Skip to content
Related Articles

Related Articles

Improve Article
Program to check if N is a Heptadecagonal Number
  • Last Updated : 08 Apr, 2021

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

Heptadecagonal Number is class of figurate number. It has 17-sided polygon called heptadecagon. The N-th heptadecagonal number counts the seventeen number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few heptadecagonal numbers are 1, 17, 48, 94, 155, 231… 

Examples:  

Input: N = 17 
Output: Yes 
Explanation: 
Second heptadecagonal number is 17.

Input: N = 30 
Output: No 



Approach:  

1. The Kth term of the heptadecagonal number is given as
K^{th} Term = \frac{15*K^{2} - 13*K}{2}
 

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

=> N = \frac{15*K^{2} - 13*K}{2}
=> K = \frac{13 + \sqrt{120*N + 169}}{30}

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

4. Else N is not a Heptadecagonal 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 heptadecagonal number
bool isheptadecagonal(int N)
{
    float n
        = (13 + sqrt(120 * N + 169))
          / 30;
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 17;
 
    // Function call
    if (isheptadecagonal(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 heptadecagonal number
static boolean isheptadecagonal(int N)
{
    float n = (float) ((13 + Math.sqrt(120 * N +
                                       169)) / 30);
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number
    int N = 17;
 
    // Function call
    if (isheptadecagonal(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 heptadecagonal number
def isheptadecagonal(N):
 
    n = (13 + np.sqrt(120 * N + 169)) / 30
 
    # Condition to check if number N
    # is a heptadecagonal number
    return (n - int(n)) == 0
 
# Driver Code
N = 17
 
# Function call
if (isheptadecagonal(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 heptadecagonal number
static bool isheptadecagonal(int N)
{
    float n = (float) ((13 + Math.Sqrt(120 * N +
                                       169)) / 30);
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given Number
    int N = 17;
 
    // Function call
    if (isheptadecagonal(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 heptadecagonal number
function isheptadecagonal(N)
{
    let n
        = (13 + Math.sqrt(120 * N + 169))
          / 30;
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
// Given Number
let N = 17;
 
// Function call
if (isheptadecagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by subham348.
</script>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes




My Personal Notes arrow_drop_up
Recommended Articles
Page :