Program to check if N is a Icositrigonal number

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

Icositrigonal number is a class of figurate number. It has 23 – sided polygon called Icositrigon. The N-th Icositrigonal number count’s the 23 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icositrigonol numbers are 1, 23, 66, 130, 215, 321, 448 …

Examples:

Input: N = 23 
Output: Yes 
Explanation:
Second icositrigonal number is 23.

Input: N = 30 
Output: No 

Approach:

1. The K^th term of the icositrigonal number is given as

2. As we have to check whether the given number can be expressed as an icositrigonal number or not. This can be checked as follows – 

=> 
=> 

3. Finally, check the value computed using this formula is an integer, which means that N is an icositrigonal number.

Below is the implementation of the above approach:

C++

// C++ implementation to check that
// a number is a icositrigonal number or not

#include <bits/stdc++.h>

using namespace std;

// Function to check that the
// number is a icositrigonal number
bool isicositrigonal(int N)
{
    float n
        = (19 + sqrt(168 * N + 361))
          / 42;

    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}

// Driver Code
int main()
{
    int i = 23;

    // Function call
    if (isicositrigonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation to check that
// a number is a icositrigonal number or not
import java.util.*;
class GFG{

// Function to check that the
// number is a icositrigonal number
static boolean isicositrigonal(int N)
{
    float n = (float)(19 + Math.sqrt(168 * N + 361)) / 42;

    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}

// Driver Code
public static void main(String args[])
{
    int i = 23;

    // Function call
    if (isicositrigonal(i))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}

// This code is contributed by Akanksha_Rai

Python3

# Python3 implementation to check that a
# number is a icositrigonal number or not
import math

# Function to check that the number
# is a icositrigonal number
def isicositrigonal(N):

    n = (19 + math.sqrt(168 * N + 361)) / 42

    # Condition to check if the number
    # is a icositrigonal number
    return (n - int(n)) == 0

# Driver Code
i = 23

# Function call
if (isicositrigonal(i)):
    print("Yes")
else:
    print("No")

# This code is contributed by divyamohan123

C#

// C# implementation to check that
// a number is a icositrigonal number or not
using System;
class GFG{

// Function to check that the
// number is a icositrigonal number
static bool isicositrigonal(int N)
{
    float n = (float)(19 + Math.Sqrt(168 * N + 361)) / 42;

    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}

// Driver Code
public static void Main()
{
    int i = 23;

    // Function call
    if (isicositrigonal(i))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}

// This code is contributed by Nidhi_Biet

Javascript

<script>

// JavaScript implementation to check that
// a number is a icositrigonal number or not

// Function to check that the
// number is a icositrigonal number
function isicositrigonal(N)
{
    var n
        = (19 + Math.sqrt(168 * N + 361))
          / 42;

    // Condition to check if the
    // number is a icositrigonal number
    return (n - parseInt(n)) == 0;
}

// Driver Code
var i = 23;
// Function call
if (isicositrigonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}

</script>

Output

Yes

Time Complexity: O(log N) because sqrt() function is being used
Auxiliary Space: O(1)