Program to check if N is a Decagonal Number

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

=> [Tex]N = 4*K^{2} – 3*K       [/Tex]
=> [Tex]K = \frac{3 + \sqrt{16*N + 9}}{8}       [/Tex]

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(logN) since sqrt function is being used

Auxiliary Space: O(1)