Program to check if N is a Icosidigonal Number

Given an integer N, the task is to check if it is a Icosidigonal Number or not. If the number N is an Icosidigonal Number then print “Yes” else print “No”.

Icosidigonal number: 
The polygon has many gons, depends on their gonal number series. In mathematics, there are a number of gonal numbers and the Icosidigonal Number is one of them and these numbers have 22 -sided polygon(icosidigon). An Icosidigonal Number belong to the class of figurative number. They have one common dots points and other dots pattern is arranged in an n-th nested Icosidigon pattern. 
The first few Icosidigonal numbers are 1, 22, 63, 124, 205, 306…

Examples:

Input: N = 22 
Output: Yes 
Explanation:
Second Icosidigonal number is 22.
Input: 30 
Output: No  

Approach:

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

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

=> 
=> 

3. If the value of K calculated using the above formula is an integer, then N is an Icosidigonal Number.

4. Else N is not an Icosidigonal Number.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the number N
// is a Icosidigonal number
bool isIcosidigonal(int N)
{
    float n
        = (18 + sqrt(160 * N + 324))
          / 40;

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

// Driver Code
int main()
{
    // Given Number
    int N = 22;

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

Java

// Java program for the above approach
class GFG{

// Function to check if the number N
// is a icosidigonal number
static boolean isIcosidigonal(int N)
{
    float n = (float) ((18 + Math.sqrt(160 * N +
                                       324)) / 40);

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

// Driver Code
public static void main(String[] args)
{

    // Given number
    int N = 22;

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

// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
import numpy as np

# Function to check if the number N
# is a icosidigonal number
def isIcosidigonal(N):

    n = (18 + np.sqrt(160 * N + 324)) / 40

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

# Driver Code
N = 22

# Function call
if (isIcosidigonal(N)):
    print ("Yes")
else:
    print ("No")

# This code is contributed by PratikBasu

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if the number N
// is a icosidigonal number
static bool isIcosidigonal(int N)
{
    float n = (float) ((18 + Math.Sqrt(160 * N +
                                    324)) / 40);

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

// Driver Code
public static void Main(string[] args)
{

    // Given number
    int N = 22;

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

// This code is contributed by rutvik_56

Javascript

<script>

// javascript program for the above approach


// Function to check if the number N
// is a Icosidigonal number
function isIcosidigonal( N)
{
    let n
        = (18 + Math.sqrt(160 * N + 324))
          / 40;

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

// Driver Code

    // Given Number
    let N = 22;

    // Function call
    if (isIcosidigonal(N)) {
         document.write( "Yes");
    }
    else {
        document.write( "No");
    }


    // This code contributed by aashish1995

</script>

Output:

Yes

Time Complexity: O(logN) because the inbuilt sqrt function has been used
Auxiliary Space: O(1)