Program to check if N is a Icositetragonal number

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

icositetragonal number: 
s a class of figurate number. It has a 24-sided polygon called Icositetragon. The N-th Icositetragonal number count’s the number of dots and all others dots are surrounding with a common sharing corner and make a pattern 
The first few icositetragonal numbers are 1, 24, 69, 136, 225, 336, …

Examples:

Input: N = 24 
Output: Yes 
Explanation:
Second icositetragonal number is 24.

Input: N = 30 
Output: No 

Approach:

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

2. As we have to check that the given number can be expressed as a icositetragonal number or not. This can be checked as follows – 

=> 
=> 

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

Below is the implementation of the above approach:

C++

// C++ implementation to check that
// a number is icositetragonal number or not

#include <bits/stdc++.h>

using namespace std;

// Function to check that the
// number is a icositetragonal number
bool isicositetragonal(int N)
{
    float n
        = (10 + sqrt(44 * N + 100))
          / 22;

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

// Driver Code
int main()
{
    int i = 24;

    // Function call
    if (isicositetragonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation to check that
// a number is icositetragonal number or not
class GFG{

// Function to check that the
// number is a icositetragonal number
static boolean isicositetragonal(int N)
{
    float n = (float)((10 + Math.sqrt(44 * N +
                                      100)) / 22);

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

// Driver Code
public static void main(String[] args)
{
    int i = 24;

    // Function call
    if (isicositetragonal(i))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to check that
# a number is icositetragonal number
# or not
import math

# Function to check that the number
# is a icositetragonal number
def isicositetragonal(N):

    n = (10 + math.sqrt(44 * N + 100)) / 22

    # Condition to check if the number
    # is a icositetragonal number
    return (n - int(n)) == 0

# Driver Code
i = 24

# Function call
if (isicositetragonal(i)):
    print("Yes")
else:
    print("No")

# This code is contributed by divyamohan123

C#

// C# implementation to check that
// a number is icositetragonal number or not
using System;
class GFG{

// Function to check that the
// number is a icositetragonal number
static bool isicositetragonal(int N)
{
    float n = (float)((10 + Math.Sqrt(44 * N +
                                      100)) / 22);

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

// Driver Code
public static void Main()
{
    int i = 24;

    // Function call
    if (isicositetragonal(i))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}

// This code is contributed by Akanksha_Rai

Javascript

<script>

// JavaScript implementation to check that
// a number is icositetragonal number or not

// Function to check that the
// number is a icositetragonal number
function isicositetragonal(N)
{
    var n = (10 + Math.sqrt(44 * N + 100))
          / 22;

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

// Driver Code
var i = 24;
// Function call
if (isicositetragonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}

</script>

Output

Yes

Time Complexity: O(log N), it is time taken by inbuilt sqrt() function
Auxiliary Space: O(1)