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n’th Pentagonal Number

Given an integer n, find the nth Pentagonal number. The first three pentagonal numbers are 1, 5, and 12 (Please see the below diagram).
The n’th pentagonal number Pn is the number of distinct dots in a pattern of dots consisting of the outlines of regular pentagons with sides up to n dots when the pentagons are overlaid so that they share one vertex [Source Wiki]
Examples :

```Input: n = 1
Output: 1

Input: n = 2
Output: 5

Input: n = 3
Output: 12```
Recommended Practice

In general, a polygonal number (triangular number, square number, etc) is a number represented as dots or pebbles arranged in the shape of a regular polygon. The first few pentagonal numbers are: 1, 5, 12, etc.
If s is the number of sides in a polygon, the formula for the nth s-gonal number P (s, n) is

```nth s-gonal number P(s, n) = (s - 2)n(n-1)/2 + n

If we put s = 5, we get

n'th Pentagonal number Pn = 3*n*(n-1)/2 + n```

Examples:

Pentagonal Number

Below are the implementations of the above idea in different programming languages.

C++

 `// C++ program for above approach` `#include` `using` `namespace` `std;`   `// Finding the nth pentagonal number` `int` `pentagonalNum(``int` `n)` `{` `    ``return` `(3 * n * n - n) / 2;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 10;` `    `  `    ``cout << ``"10th Pentagonal Number is = "` `         ``<< pentagonalNum(n);`   `    ``return` `0;` `}`   `// This code is contributed by Code_Mech`

C

 `// C program for above approach` `#include ` `#include `   `// Finding the nth Pentagonal Number` `int` `pentagonalNum(``int` `n)` `{` `    ``return` `(3*n*n - n)/2;` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``int` `n = 10;` `    ``printf``(``"10th Pentagonal Number is = %d \n \n"``, ` `                             ``pentagonalNum(n));`   `    ``return` `0;` `}`

Java

 `// Java program for above approach` `class` `Pentagonal` `{` `    ``int` `pentagonalNum(``int` `n)` `    ``{` `        ``return` `(``3``*n*n - n)/``2``;` `    ``}` `}`   `public` `class` `GeeksCode` `{` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``Pentagonal obj = ``new` `Pentagonal();` `        ``int` `n = ``10``;    ` `        ``System.out.printf(``"10th petagonal number is = "` `                          ``+ obj.pentagonalNum(n));` `    ``}` `}`

Python3

 `# Python program for finding pentagonal numbers` `def` `pentagonalNum( n ):` `    ``return` `(``3``*``n``*``n ``-` `n)``/``2` `#Script Begins`   `n ``=` `10` `print` `(``"10th Pentagonal Number is = "``, pentagonalNum(n))` ` `  `#Scripts Ends`

C#

 `// C# program for above approach` `using` `System;`   `class` `GFG {` `    `  `    ``static` `int` `pentagonalNum(``int` `n)` `    ``{` `        ``return` `(3 * n * n - n) / 2;` `    ``}`   `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 10; ` `        `  `        ``Console.WriteLine(``"10th petagonal"` `        ``+ ``" number is = "` `+ pentagonalNum(n));` `    ``}` `}`   `// This code is contributed by vt_m.`

PHP

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Javascript

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Output

`10th Pentagonal Number is = 145`

Time Complexity: O(1) // since no loop or recursion is used the algorithm takes up constant time to perform the operations
Auxiliary Space: O(1) // since no extra array or data structure is used so the space taken by the algorithm is constant

Another Approach:

The formula indicates that the n-th pentagonal number depends quadratically on n. Therefore, try to find the positive integral root of N = P(n) equation.
P(n) = nth pentagonal number
N = Given Number
Solve for n:
P(n) = N
or (3*n*n – n)/2 = N
or 3*n*n – n – 2*N = 0 … (i)
The positive root of equation (i)
n = (1 + sqrt(24N+1))/6
After obtaining n, check if it is an integer or not. n is an integer if n – floor(n) is 0.

C++

 `// C++ Program to check a` `// pentagonal number` `#include ` `using` `namespace` `std;`   `// Function to determine if` `// N is pentagonal or not.` `bool` `isPentagonal(``int` `N)` `{    ` `    ``// Get positive root of` `    ``// equation P(n) = N.` `    ``float` `n = (1 + ``sqrt``(24*N + 1))/6;` `    `  `    ``// Check if n is an integral` `    ``// value of not. To get the` `    ``// floor of n, type cast to int.` `    ``return` `(n - (``int``) n) == 0;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 145;    ` `    ``if` `(isPentagonal(N)) ` `        ``cout << N << ``" is pentagonal "` `<< endl;    ` `    ``else` `        ``cout << N << ``" is not pentagonal"` `<< endl;    ` `    ``return` `0;` `}`

Java

 `// Java Program to check a` `// pentagonal number` `import` `java.util.*;`   `public` `class` `Main {`   `    ``// Function to determine if` `    ``// N is pentagonal or not.` `    ``public` `static` `boolean` `isPentagonal(``int` `N)` `    ``{` `        ``// Get positive root of` `        ``// equation P(n) = N.` `        ``float` `n = (``1` `+ (``float``)Math.sqrt(``24` `* N + ``1``)) / ``6``;`   `        ``// Check if n is an integral` `        ``// value of not. To get the` `        ``// floor of n, type cast to int.` `        ``return` `(n - (``int``)n) == ``0``;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int` `N = ``145``;` `        ``if` `(isPentagonal(N))` `            ``System.out.println(N + ``" is pentagonal "``);` `        ``else` `            ``System.out.println(N + ``" is not pentagonal"``);` `    ``}` `}`

Python3

 `import` `math`   `# Function to determine if N is pentagonal or not.` `def` `isPentagonal(N):` `    ``# Get positive root of equation P(n) = N` `    ``n ``=` `(``1` `+` `math.sqrt(``24``*``N ``+` `1``))``/``6` `    `  `    ``# Check if n is an integral value or not.` `    ``# To get the floor of n, use the int() function` `    ``return` `(n ``-` `int``(n)) ``=``=` `0`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    ``N ``=` `145` `    `  `    ``if` `isPentagonal(N):` `        ``print``(N, ``"is pentagonal"``)` `    ``else``:` `        ``print``(N, ``"is not pentagonal"``)`

C#

 `using` `System;`   `public` `class` `Program` `{`   `  ``// Function to determine if` `  ``// N is pentagonal or not.` `  ``public` `static` `bool` `IsPentagonal(``int` `n)` `  ``{` `    ``// Get positive root of equation P(n) = N.` `    ``float` `x = (1 + MathF.Sqrt(24 * n + 1)) / 6;`   `    ``// Check if x is an integral value or not` `    ``return` `MathF.Floor(x) == x;` `  ``}`   `  ``public` `static` `void` `Main()` `  ``{` `    ``int` `n = 145;` `    ``if` `(IsPentagonal(n))` `      ``Console.WriteLine(n + ``" is pentagonal"``);` `    ``else` `      ``Console.WriteLine(n + ``" is not pentagonal"``);`   `    ``// Pause the console so that we can see the output` `    ``Console.ReadLine();` `  ``}` `}` `// This code is contributed by divyansh2212`

Javascript

 `// js equivalent `   `// import math functions` `function` `isPentagonal(N) {` `    ``// Get positive root of equation P(n) = N` `    ``let n = (1 + Math.sqrt(24 * N + 1)) / 6;` `    `  `    ``// Check if n is an integral value or not.` `    ``// To get the floor of n, use the Math.floor() function` `    ``return` `(n - Math.floor(n)) === 0;` `}`   `// Driver code ` `let N = 145;`   `if` `(isPentagonal(N)) {` `    ``console.log(`\${N} is pentagonal`);` `} ``else` `{` `    ``console.log(`\${N} is not pentagonal`);` `}`

Output

`145 is pentagonal `

Time Complexity: O(log n) //the inbuilt sqrt function takes logarithmic time to execute
Auxiliary Space: O(1) // since no extra array or data structure is used so the space taken by the algorithm is constant

Reference:
https://en.wikipedia.org/wiki/Polygonal_number
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