Problem 14

Longest Collatz Sequence

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

$n \to n / 2$ ( $n$ is even)

$n \to 3 n + 1$ ( $n$ is odd)

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

$13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$

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.

最长考拉兹序列

考虑如下定义在正整数集上的迭代规则：

$n \to n / 2$ （若 $n$ 为偶数）

$n \to 3 n + 1$ （若 $n$ 为奇数）

从 $13$ 开始，可以迭代生成如下的序列：

$13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$

可以看出这个序列（从 $13$ 开始到 $1$ 结束）共有 $10$ 项。尽管还未被证明，但普遍认为，从任何数开始最终都能抵达 $1$ 并结束（这被称为“考拉兹猜想”）。

在小于一百万的数中，从哪个数开始迭代生成的序列最长？

注：在迭代过程中允许出现超过一百万的项。