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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)

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