Even number pattern 2 in C

Write a C program to print the given even number pattern using loop. How to print the given even number pattern using for loop in C programming. Logic to print the given number pattern in C program.

Required knowledge

Basic C programming , Loop

Logic to print the given even number pattern

To get the logic of the pattern have a close look on to the pattern. You can observe following things about the pattern.

  1. The pattern consists of total N rows (where N is the total number of rows to be printed).
  2. To make things little easier let's divide the pattern in two internal parts.
  3. Loop formation to print the first part of the pattern will be for(j=2; j<=i*2; i+=2).
  4. Loop formation to print the second part of the pattern will be for(j=(i-1)*2; j>=2; j-=2).

Let's combine the logic of both parts in single program.

Program to print the given even number pattern

/**
 * C program to print the given number pattern
 */

#include <stdio.h>

int main()
{
    int i, j, N;

    printf("Enter rows: ");
    scanf("%d", &N);

    for(i=1; i<=N; i++)
    {
        // Prints first part of the pattern
        for(j=2; j<=i*2; j+=2)
        {
            printf("%d", j);
        }

        // Prints second part of the pattern
        for(j=(i-1)*2; j>=2; j-=2)
        {
            printf("%d", j);
        }

        printf("\n");
    }

    return 0;
}

Output

Enter rows: 5
2
242
24642
2468642
2468108642

Once you get the logic of above pattern you can easily print the second pattern. You just need to add extra space before printing the numbers.

Program to print the above pattern

/**
 * C program to print the given number pattern
 */

#include <stdio.h>

int main()
{
    int i, j, N;

    printf("Enter rows: ");
    scanf("%d", &N);

    for(i=1; i<=N; i++)
    {
        // Prints space
        for(j=i; j<N; j++)
        {
            printf(" ");
        }

        // Prints first part of the pattern
        for(j=2; j<=i*2; j+=2)
        {
            printf("%d", j);
        }

        // Prints second part of the pattern
        for(j=(i-1)*2; j>=2; j-=2)
        {
            printf("%d", j);
        }

        printf("\n");
    }

    return 0;
}

Happy coding 😉

About Pankaj

Pankaj Prakash is the founder, editor and blogger at Codeforwin. He loves to learn new techs and write programming articles especially for beginners. He works at Vasudhaika Software Sols as a Software Design Engineer and manages Codeforwin. In short Pankaj is Web developer, Blogger, Learner, Tech and Music lover.

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