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# Program to check if N is a Octagonal Number

• Last Updated : 23 Jun, 2021

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

Octagonal Number is the figure number that represent octagonal. Octagonal Numbers can be formed by placing triangular numbers on the four sides of a square. The first few Octagonal Numbers are 1, 8, 21, 40, 65, 96 …

Examples:

Input: N = 8
Output: Yes
Explanation:
Second octagonal number is 8.
Input: N = 30
Output: No

Approach:

1. The Kth term of the octagonal number is given as

2. As we have to check that the given number can be expressed as a Octagonal Number or not. This can be checked as:

=>
=>

1.
2. If the value of K calculated using the above formula is an integer, then N is a Octagonal Number.
3. Else N is not a Octagonal Number.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Function to check if N is a// Octagonal Numberbool isoctagonal(int N){    float n        = (2 + sqrt(12 * N + 4))          / 6;     // Condition to check if the    // number is a octagonal number    return (n - (int)n) == 0;} // Driver Codeint main(){    // Given Number    int N = 8;     // Function call    if (isoctagonal(N)) {        cout << "Yes";    }    else {        cout << "No";    }    return 0;}

## Java

 // Java program for the above approachimport java.io.*;import java.util.*; class GFG {     // Function to check if N is a// octagonal numberpublic static boolean isoctagonal(int N){    double n = (2 + Math.sqrt(12 * N + 4)) / 6;         // Condition to check if the    // number is a octagonal number    return (n - (int)n) == 0;}     // Driver codepublic static void main(String[] args){         // Given Number    int N = 8;         // Function call    if (isoctagonal(N))    {        System.out.println("Yes");    }    else    {        System.out.println("No");    }}} // This code is contributed by coder001

## Python3

 # Python3 program for the above approachfrom math import sqrt # Function to check if N is a# octagonal numberdef isoctagonal(N):     n = (2 + sqrt(12 * N + 4)) / 6;     # Condition to check if the    # number is a octagonal number    return (n - int(n)) == 0;     # Driver Codeif __name__ == "__main__":     # Given number    N = 8;     # Function call    if (isoctagonal(N)):        print("Yes");         else:        print("No"); # This code is contributed by AnkitRai01

## C#

 // C# program for the above approachusing System; class GFG {     // Function to check if N is a// octagonal numberpublic static bool isoctagonal(int N){    double n = (2 + Math.Sqrt(12 * N +                              4)) / 6;         // Condition to check if the    // number is a octagonal number    return (n - (int)n) == 0;}     // Driver codepublic static void Main(String[] args){         // Given number    int N = 8;         // Function call    if (isoctagonal(N))    {        Console.WriteLine("Yes");    }    else    {        Console.WriteLine("No");    }}} // This code is contributed by Rohit_ranjan

## Javascript

 
Output:
Yes

Time Complexity: O(1)

Auxiliary Space: O(1)

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