# Stella Octangula Number

Given a number n, check it is the Stella Octangula number or not. A number of the form where n is a whole number(0, 1, 2, 3, 4, …) is called Stella Octangula. First few Stella Octangula numbers are 0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, …

Stella octangula numbers which are perfect squares are 1 and 9653449.

Given a number x, check if it is Stella octangula.

Examples:

```Input: x = 51
Output: Yes
For n = 3, the value of expression
n(2n2 - 1) is 51

Input: n = 53
Output: No
```

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

A simple solution is to run a loop starting from n = 0. For every n, check if n(2n2 – 1) is equal to x. We run the loop while value of n(2n2 – 1) is smaller than or equal to x.

An efficient solution is to use Unbounded Binary Search. We first find a value of n such that n(2n2 – 1) is greater than x using repeated doubling. Then we apply Binary Search.

## C++

 `// Program to check if a number is Stella ` `// Octangula Number ` `#include ` `using` `namespace` `std; ` ` `  `// Returns value of n*(2*n*n - 1) ` `int` `f(``int` `n) { ` `   ``return` `n*(2*n*n - 1); ` `} ` ` `  `// Finds if a value of f(n) is equl to x ` `// where n is in interval [low..high] ` `bool` `binarySearch(``int` `low, ``int` `high, ``int` `x) ` `{ ` `    ``while` `(low <= high) { ` `        ``long` `long` `mid = (low + high) / 2; ` ` `  `        ``if` `(f(mid) < x) ` `            ``low = mid + 1; ` `        ``else` `if` `(f(mid) > x) ` `            ``high = mid - 1; ` `        ``else` `            ``return` `true``; ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `// Returns true if x isStella Octangula Number. ` `// Else returns false. ` `bool` `isStellaOctangula(``int` `x) ` `{ ` `    ``if` `(x == 0) ` `      ``return` `true``; ` ` `  `    ``// Find 'high' for binary search by ` `    ``// repeated doubling ` `    ``int` `i = 1; ` `    ``while` `(f(i) < x) ` `        ``i = i*2; ` ` `  `    ``// If condition is satisfied for a ` `    ``// power of 2. ` `    ``if` `(f(i) == x) ` `        ``return` `true``; ` ` `  `    ``//  Call binary search ` `    ``return` `binarySearch(i/2, i, x); ` `} ` ` `  `// driver code ` `int` `main() ` `{ ` `    ``int` `n = 51; ` `    ``if` `(isStellaOctangula(n)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Program to check if  ` `// a number is Stella  ` `// Octangula Number ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Returns value of ` `// n*(2*n*n - 1) ` `static` `int` `f(``int` `n)  ` `{ ` `    ``return` `n * (``2` `* n *  ` `                ``n - ``1``); ` `} ` ` `  `// Finds if a value of  ` `// f(n) is equl to x  ` `// where n is in  ` `// interval [low..high] ` `static` `boolean` `binarySearch(``int` `low,  ` `                            ``int` `high,  ` `                            ``int` `x) ` `{ ` `    ``while` `(low <= high)  ` `    ``{ ` `        ``int` `mid = (low + high) / ``2``; ` ` `  `        ``if` `(f(mid) < x) ` `            ``low = mid + ``1``; ` `        ``else` `if` `(f(mid) > x) ` `            ``high = mid - ``1``; ` `        ``else` `            ``return` `true``; ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `// Returns true if x  ` `// is Stella Octangula  ` `// Number.Else returns  ` `// false. ` `static` `boolean` `isStellaOctangula(``int` `x) ` `{ ` `    ``if` `(x == ``0``) ` `    ``return` `true``; ` ` `  `    ``// Find 'high' for  ` `    ``// binary search by ` `    ``// repeated doubling ` `    ``int` `i = ``1``; ` `    ``while` `(f(i) < x) ` `        ``i = i * ``2``; ` ` `  `    ``// If condition is  ` `    ``// satisfied for a ` `    ``// power of 2. ` `    ``if` `(f(i) == x) ` `        ``return` `true``; ` ` `  `    ``// Call binary search ` `    ``return` `binarySearch(i / ``2``,  ` `                        ``i, x); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `n = ``51``; ` `    ``if` `(isStellaOctangula(n)) ` `        ``System.out.print(``"Yes"``); ` `    ``else` `        ``System.out.print(``"No"``); ` `} ` `} ` ` `  `// This code is contributed  ` `// by anuj_67. `

## Python3

 `# Python3 program to check if a number  ` `# is Stella octangula number ` ` `  `# Returns value of n*(2*n*n - 1) ` `def` `f(n): ` `     `  `    ``return` `n ``*` `(``2` `*` `n ``*` `n ``-` `1``); ` ` `  `# Finds if a value of f(n) is equl to x ` `# where n is in interval [low..high] ` `def` `binarySearch(low, high, x): ` ` `  `    ``while` `(low <``=` `high): ` `        ``mid ``=` `int``((low ``+` `high) ``/``/` `2``); ` ` `  `        ``if` `(f(mid) < x): ` `            ``low ``=` `mid ``+` `1``; ` `        ``elif` `(f(mid) > x): ` `            ``high ``=` `mid ``-` `1``; ` `        ``else``: ` `            ``return` `True``; ` `             `  `    ``return` `False``; ` ` `  `# Returns true if x isStella octangula ` `#  number. Else returns false. ` `def` `isStellaOctangula(x): ` ` `  `    ``if` `(x ``=``=` `0``): ` `        ``return` `True``; ` ` `  `    ``# Find 'high' for binary search  ` `    ``# by repeated doubling ` `    ``i ``=` `1``; ` `    ``while` `(f(i) < x): ` `        ``i ``=` `i ``*` `2``; ` ` `  `    ``# If condition is satisfied for a ` `    ``# power of 2. ` `    ``if` `(f(i) ``=``=` `x): ` `        ``return` `True``; ` ` `  `    ``# Call binary search ` `    ``return` `binarySearch(i ``/` `2``, i, x); ` ` `  `# Driver code ` `n ``=` `51``; ` ` `  `if` `(isStellaOctangula(n) ``=``=` `True``): ` `    ``print``(``"Yes"``); ` `else``: ` `    ``print``(``"No"``); ` ` `  `# This code is contributed by Code_Mech `

## C#

 `// Program to check if  ` `// a number is Stella  ` `// Octangula Number ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Returns value of ` `// n*(2*n*n - 1) ` `static` `int` `f(``int` `n)  ` `{ ` `    ``return` `n * (2 * n *  ` `                ``n - 1); ` `} ` ` `  `// Finds if a value of  ` `// f(n) is equl to x  ` `// where n is in  ` `// interval [low..high] ` `static` `bool` `binarySearch(``int` `low,  ` `                         ``int` `high,  ` `                         ``int` `x) ` `{ ` `    ``while` `(low <= high)  ` `    ``{ ` `        ``int` `mid = (low + high) / 2; ` ` `  `        ``if` `(f(mid) < x) ` `            ``low = mid + 1; ` `        ``else` `if` `(f(mid) > x) ` `            ``high = mid - 1; ` `        ``else` `            ``return` `true``; ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `// Returns true if x  ` `// is Stella Octangula  ` `// Number.Else returns  ` `// false. ` `static` `bool` `isStellaOctangula(``int` `x) ` `{ ` `    ``if` `(x == 0) ` `    ``return` `true``; ` ` `  `    ``// Find 'high' for  ` `    ``// binary search by ` `    ``// repeated doubling ` `    ``int` `i = 1; ` `    ``while` `(f(i) < x) ` `        ``i = i * 2; ` ` `  `    ``// If condition is  ` `    ``// satisfied for a ` `    ``// power of 2. ` `    ``if` `(f(i) == x) ` `        ``return` `true``; ` ` `  `    ``// Call binary search ` `    ``return` `binarySearch(i / 2,  ` `                        ``i, x); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `n = 51; ` `    ``if` `(isStellaOctangula(n)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` `} ` `} ` ` `  `// This code is contributed  ` `// by anuj_67. `

Output:

```Yes
```

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