# Program to check if N is a Icosagonal Number

• Last Updated : 22 Apr, 2021

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 Kth 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 using namespace std; // Function to check if the number// N is a icosagonal numberbool 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 Codeint main(){    // Given Number    int N = 20;     // Function call    if (iicosagonal(N)) {        cout << "Yes";    }    else {        cout << "No";    }    return 0;}

## Java

 // Java program for the above approachimport java.util.*; class GFG{ // Function to check if the number// N is a icosagonal numberstatic 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 Codepublic 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 approachimport numpy as np # Function to check if the number# N is a icosagonal numberdef 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 CodeN = 20 # Function callif (iicosagonal(N)):    print ("Yes")else:    print ("No") # This code is contributed by PratikBasu

## C#

 // C# program for the above approachusing System; class GFG{ // Function to check if the number// N is a icosagonal numberstatic 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 Codepublic 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

 

Output:

Yes

Time Complexity: O(1)

Auxiliary Space: O(1)

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