Program to check if N is a Icosagonal Number

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

Icosagonal Number is a twenty-sided polygon. The number derived from the figurative class. There are different patterns observed in this series. The dots are countable, arrange in a specific way of position, and create a diagram. All the dots have common dots points, all other dots are connected to these points and except this common point the dots connected to their ith dots with their respective successive layer… The first few Icosagonal numbers are 1, 20, 57, 112, 185, 276…

Examples:

Input: N = 20 
Output: Yes 
Explanation:
Second Icosagonal Number is 20.

Input: N = 30 
Output: No 

Approach:

1. The K^th term of the Icosagonal Number is given as

2. As we have to check that the given number can be expressed as a icosagonal 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 Icosagonal Number.

4. Else the number N is not an Icosagonal 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 icosagonal number
bool iicosagonal(int N)
{
    float n
        = (16 + sqrt(144 * N + 256))
          / 36;

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

// Driver Code
int main()
{
    // Given Number
    int N = 20;

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

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to check if the number
// N is a icosagonal number
static boolean iicosagonal(int N)
{
    float n = (float)((16 + Math.sqrt(144 * N +
                                      256)) / 36);

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

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

    // Given Number
    int N = 20;

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

// This code is contributed by Rohit_ranjan

Python3

# Python3 program for the above approach
import numpy as np

# Function to check if the number
# N is a icosagonal number
def iicosagonal(N):

    n = (16 + np.sqrt(144 * N + 256)) / 36

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

# Driver Code
N = 20

# Function call
if (iicosagonal(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 icosagonal number
static bool iicosagonal(int N)
{
    float n = (float)((16 + Math.Sqrt(144 * N +
                                      256)) / 36);

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

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

    // Given Number
    int N = 20;

    // Function call
    if (iicosagonal(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 icosagonal number
function iicosagonal(N)
{
    var n
        = (16 + Math.sqrt(144 * N + 256))
          / 36;

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

// Driver Code

// Given Number
var N = 20;
// Function call
if (iicosagonal(N)) {
    document.write( "Yes");
}
else {
    document.write("No");
}


</script>

Output:

Yes

Time Complexity: O(logN) because inbuilt sqrt function is being used

Auxiliary Space: O(1)