[Project Euler] Problem 14 in C

Problem: Project Euler Problem 14

The following iterative sequence is defined for the set of positive integers:

n  n/2 (n is even)
n  3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13  40  20  10  5  16  8  4  2  1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. 

Which starting number, under one million, produces the longest chain? 

NOTE: Once the chain starts the terms are allowed to go above one million.

About the Problem
The problem asks us to find the number which produces the longest chain.
From reading upon Collatz conjecture, no matter what the starting positive number is, the final result will always be 1.
This astonished me because despite the function based up on conditional statements, the function converges onto one point.

Solving the Problem
I remember solving a similar problem on uVa.
So, I had to dig through project after project related to uVa to find the code I want.
I finally found what I want and all I needed to do was tweak the main() function to make it work.

Naturally for this problem, I had used a recursive function to count the number of iterations the code had to take in order to reach 1 and I had used brute force method to solve this problem, which has greatly hindered the performance (reaching almost 1.2 seconds!)

You'll know what I mean when you view my code.
Code:

#include <stdio.h>
#include <stdlib.h>

unsigned long int countIterations(unsigned long int number){
    if (number == 1) return 1;
    if (number % 2 == 0){
        return 1 + countIterations(number / 2);
    } else {
        return 1 + countIterations(number * 3 + 1);
    }
}

int main()
{
    unsigned long int maxIter = 0, maxNum;
    unsigned long int i, n;
    for (i = 1; i < 1000000; i++){
        if (maxIter < (n = countIterations(i))){
            maxIter = n;
            maxNum = i;
        }
    }

    printf("%lu\n", maxNum);

    return 0;
}

Challenges Faced/ Lessons Learned
It was definitely like uVa problems revisited and I learned that keeping good documentation of all the codes you have created is a good idea, considering it took me nearly 10 minutes to find the small piece of code.
It would have been easier for me to just write the code from scratch!

Next Step

• To increase the performance of the code, I can use the optimization process called memoization. This method is possible because there is only one flow of the series. I mean if we see the same integer we have calculated before during the series, if we store the number of steps taken before, we can easily just add to the number of steps taken up until the familiar integer is shown.

Questions
N/A

__________________________________________________________________
Best Execution Time: 1.169s
Answer:

837799