C（范围 1-100,000,000）最大循环的 Collatz 猜想

问：

我正在制作一个程序，用于计算数字范围 （1 - 100,000,000） 之间 Collatz 猜想的最多循环，我需要它在 4 分钟内在 linux 系统上运行，并且

command gcc -O0 -m32 -Wall -Wextra -Werror -pedantic -o collatz collatz.c.

关于如何改进算法的任何想法，因为在 10,000,000 之后，程序需要很长时间才能完成，请记住，我不能使用多线程或预先计算的数据。

欢迎所有回答，感谢您抽出宝贵时间接受采访。 😀

// Lets find the longest collatz
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

// This function finds how many times it took for a number to reach 1
unsigned long int collatz(uint32_t First_num) {
    // steps measures how many times a number needs to reach 1
    uint32_t steps = 1;
    // This is the loop that applies the rules of the collatz conjecture
    while (First_num != 1) {
        (First_num % 2 == 0) ? (First_numm >>= 1) : (First_num = (3 * First_num + 1) / 2);
        steps++;
    }
    return steps;
}

// This is the function that calculates the maximum loops a number needs to reach 1
int max(uint32_t First_num, uint32_t Second_num) {
    // This "if" determines if the number is correct
    if (First_num <= 0 || Second_num <= 0) {
        printf("0\n");
    }
    uint32_t max = 0;
    uint32_t max_num = 0;
    uint32_t i = First_num;
    // This loop is used to find the maximum times the collatz function needs to be used
    for (i; i <= Second_num; i++) {
        printf("The num %u took the most %ld!!!\n", i, collatz(i));
        // This block finds the number with the most loops
        if ((collatz(i)) >= max) {                                                                                                                                                                                                               
            max = collatz(i);                                                                                              
            max_num = i;
        }
    }
    printf("The num %u took the most %u !!!\n", max_num, max);                                                                                                                                                                        
    return 0;                                                                                                                                                                                                       
}
                                                                                                                                                                                                                               
// This is the main function in which the "max" function is used and the user gives his input and gets his output                                                                                                               
int main(int argc, char **argv) {                                                                                                                                                                                                        
    if (argc != 3) {                                                                                                                                                                                                                         
        printf("Program needs to be called as './prog num1 num2'\n");                                                                                                                                                                   
        return 1;                                                                                                                                                                                                               
    }                                                                                                                                                                                                                       
    //Users input                                                                                                                                                                                                                   
    uint32_t num1 = atoi(argv[1]);                                                                                                                                                                                                  
    uint32_t num2 = atoi(argv[2]);                                                                                                                                                                                                      
    // Users input                                                                                                                                                                                                                  
    max(num1, num2);                                                                                                                                                                                                                 
    return 0;                                                                                                                                                                                                                   
}                                                                                                                                                                                                                               
//End of the program