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Enneacontagon Number

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  • Last Updated : 22 Jun, 2021

Given a number N, the task is to find Nth Enneacontagon number.
 

A Enneacontagon number. is class of figurate number. It has 90 – sided polygon called enneacontagon. The N-th enneacontagon number count’s the 90 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few enneacontagonol numbers are 1, 90, 267, 532 … 
 

Examples: 
 

Input: N = 2 
Output: 90 
Explanation: 
The second enneacontagonol number is 90. 
Input: N = 3 
Output: 267 
 

 

Approach: The N-th enneacontagon 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 90 sided polygon is
     

Tn =\frac{((90-2)n^2 - (90-4)n)}{2} =\frac{(88n^2 - 86)}{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 enneacontagon Number
int enneacontagonNum(int n)
{
    return (88 * n * n - 86 * n) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
    cout << "3rd enneacontagon Number is = "
         << enneacontagonNum(n);
 
    return 0;
}
 
// This code is contributed by Akanksha_Rai

C




// C program for above approach
#include <stdio.h>
#include <stdlib.h>
 
// Finding the nth enneacontagon Number
int enneacontagonNum(int n)
{
    return (88 * n * n - 86 * n) / 2;
}
 
// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd enneacontagon Number is = %d",
           enneacontagonNum(n));
 
    return 0;
}

Java




// Java program for above approach
class GFG{
 
// Finding the nth enneacontagon number
public static int enneacontagonNum(int n)
{
    return (88 * n * n - 86 * n) / 2;
}
 
// Driver code   
public static void main(String[] args)
{
    int n = 3;
 
    System.out.println("3rd enneacontagon Number is = " +
                                    enneacontagonNum(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 program for above approach
 
# Finding the nth enneacontagon Number
def enneacontagonNum(n):
 
    return (88 * n * n - 86 * n) // 2
 
# Driver Code
n = 3
print("3rd enneacontagon Number is = ",
                   enneacontagonNum(n))
 
# This code is contributed by divyamohan123

C#




// C# program for above approach
using System;
class GFG{
 
// Finding the nth enneacontagon number
public static int enneacontagonNum(int n)
{
    return (88 * n * n - 86 * n) / 2;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3;
 
    Console.WriteLine("3rd enneacontagon Number is = " +
                                   enneacontagonNum(n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript program for above approach
 
// Finding the nth enneacontagon Number
function enneacontagonNum(n)
{
    return (88 * n * n - 86 * n) / 2;
}
 
// Driver Code
var n = 3;
document.write("3rd enneacontagon Number is = " + enneacontagonNum(n));
 
</script>
Output: 
3rd enneacontagon Number is = 267

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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

 


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