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Program to check if N is a Decagonal Number

  • Last Updated : 23 Jun, 2021

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

Decagonal Number is a figurate number that extends the concept of triangular and square numbers to the decagon (10-sided polygon). The nth decagonal numbers count the number of dots in a pattern of n nested decagons, all sharing a common corner, where the ith decagon in the pattern has sides made of i dots spaced one unit apart from each other. The first few decagonal numbers are 1, 10, 27, 52, 85, 126, 175, … 
 

Examples: 
 

Input: N = 10 
Output: Yes 
Explanation: 
Second decagonal number is 10.
Input: N = 30 
Output: No 
 

 



Approach: 
 

  1. The Kth term of the decagonal number is given as
    K^{th} Term = 4*K^{2} - 3*K
     
  2. As we have to check that the given number can be expressed as a Decagonal Number or not. This can be checked as: 
     

=> N = 4*K^{2} - 3*K
=> K = \frac{3 + \sqrt{16*N + 9}}{8}
 

  1.  
  2. If the value of K calculated using the above formula is an integer, then N is a Decagonal Number.
  3. Else N is not a Decagonal 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 N is a
// Decagonal Number
bool isdecagonal(int N)
{
    float n
        = (3 + sqrt(16 * N + 9))
          / 8;
 
    // Condition to check if the
    // number is a decagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 10;
 
    // Function call
    if (isdecagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java program for the above approach
import java.lang.Math;
 
class GFG{
     
// Function to check if N is a
// decagonal number
public static boolean isdecagonal(int N)
{
    double n = (3 + Math.sqrt(16 * N + 9)) / 8;
     
    // Condition to check if the
    // number is a decagonal number
    return (n - (int)n) == 0;
}
 
// Driver code   
public static void main(String[] args)
{
         
    // Given number
    int N = 10;
     
    // Function call
    if (isdecagonal(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by divyeshrabadiya07   

Python3




# Python3 program for the above approach
import math
 
# Function to check if N is a
# decagonal number
def isdecagonal(N):
 
    n = (3 + math.sqrt(16 * N + 9)) / 8
     
    # Condition to check if the
    # number is a decagonal number
    return (n - int(n)) == 0
     
# Driver Code
if __name__=='__main__':
     
    # Given number
    N = 10
     
    # Function Call
    if isdecagonal(N):
        print('Yes')
    else:
        print('No')
 
# This code is contributed by rutvik_56

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if N
// is a decagonal Number
static bool isdecagonal(int N)
{
    double n = (3 + Math.Sqrt(16 * N + 9)) / 8;
     
    // Condition to check if the
    // number is a decagonal number
    return (n - (int)n) == 0;
}
     
// Driver Code
static public void Main ()
{
     
    // Given Number
    int N = 10;
     
    // Function call
    if (isdecagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by ShubhamCoder

Javascript




<script>
 
// javascript program for the above approach
 
 
// Function to check if N is a
// Decagonal Number
function isdecagonal( N)
{
    let n
        = (3 + Math.sqrt(16 * N + 9))
          / 8;
 
    // Condition to check if the
    // number is a decagonal number
    return (n - parseInt(n)) == 0;
}
 
 
// Driver Code
 
    // Given Number
    let N = 10;
 
    // Function Call
    if (isdecagonal(N)) {
        document.write( "Yes");
    }
    else {
        document.write( "No");
    }
 
// This code contributed by gauravrajput1
 
</script>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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