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Program to check if N is a Centered Pentagonal Number or not
  • Last Updated : 05 Apr, 2021

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

Centered Pentagonal Number is a centered figurate number that represents a pentagon with a dot in the centre and other dots surrounding it in pentagonal layers successively. The first few Centered Pentagonal Number are 1, 6, 16, 31, 51, 76, 106 … 

Examples:  

Input: N = 6 
Output: Yes 
Explanation: 
Second Centered pentagonal number is 6.

Input: N = 20 
Output: No 



Approach: 
 

1. The Kth term of the Centered Pentagonal Number is given as
K^{th} Term = \frac{5*K^{2} - 5*K + 2}{2}
 

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

=> N = \frac{5*K^{2} - 5*K + 2}{2}
=> K = \frac{5 + \sqrt{40*N - 15}}{10}

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

4. Else the number N is not a Centered Pentagonal 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 number N
// is a Centered pentagonal number
bool isCenteredpentagonal(int N)
{
    float n
        = (5 + sqrt(40 * N - 15))
          / 10;
 
    // Condition to check if N is a
    // Centered pentagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 6;
 
    // Function call
    if (isCenteredpentagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if number N
// is a centered pentagonal number
static boolean isCenteredpentagonal(int N)
{
    float n = (float) ((5 + Math.sqrt(40 * N -
                                      15)) / 10);
 
    // Condition to check if N is a
    // centered pentagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number
    int N = 6;
 
    // Function call
    if (isCenteredpentagonal(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 program for the above approach
import numpy as np
 
# Function to check if number N
# is a centered pentagonal number
def isCenteredpentagonal(N):
 
    n = (5 + np.sqrt(40 * N - 15)) / 10
 
    # Condition to check if N is a
    # centered pentagonal number
    return (n - int(n)) == 0
 
# Driver Code
N = 6
 
# Function call
if (isCenteredpentagonal(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 number N
// is a centered pentagonal number
static bool isCenteredpentagonal(int N)
{
    float n = (float) ((5 + Math.Sqrt(40 * N -
                                      15)) / 10);
 
    // Condition to check if N is a
    // centered pentagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given number
    int N = 6;
 
    // Function call
    if (isCenteredpentagonal(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 number N
// is a centered pentagonal number
function isCenteredpentagonal(N)
{
    var n =  ((5 + Math.sqrt(40 * N -
                                      15)) / 10);
 
    // Condition to check if N is a
    // centered pentagonal number
    return (n - parseInt(n) == 0);
}
 
// Driver Code
//Given Number
var N = 6;
 
// Function call
if (isCenteredpentagonal(N))
{
    document.write("Yes");
}
else
{
    document.write("No");
}
 
// This code is contributed by Amit Katiyar
</script>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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