Program to check if N is an Icosikaioctagonal Number

Given an integer N, the task is to check if it is a icosikaioctagonal number or not.

An icosikaioctagonal number is class of figurate number. It has 28 – sided polygon called icosikaioctagon. The N-th icosikaioctagonal number count’s the 28 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few icosikaioctagonol numbers are 1, 28, 81, 160 ... 

Examples:  

Input: N = 28 
Output: Yes 
Explanation: 
Second icosikaioctagonal number is 28.

Input: 30 
Output: No 

Approach: 
 

1. The K^th term of the icosikaioctagonal number is given as:

K^{th} Term = \frac{26*K^{2} - 24*K}{2}  

2. As we have to check that the given number can be expressed as an icosikaioctagonal number or not. This can be checked as follows - 

=> N = \frac{26*K^{2} - 24*K}{2}       
=> K = \frac{24 + \sqrt{208*N + 576}}{52}  

3. Finally, check the value computed using these formulae is an integer, which means that N is an icosikaioctagonal number.

Below is the implementation of the above approach: 

// C++ program to check whether a
// number is an icosikaioctagonal 
// number or not

#include <bits/stdc++.h>

using namespace std;

// Function to check whether a number
// is an icosikaioctagonal number or not
bool isicosikaioctagonal(int N)
{
    float n
        = (24 + sqrt(208 * N + 576))
          / 52;

    // Condition to check if the
    // number is an 
    // icosikaioctagonal number
    return (n - (int)n) == 0;
}

// Driver code
int main()
{
    int i = 28;

    if (isicosikaioctagonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

// Java program to check whether a
// number is an icosikaioctagonal 
// number or not
class GFG{ 

// Function to check whether a
// number is an icosikaioctagonal 
// number or not
static boolean isicosikaioctagonal(int N) 
{ 
    float n = (float) ((24 + Math.sqrt(208 * N + 
                                       576)) / 52);
    
    // Condition to check whether a
    // number is an icosikaioctagonal 
    // number or not
    return (n - (int)n) == 0; 
} 

// Driver Code 
public static void main(String[] args) 
{ 
    
    // Given number 
    int N = 28; 
    
    // Function call 
    if (isicosikaioctagonal(N)) 
    { 
        System.out.print("Yes"); 
    } 
    else
    { 
        System.out.print("No"); 
    } 
} 
} 

// This code is contributed by shubham

# Python3 program to check whether a
# number is an icosikaioctagonal 
# number or not
import math

# Function to check whether a number
# is an icosikaioctagonal number or not
def isicosikaioctagonal(N):

    n = (24 + math.sqrt(208 * N +
                        576)) // 52;

    # Condition to check if the
    # number is an 
    # icosikaioctagonal number
    return (n - int(n)) == 0;

# Driver code
i = 28;

if (isicosikaioctagonal(i)):
    print("Yes");
else:
    print("No");

# This code is contributed by Code_Mech

// C# program to check whether a
// number is an icosikaioctagonal 
// number or not
using System;
class GFG{ 

// Function to check whether a
// number is an icosikaioctagonal 
// number or not
static bool isicosikaioctagonal(int N) 
{ 
    float n = (float)((24 + Math.Sqrt(208 * N + 
                                      576)) / 52);
    
    // Condition to check whether a
    // number is an icosikaioctagonal 
    // number or not
    return (n - (int)n) == 0; 
} 

// Driver Code 
public static void Main() 
{ 
    
    // Given number 
    int N = 28; 
    
    // Function call 
    if (isicosikaioctagonal(N)) 
    { 
        Console.Write("Yes"); 
    } 
    else
    { 
        Console.Write("No"); 
    } 
} 
} 

// This code is contributed by Code_Mech

<script>
// Javascript program to check whether a
// number is an icosikaioctagonal 
// number or not

// Function to check whether a number
// is an icosikaioctagonal number or not
function isicosikaioctagonal(N)
{
    let n
        = (24 + Math.sqrt(208 * N + 576))
          / 52;

    // Condition to check if the
    // number is an 
    // icosikaioctagonal number
    return (n - parseInt(n)) == 0;
}

// Driver code
let i = 28;

if (isicosikaioctagonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}

// This code is contributed by rishavmahato348.
</script>

Yes

Time Complexity: O(logN) for given N, as it is using inbuilt sqrt function
Auxiliary Space: O(1)

spp____

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Article Tags :

• Mathematical
• DSA

Practice Tags :

• Mathematical