C Program for n-th odd number

Given a number N, we have to find the N-th odd number. Odd numbers are the numbers which are not completely divisible by 2 and their remainder is not zero, like 1, 3, 5, 7, 9, etc.

If we closely observe the sequence of odd numbers, we can represent them mathematically as −

1st odd number: (2*1)-1 = 1
2nd odd number: (2*2)-1 = 3  
3rd odd number: (2*3)-1 = 5
4th odd number: (2*4)-1 = 7
...
Nth odd number: (2*N)-1

So, to solve the problem we can simply multiply the number N with 2 and subtract 1 from the result.

Syntax

nth_odd_number = (2 * n) - 1

Examples

Input: 4
Output: 7
Explanation: The 4th odd number in sequence 1, 3, 5, 7...

Input: 10
Output: 19
Explanation: The 10th odd number using formula (2*10)-1 = 19

Algorithm

START
   STEP 1 -> Declare and assign an integer 'n'
   STEP 2 -> Calculate nth odd number using formula: n*2-1
   STEP 3 -> Print the result
STOP

Example

#include <stdio.h>

int main() {
    int n = 10;
    int nth_odd;
    
    // Calculate nth odd number using formula
    nth_odd = (n * 2) - 1;
    
    printf("The %dth odd number is: %d<br>", n, nth_odd);
    
    // Let's verify with first few odd numbers
    printf("\nFirst 5 odd numbers:<br>");
    for(int i = 1; i <= 5; i++) {
        printf("%d ", (i * 2) - 1);
    }
    printf("<br>");
    
    return 0;
}

Output

The 10th odd number is: 19

First 5 odd numbers:
1 3 5 7 9 

Key Points

• The formula (2*N)-1 directly gives the Nth odd number
• Time complexity is O(1) as it uses a direct mathematical formula
• Space complexity is O(1) as no extra space is required

Conclusion

Finding the Nth odd number is efficiently solved using the mathematical formula (2*N)-1. This approach provides constant time complexity and is much faster than iterative methods.