# 考拉兹猜想

$f(n) = \begin{cases} n/2 &\mbox{if } n \equiv 0 \\ 3n+1 & \mbox{if } n\equiv 1 \end{cases} \pmod{2}.$

## 例子

• 如n = 6，根据上述数式，得出序列6, 3, 10, 5, 16, 8, 4, 2, 1。(步驟中最高的數是16，共有8個步驟)
• 如n = 11，根据上述数式，得出序列11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1。(步驟中最高的數是52，共有14個步驟)
• 如n = 27，根据上述数式，得出序列
{ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 }（步驟中最高的數是9232，共有111個步驟）

n = 27时的序列分布（横轴－步数；纵轴－运算结果）

## 计算机验证

### Python

def collatz(n):
print(n)
if n % 2 == 1 and n > 1:
collatz(3*n + 1)
elif n % 2 == 0:
collatz(n / 2)


### C

#include <stdio.h>

void collatz(unsigned int n)
{
printf("%lu\n", n);
if(n == 1)            // Terminate when n == 1;
return ;

n += ((n << 1) + 1);  // n = 2n + 1; (2n + 1) + n = 3n + 1;

while(!(n & 1))       // remove all trailing '0's
n = n >> 1;

return collatz(n);
}