if (j == 1 || j == i || i == n)
      {
        printf("%2d ", x++);
      }
      else
      {
        printf("   "); // 3spaces
      }
    }
    printf("\n");
  }
  return 0;
}

#include <iostream.h>

int main()
{
  int n = 5, x = 1;
  
  for(int i = 1; i <= n; i++)
  {
    for(int j = 1; j <= i; j++)
    {
      if (j == 1 || j == i || i == n)
      {
        cout<<x++<<" ";
      }
      else
      {
        cout<<"  "; // 2ws
      }
    }
    cout<<endl;
  }
  return 0;
}

class PatternProg
{
	public static void main(String args[])
	{
	  int n = 5;
	  int x = 1;

	  for (int i = 1; i <= n; i++)
	  {
		for (int j = 1; j <= i; j++)
		{
		  if (j == 1 || j == i || i == n)
		  {
			System.out.printf("%3d",x);
			x++;
		  }
		  else
		  {
			System.out.print("   "); //3ws
		  }
		}
		System.out.println();
	  }
	  
	}
}

using System;

class PatternProg
{
  public static void Main()
  {
    int n = 5;
    int x = 1;

    for (int i = 1; i <= n; i++)
    {
      for (int j = 1; j <= i; j++)
      {
        if (j == 1 || j == i || i == n)
        {
          Console.Write("{0,3:D}", x);
          x++;
        }
        else
        {
          Console.Write("   "); //3ws
        }
      }
      Console.WriteLine();
    }

    Console.ReadKey(true);

  }
}

n = 5
d = 1

for x in range(1, n + 1):
  for y in range(1, x + 1):
    if y == 1 or y == x or x == n:
       print("{:2d} ".format(d), end="")
       d += 1
    else:
       print("   ", end="")  # 3ws
  print()