Count of Octal numbers upto N digits

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

Examples:
Input: N = 1 
Output: 7 
Explanation:
1, 2, 3, 4, 5, 6, 7 are 1 digit Natural Octal numbers.
Input: N = 2 
Output: 63 
Explanation:
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 * 8N - 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*logN) because using inbuilt pow function in loop, time complexity of pow function is logN

Auxiliary Space: O(1)