Program to check if N is a Icosihexagonal Number

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

Icosihexagonal 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 = 26
Output: Yes
Explanation:
Second icosihexagonal number is 26.

Input: 30
Output: No



Approach:

  1. The Kth term of the icosihexagonal number is given as
    K^{th} Term =  \frac{24*K^{2} - 22*K}{2}
  2. As we have to check that the given number can be expressed as a icosihexagonal number or not. This can be checked as follows –

    => N =  \frac{24*K^{2} - 22*K}{2}
    => K = \frac{22 + \sqrt{192*N + 484}}{48}

  3. Finally, check the value of computed using this formulae is an integer, which means that N is a icosihexagonal number.

Below is the implementation of the above approach:

C++

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// C++ program to check whether
// a number is a icosihexagonal number 
// or not
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to check whether the
// number is a icosihexagonal number
bool isicosihexagonal(int N)
{
    float n
        = (22 + sqrt(192 * N + 484))
          / 48;
  
    // Condition to check if the
    // number is a icosihexagonal number
    return (n - (int)n) == 0;
}
  
// Driver code
int main()
{
    int i = 26;
  
    if (isicosihexagonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

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Java

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// Java program to check whether the
// number is a icosihexagonal number
// or not
class GFG{ 
  
// Function to check whether the
// number is a icosihexagonal number
static boolean isicosihexagonal(int N) 
    float n = (float) ((22 + Math.sqrt(192 * N +
                                       484)) / 48);
      
    // Condition to check if the number 
    // is a icosihexagonal number 
    return (n - (int)n) == 0
  
// Driver Code 
public static void main(String[] args) 
      
    // Given number 
    int N = 26
      
    // Function call 
    if (isicosihexagonal(N)) 
    
        System.out.print("Yes"); 
    
    else
    
        System.out.print("No"); 
    
  
// This code is contributed by shubham

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Python3

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# Python3 program to check whether 
# a number is a icosihexagonal number 
# or not 
import numpy as np
  
# Function to check whether the 
# number is a icosihexagonal number 
def isicosihexagonal(N):
  
    n = (22 + np.sqrt(192 * N + 484)) / 48
  
    # Condition to check if the 
    # number is a icosihexagonal number 
    return (n - (int(n))) == 0
  
# Driver code 
i = 26
  
if (isicosihexagonal(i)): 
    print ("Yes")
else
    print ("No")
  
# This code is contributed by PratikBasu

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C#

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// C# program to check whether the
// number is a icosihexagonal number
// or not
using System;
class GFG{ 
  
// Function to check whether the
// number is a icosihexagonal number
static bool isicosihexagonal(int N)
    float n = (float)((22 + Math.Sqrt(192 * N +
                                      484)) / 48);
      
    // Condition to check if the number 
    // is a icosihexagonal number 
    return (n - (int)n) == 0; 
  
// Driver Code 
public static void Main() 
      
    // Given number 
    int N = 26; 
      
    // Function call 
    if (isicosihexagonal(N)) 
    
        Console.Write("Yes"); 
    
    else
    
        Console.Write("No"); 
    
  
// This code is contributed by Code_Mech

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Output:

Yes

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