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Program to check if N is a Centered Triangular Number

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Given an integer N, the task is to check if it is a Centered triangular number or not.
 

Centered triangular number is a centered polygonal number that represents a triangle with a dot in the centre and all other dots surrounding the centre in successive triangular layers . The first few Centered triangular numbers are 1, 4, 10, 19, 31, 46, 64, 85, 109, 136, …


Examples: 
 

Input: N = 4 
Output: Yes
Input: 20 
Output: No 
 


 


Approach: 
 

  • The Kth Centered triangular number can be expressed as: 
    K^{th} Term = \frac{3*K^{2} + 3*K + 2}{2}
     
  • In order to check if the given number N can be expressed as a Centered triangular number or not, we need to check if \frac{-3 + \sqrt{24*N - 15}}{6}     gives an integer or not. 
     


Below is the implementation of the above approach: 
 

C++

// C++ implementation to check
// whether a given number is a
// Centered triangular number or not
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the
// number is a Centered
// Triangular Number
bool isCenteredtriangular(int N)
{
    float K = (-3
               + sqrt(24 * N - 15))
              / 6;
 
    // Condition for K to be
    // an integer
    return (K - (int)K) == 0;
}
 
// Driver Code
int main()
{
    int N = 85;
    if (isCenteredtriangular(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

                    

Java

// Java implementation to check that
// a number is a centered triangular
// number or not
import java.lang.Math;
 
class GFG{
     
// Function to check that the number
// is a centered triangular number
public static boolean isCenteredTriangular(int N)
{
    double K = (-3 + Math.sqrt(24 * N - 15)) / 6;
 
    // Condition to check if the number
    // is a centered triangular number
    return (K - (int)K) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 85;
 
    // Function call
    if (isCenteredTriangular(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by ShubhamCoder

                    

Python3

# Python3 implementation to check
# whether a given number is a
# Centered triangular number or not
import math
 
# Function to check if the
# number is a Centered
# Triangular Number
def isCenteredtriangular(N):
     
    K = (-3 + math.sqrt(24 * N - 15)) / 6
     
    # Condition for K to be
    # an integer
    if (K - int(K)) == 0:
        return True
         
    return False
 
# Driver Code
 
# Given Number
N = 85
 
# Function call
if (isCenteredtriangular(N)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by shubhamsingh10

                    

C#

// C# implementation to check whether
// a given number is a centered
// triangular number or not
using System;
 
class GFG{
 
// Function to check if the number
// is a centered triangular number
static bool isCenteredtriangular(int N)
{
    double K = (-3 + Math.Sqrt(24 * N - 15)) / 6;
     
    // Condition for K to be
    // an integer
    return (K - (int)K) == 0;
}
     
// Driver Code
static public void Main ()
{
    int N = 85;
    if (isCenteredtriangular(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by shubhamsingh10

                    

Javascript

<script>
 
// JavaScript implementation to check
// whether a given number is a
// Centered triangular number or not
 
// Function to check if the
// number is a Centered
// Triangular Number
function isCenteredtriangular(N)
{
    var K = (-3
               + Math.sqrt(24 * N - 15))
              / 6;
 
    // Condition for K to be
    // an integer
    return (K - parseInt(K)) == 0;
}
 
// Driver Code
var N = 85;
if (isCenteredtriangular(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
</script>

                    

Output: 
Yes

 

Time Complexity: O(logN) because inbuilt sqrt function is being used

Auxiliary Space: O(1)



Last Updated : 19 Sep, 2022
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