Given an integer **N**, the task is to find the count of natural octal numbers up to **N** digits.

Examples:Input:N = 1Output:7Explanation:

1, 2, 3, 4, 5, 6, 7 are 1 digit Natural Octal numbers.Input:N = 2Output:63Explanation:

There are a total of 56 two digit octal numbers and 7 one digit octal numbers. Therefore, 56 + 7 = 63.

**Approach:** On observing carefully, a geometric progression series is formed [ **7 56 448 3584 28672 229376…** ] with the first term being **7** and a common ratio of **8**.

Therefore,

Nth term = Number of Octal numbers of N digits = 7 * 8^{N - 1}

Finally, count of all octal numbers up to N digits can be found out by iterating a loop from 1 to N and calculating the sum of ith term using the above formula.

Below is the implementation of the above approach:

## C++

`// C++ program to find the count of` `// natural octal numbers upto N digits` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the count of` `// natural octal numbers upto N digits` `int` `count(` `int` `N)` `{` ` ` `int` `sum = 0;` ` ` `// Loop to iterate from 1 to N` ` ` `// and calculating number of` ` ` `// octal numbers for every 'i'th digit.` ` ` `for` `(` `int` `i = 1; i <= N; i++) {` ` ` `sum += 7 * ` `pow` `(8, i - 1);` ` ` `}` ` ` `return` `sum;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 4;` ` ` `cout << count(N);` ` ` `return` `0;` `}` |

## Java

`// Java program to find the count of` `// natural octal numbers upto N digits` `public` `class` `GFG {` ` ` ` ` `// Function to return the count of` ` ` `// natural octal numbers upto N digits` ` ` `static` `int` `count(` `int` `N)` ` ` `{` ` ` `int` `sum = ` `0` `;` ` ` ` ` `// Loop to iterate from 1 to N` ` ` `// and calculating number of` ` ` `// octal numbers for every 'i'th digit.` ` ` `for` `(` `int` `i = ` `1` `; i <= N; i++) {` ` ` `sum += ` `7` `* Math.pow(` `8` `, i - ` `1` `);` ` ` `}` ` ` `return` `sum;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `N = ` `4` `;` ` ` `System.out.println(count(N));` ` ` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Python3

`# Python3 program to find the count of` `# natural octal numbers upto N digits` `# Function to return the count of` `# natural octal numbers upto N digits` `def` `count(N) :` ` ` `sum` `=` `0` `;` ` ` `# Loop to iterate from 1 to N` ` ` `# and calculating number of` ` ` `# octal numbers for every 'i'th digit.` ` ` `for` `i ` `in` `range` `(N ` `+` `1` `) :` ` ` `sum` `+` `=` `7` `*` `(` `8` `*` `*` `(i ` `-` `1` `));` ` ` ` ` `return` `int` `(` `sum` `);` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N ` `=` `4` `;` ` ` `print` `(count(N));` `# This code is contributed by AnkitRai01` |

## C#

`// C# program to find the count of` `// natural octal numbers upto N digits` `using` `System;` `class` `GFG` `{` ` ` ` ` `// Function to return the count of` ` ` `// natural octal numbers upto N digits` ` ` `static` `int` `count(` `int` `N)` ` ` `{` ` ` `int` `sum = 0;` ` ` ` ` `// Loop to iterate from 1 to N` ` ` `// and calculating number of` ` ` `// octal numbers for every 'i'th digit.` ` ` `for` `(` `int` `i = 1; i <= N; i++)` ` ` `{` ` ` `sum += (` `int` `)(7 * Math.Pow(8, i - 1));` ` ` `}` ` ` `return` `sum;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `N = 4;` ` ` `Console.WriteLine(count(N));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` `// Javascript program to find the count of` `// natural octal numbers upto N digits` `// Function to return the count of` `// natural octal numbers upto N digits` `function` `count(N)` `{` ` ` `var` `sum = 0;` ` ` `// Loop to iterate from 1 to N` ` ` `// and calculating number of` ` ` `// octal numbers for every 'i'th digit.` ` ` `for` `(` `var` `i = 1; i <= N; i++) {` ` ` `sum += 7 * Math.pow(8, i - 1);` ` ` `}` ` ` `return` `sum;` `}` `// Driver code` ` ` `var` `N = 4;` ` ` `document.write(count(N));` `</script>` |

**Output:**

4095

**Time Complexity:** O(N)

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