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Program to check if N is a Icositrigonal number

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

Icositrigonal number is a class of figurate number. It has 23 – sided polygon called Icositrigon. The N-th Icositrigonal number count’s the 23 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icositrigonol numbers are 1, 23, 66, 130, 215, 321, 448 …

Examples: 

Input: N = 23 
Output: Yes 
Explanation: 
Second icositrigonal number is 23.

Input: N = 30 
Output: No 

Approach: 

1. The Kth term of the icositrigonal number is given as
K^{th} Term = \frac{21*K^{2} - 19*K}{2}

2. As we have to check whether the given number can be expressed as an icositrigonal number or not. This can be checked as follows – 

=> N = \frac{21*K^{2} - 19*K}{2}
=> K = \frac{19 + \sqrt{168*N + 361}}{42}
 

3. Finally, check the value computed using this formula is an integer, which means that N is an icositrigonal number.

Below is the implementation of the above approach:

C++

// C++ implementation to check that
// a number is a icositrigonal number or not
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to check that the
// number is a icositrigonal number
bool isicositrigonal(int N)
{
    float n
        = (19 + sqrt(168 * N + 361))
          / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

                    

Java

// Java implementation to check that
// a number is a icositrigonal number or not
import java.util.*;
class GFG{
 
// Function to check that the
// number is a icositrigonal number
static boolean isicositrigonal(int N)
{
    float n = (float)(19 + Math.sqrt(168 * N + 361)) / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String args[])
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Akanksha_Rai

                    

Python3

# Python3 implementation to check that a 
# number is a icositrigonal number or not
import math
 
# Function to check that the number
# is a icositrigonal number
def isicositrigonal(N):
 
    n = (19 + math.sqrt(168 * N + 361)) / 42
 
    # Condition to check if the number 
    # is a icositrigonal number
    return (n - int(n)) == 0
 
# Driver Code
i = 23
 
# Function call
if (isicositrigonal(i)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by divyamohan123

                    

C#

// C# implementation to check that
// a number is a icositrigonal number or not
using System;
class GFG{
 
// Function to check that the
// number is a icositrigonal number
static bool isicositrigonal(int N)
{
    float n = (float)(19 + Math.Sqrt(168 * N + 361)) / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Nidhi_Biet

                    

Javascript

<script>
 
// JavaScript implementation to check that
// a number is a icositrigonal number or not
 
// Function to check that the
// number is a icositrigonal number
function isicositrigonal(N)
{
    var n
        = (19 + Math.sqrt(168 * N + 361))
          / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
var i = 23;
// Function call
if (isicositrigonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
    
</script>

                    

Output
Yes

Time Complexity: O(log N) because sqrt() function is being used
Auxiliary Space: O(1)



Last Updated : 02 Dec, 2022
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