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Icosihexagonal Number

  • Last Updated : 22 Jun, 2021

Given a number N, the task is to find Nth 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 = \frac{((s-2)n^2 - (s-4)n)}{2}
     
  • Therefore Nth term of 26 sided polygon is
     

Tn =\frac{((26-2)n^2 - (26-4)n)}{2} =\frac{(24n^2 - 22n)}{2}

  •  

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

 

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