# Leyland Number

In number theory, a Leyland number is a number of the form xy + yx, where x and y are integers greater than 1 and 1 <y <= x.
Given a positive integer N. The task is to print first N Leyland number in ascending order. The first few Leyland numbers are 8, 17, 32, 54, 57, 100, …

Examples:

Input : N = 1
Output : 8
22 + 22 = 4 + 4 = 8.

Input : N = 6
Output : 100

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea to run two loop, one for x and other for y. The outer loop start with 2 to n and for each iteration of outer loop, run inner loop start from 2 t x. And store xy + yx in an array. After calculating all the value sort them and print first n numbers.

Below is the implementation of this approach:

## C++

 // CPP program to print first N Leyland Numbers. #include #define MAX 100 using namespace std;    // Print first n Leyland Number. void leyland(int n) {     vector ans;        // Outer loop for x from 2 to n.     for (int x = 2; x <= n; x++) {            // Inner loop for y from 2 to x.         for (int y = 2; y <= x; y++) {                // Calculating x^y + y^x             int temp = pow(x, y) + pow(y, x);                ans.push_back(temp);         }     }        // Sorting the all Leyland Number.     sort(ans.begin(), ans.end());        // Printing first n Leyland number.     for (int i = 0; i < n; i++)         cout << ans[i] << " "; }    // Driven Program int main() {     int n = 6;     leyland(n);     return 0; }

## Java

 // Java program to print first N  // Leyland Numbers. import java.util.*; import java.lang.*;    public class GFG{           private static final int MAX = 0;           // Print first n Leyland Number.     public static void leyland(int n)     {         List ans = new ArrayList();                           // Outer loop for x from 2 to n.         for (int x = 2; x <= n; x++) {                    // Inner loop for y from 2 to x.             for (int y = 2; y <= x; y++) {                        // Calculating x^y + y^x                 int temp = (int)Math.pow(x, y) +                             (int)Math.pow(y, x);                        ans.add(temp);             }         }                // Sorting the all Leyland Number.         Collections.sort(ans);                // Printing first n Leyland number.         for (int i = 0; i < n; i++)             System.out.print(ans.get(i) + " ");     }            // Driven Program     public static void main(String args[])     {         int n = 6;         leyland(n);     } }    // This code is contributed by Sachin Bisht

## Python3

 # Python3 program to print first N  # Leyland Numbers. import math    # Print first n Leyland Number. def leyland(n):     ans = []     x = 2     y = 2        # Outer loop for x from 2 to n.     while x <= n :            # Inner loop for y from 2 to x.         y = 2         while y <= x :                # Calculating x^y + y^x             temp = pow(x, y) + pow(y, x)                ans.append(temp);             y = y + 1         x = x + 1        # Sorting the all Leyland Number.     ans.sort();        i = 0        # Printing first n Leyland number.     while i < n :         print(ans[i], end = " ")         i = i + 1    # Driver Code n = 6 leyland(n)    # This code is contributed by rishabh_jain

## C#

 // C# program to print  // first N Leyland Numbers. using System; using System.Collections;    class GFG {            // Print first n      // Leyland Number.     public static void leyland(int n)     {         ArrayList ans = new ArrayList();                // Outer loop for x         // from 2 to n.         for (int x = 2; x <= n; x++)          {                    // Inner loop for             // y from 2 to x.             for (int y = 2; y <= x; y++)              {                        // Calculating x^y + y^x                 int temp = (int)Math.Pow(x, y) +                             (int)Math.Pow(y, x);                        ans.Add(temp);             }         }                // Sorting the all          // Leyland Number.         ans.Sort();                // Printing first         // n Leyland number.         for (int i = 0 ; i < n; i++)         {             Console.Write(ans[i] + " ");         }     }            // Driver Code     public static void Main()     {         int n = 6;         leyland(n);     } }    // This code is contributed by Sam007

## PHP



Output:

8 17 32 54 57 100

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