# Find nth Term of the Series 1 2 2 4 4 4 4 8 8 8 8 8 8 8 8 …

Given a number n, the task is to find the nth term of the Series

1 2 2 4 4 4 4 8 8 8 8 8 8 8 8 …

Example:

```Input: n = 9
Output: 8

Input: n = 1025
Output: 1024
```

Naive Approach:

1. Run a loop from i = 0 to n
2. Inside loop increment i by i+k
3. Inside loop increment k by 2*k
4. Run above loop while i is less than n
5. Return k/2 as the result

Complexity: log(n)

Below is the implementation of the above approach:

## C++

 `// CPP Program to find Nth term ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function that will return nth term ` `int` `getValue(``int` `n) ` `{ ` `    ``int` `i = 0, k = 1; ` ` `  `    ``while` `(i < n) { ` `        ``i = i + k; ` `        ``k = k * 2; ` `    ``} ` ` `  `    ``return` `k / 2; ` `} ` ` `  `// Driver Code ` `int` `main(``void``) ` `{ ` ` `  `    ``// Get n ` `    ``int` `n = 9; ` ` `  `    ``// Get the value ` `    ``cout << getValue(n) << endl; ` ` `  `    ``// Get n ` `    ``n = 1025; ` ` `  `    ``// Get the value ` `    ``cout << getValue(n) << endl; ` `} `

## Java

 `// Java Program to find Nth term ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function that will return nth term ` `    ``static` `int` `getValue(``int` `n) ` `    ``{ ` `        ``int` `i = ``0``, k = ``1``; ` `     `  `        ``while` `(i < n) { ` `            ``i = i + k; ` `            ``k = k * ``2``; ` `        ``} ` `     `  `        ``return` `k / ``2``; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `     `  `        ``// Get n ` `        ``int` `n = ``9``; ` `     `  `        ``// Get the value ` `        ``System.out.println(getValue(n)); ` `         `  `        ``// Get n ` `        ``n = ``1025``; ` `     `  `        ``// Get the value ` `        ``System.out.println(getValue(n)); ` `    ``} ` `} `

## Python3

 `# Python3 Program to find Nth term  ` ` `  `# Function that will return nth term  ` `def` `getValue(n):  ` ` `  `    ``i ``=` `0``;  ` `    ``k ``=` `1``;  ` ` `  `    ``while` `(i < n): ` `        ``i ``=` `i ``+` `k;  ` `        ``k ``=` `k ``*` `2``;  ` ` `  `    ``return` `int``(k ``/` `2``);  ` ` `  `# Driver Code  ` ` `  `# Get n  ` `n ``=` `9``;  ` ` `  `# Get the value  ` `print``(getValue(n));  ` ` `  `# Get n  ` `n ``=` `1025``;  ` ` `  `# Get the value  ` `print``(getValue(n));  ` ` `  `# This code is contributed by mits `

## C#

 `// C# Program to find Nth term ` ` `  `using` `System; ` `class` `GFG ` `{ ` `     `  `    ``// Function that will return nth term ` `    ``static` `int` `getValue(``int` `n) ` `    ``{ ` `        ``int` `i = 0, k = 1; ` `     `  `        ``while` `(i < n) { ` `            ``i = i + k; ` `            ``k = k * 2; ` `        ``} ` `     `  `        ``return` `k / 2; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `     `  `        ``// Get n ` `        ``int` `n = 9; ` `     `  `        ``// Get the value ` `        ``Console.WriteLine(getValue(n)); ` `         `  `        ``// Get n ` `        ``n = 1025; ` `     `  `        ``// Get the value ` `        ``Console.WriteLine(getValue(n)); ` `    ``} ` `} `

## PHP

 ` `

Output:

```8
1024
```

Efficient Approach: This Problem can be solved in O(1) time complexity.
Let nth term of the sequence be equal to 2m Below is the implementation of the above approach:

## C++

 `// CPP Program to find Nth term ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function that will return nth term ` `int` `getValue(``int` `n) ` `{ ` `    ``// Find log of n+1 on base 2 ` `    ``int` `result = (``floor``)(``log``(n + 1) / ``log``(2)); ` ` `  `    ``return` `pow``(2, result); ` `} ` ` `  `// Driver Code ` `int` `main(``void``) ` `{ ` `    ``// Get n ` `    ``int` `n = 9; ` ` `  `    ``// Get the value ` `    ``cout << getValue(n) << endl; ` ` `  `    ``// Get n ` `    ``n = 1025; ` ` `  `    ``// Get the value ` `    ``cout << getValue(n) << endl; ` `}  `

## Java

 `// Java Program to find Nth term ` ` `  `import` `java.lang.*; ` `import` `java.lang.Math; ` `import` `java.io.*; ` ` `  `class` `GFG { ` `    ``// Function that will return nth term ` `static` `double` `getValue(``double` `n) ` `{ ` `    ``// Find log of n+1 on base 2 ` ` ``double` `result =(Math.floor(Math.log(n + ``1``) / Math.log(``2``))); ` ` `  `    ``return` `Math.pow(``2``, result); ` `} ` ` `  `// Driver Code ` `    ``public` `static` `void` `main (String[] args) { ` `    ``// Get n ` `    ``double` `n = ``9``; ` `    ``// Get the value ` `    ``System.out.println (getValue(n)); ` `    ``// Get n ` `    ``n = ``1025``; ` `    ``// Get the value ` `        ``System.out.println (getValue(n)); ` `    ``} ` `} `

## Python3

 `# Python3 Program to find Nth term ` `import` `math ` ` `  `# Function that will return nth term ` `def` `getValue(n): ` ` `  `    ``# Find log of n+1 on base 2 ` `    ``result ``=` `int``(math.floor(math.log(n ``+` `1``) ``/`  `                            ``math.log(``2``))) ` `    ``return` `int``(math.``pow``(``2``, result)) ` ` `  `# Driver code ` `n ``=` `9` `print``(getValue(n)) ` ` `  `n ``=` `1025` `print``(getValue(n)) ` ` `  `# This code is contributed  ` `# by Shrikant13  `

## C#

 `// C# Program to find Nth term ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function that will return nth term ` `    ``static` `double` `getValue(``double` `n) ` `    ``{ ` `        ``// Find log of n+1 on base 2 ` `        ``double` `result =(Math.Floor(Math.Log(n + 1) / Math.Log(2))); ` `        ``return` `Math.Pow(2, result); ` `    ``} ` ` `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``// Get n ` `        ``double` `n = 9; ` `         `  `        ``// Get the value ` `        ``Console.WriteLine(getValue(n)); ` `         `  `        ``// Get n ` `        ``n = 1025; ` `         `  `        ``// Get the value ` `        ``Console.WriteLine (getValue(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by SoM15242 `

## PHP

 ` `

Output:

```8
1024
```

My Personal Notes arrow_drop_up Let the code do the talking

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