Icosihexagonal Number

Given a number N, the task is to find N^th Icosihexagon number.

An Icosihexagon number is class of figurate number. It has 26 – sided polygon called Icosihexagon. The N-th Icosihexagonal number count’s the 26 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Icosihexagonol numbers are 1, 26, 75, 148 …

Examples:

Input: N = 2 
Output: 26 
Explanation:
The second Icosihexagonol number is 26. 
Input: N = 3 
Output: 75 

Approach: The N-th Icosihexagonal number is given by the formula:

• Nth term of s sided polygon = 
  
• Therefore Nth term of 26 sided polygon is
  

• 
  

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Finding the nth Icosihexagonal Number
int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}

// Driver Code
int main()
{
    int n = 3;
    cout << "3rd Icosihexagonal Number is = "
         << IcosihexagonalNum(n);

    return 0;
}

// This code is contributed by Code_Mech

C

// C program for above approach
#include <stdio.h>
#include <stdlib.h>

// Finding the nth Icosihexagonal Number
int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}

// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd Icosihexagonal Number is = %d",
           IcosihexagonalNum(n));

    return 0;
}

Java

// Java program for above approach
class GFG{

// Finding the nth icosihexagonal number
public static int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}

// Driver code
public static void main(String[] args)
{
    int n = 3;

    System.out.println("3rd Icosihexagonal Number is = " +
                                    IcosihexagonalNum(n));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for above approach

# Finding the nth Icosihexagonal Number
def IcosihexagonalNum(n):

    return (24 * n * n - 22 * n) // 2

# Driver Code
n = 3
print("3rd Icosihexagonal Number is = ",
                   IcosihexagonalNum(n))

# This code is contributed by divyamohan123

C#

// C# program for above approach
using System;

class GFG{

// Finding the nth icosihexagonal number
public static int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}

// Driver code
public static void Main(String[] args)
{
    int n = 3;

    Console.WriteLine("3rd Icosihexagonal Number is = " +
                                   IcosihexagonalNum(n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// javascript program for above approach


// Finding the nth Icosihexagonal Number
function IcosihexagonalNum( n)
{
    return (24 * n * n - 22 * n) / 2;
}

// Driver code
let n = 3;
document.write("3rd Icosihexagonal Number is " + IcosihexagonalNum(n));

// This code contributed by gauravrajput1

</script>

Output:

3rd Icosihexagonal Number is = 75

Time Complexity: O(1)

Auxiliary Space: O(1)

Reference: https://en.wikipedia.org/wiki/Icosihexagon