# Check if a given number is Pronic | Efficient Approach

A pronic number is such a number which can be represented as a product of two consecutive positive integers. By multiplying these two consecutive positive integers, there can be formed a rectangle which is represented by the product or pronic number. So it is also known as Rectangular Number.

The first few Pronic numbers are:
0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462 . . . . . .

Pronic number is a number which is the product of two consecutive integers, that is, a number n is a product of x and (x+1). The task is to check if a given number is pronic or not.

Mathematical Representation:

```If x is a pronic number, then x=n(n+1) ∀ n∈N0
Where, N0={0, 1, 2, 3, 4, ....}, (A set of Naturral Numbers)
```

Examples:

```Input : 56
Output : YES
Explanation: 56 = 7 * 8 i.e 56 is a product
of two consecutive integers 7 and 8.

Input : 65
Output : NO
Explanation: 65 cannot be represented as a
product of any two consecutive integers.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We had previously discussed an approach to check if a number is pronic or not in this article using a loop. The time Complexity of the previous algorithm is comparatively very high and in terms of Big-O asymptotic notation, it is O(√n).
In this article, we are going to explain an efficient approach with time complexity of O(log(log n). The idea is to observe that if a number can be expressed as the product of two consecutive integers then the two integers will be close to the square of root of that number. A more proper observation will lead to the fact that a number N can be represented as product of two consecutive integers only if the product of floor(sqrt(N)) and floor(sqrt(N))+1 is equal to N.

Below is the step by step algorithm of above approach:

```Step 1: Evaluate the square root value of the given number.
Step 2: Calculate the floor value of that square root.
Step 3: Calculate the product of value calculated in step-2
and its next consecutive number.
Step 4: Check the product value in step-3 with the given number.
Step 4.1: If the condition satisfies,
then the number is a pronic number.
Step 4.2: Otherwise the number is not a pronic number.
```

Below is the implementation of above algorithm:

## C

 `// C/C++ program to check if a number is pronic or not ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// function to check Pronic Number ` `bool` `pronic_check(``int` `n) ` `{ ` `    ``int` `x = (``int``)(``sqrt``(n)); ` ` `  `    ``// Checking Pronic Number by  ` `    ``// multiplying consecutive numbers ` `    ``if` `(x*(x+1)==n)  ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver Code ` `int` `main(``void``) ` `{ ` `    ``int` `n = 56;     ` `    ``pronic_check(n) == ``true``? cout << ``"YES"` `:  ` `                             ``cout << ``"NO"``; ` `     `  `    ``return` `0; ` `} `

## Java

 `// Java program to check if a number is pronic or not ` ` `  `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.math.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to check Pronic Number ` `    ``static` `boolean` `pronic_check(``int` `n)  ` `    ``{ ` `        ``int` `x = (``int``)(Math.sqrt(n)); ` `     `  `        ``// Checking Pronic Number by  ` `        ``// multiplying consecutive numbers ` `        ``if` `(x * (x + ``1``) == n) ` `            ``return` `true``; ` `        ``else` `            ``return` `false``; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `n = ``56``;         ` `        ``if` `(pronic_check(n)==``true``) ` `            ``System.out.println(``"YES"``); ` `        ``else` `            ``System.out.println(``"NO"``); ` `    ``} ` `} `

## Python3

 `# Python program to check if a number is pronic or not ` ` `  `import` `math ` ` `  `# function to check Pronic Number ` `def` `pronic_check(n) : ` `    ``x ``=` `(``int``)(math.sqrt(n)) ` ` `  `    ``# Checking Pronic Number by multiplying  ` `    ``# consecutive numbers ` `    ``if` `(x``*``(x ``+` `1``)``=``=` `n): ` `        ``return` `True` `    ``else``: ` `        ``return` `False` ` `  `# Driver Code ` `n ``=` `56` ` `  `if` `(pronic_check(n)``=``=``True``): ` `    ``print``(``"YES"``) ` `else``: ` `    ``print``(``"NO"``) `

## C#

 `// C# program to check if a number is  ` `// pronic or not ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to check Pronic Number ` `    ``static` `bool` `pronic_check(``int` `n)  ` `    ``{ ` `        ``int` `x = (``int``)(Math.Sqrt(n)); ` `     `  `        ``// Checking Pronic Number by  ` `        ``// multiplying consecutive numbers ` `        ``if` `(x * (x + 1) == n) ` `            ``return` `true``; ` `        ``else` `            ``return` `false``; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `n = 56;  ` `         `  `        ``if` `(pronic_check(n)==``true``) ` `            ``Console.Write(``"YES"``); ` `        ``else` `            ``Console.Write(``"NO"``); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```YES
```

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