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Problem 74

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Digit Factorial Chains

The number $145$ is well known for the property that the sum of the factorial of its digits is equal to $145$:

$$1!+4!+5!=1+24+120=145$$

Perhaps less well known is $169$, in that it produces the longest chain of numbers that link back to $169$; it turns out that there are only three such loops that exist:

$$\begin{array}{rl}& 169\to 363601\to 1454\to 169\\ & 871\to 45361\to 871\\ & 872\to 45362\to 872\end{array}$$

It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

$$\begin{array}{rl}& 69\to 363600\to 1454\to 169\to 363601(\to 1454)\\ & 78\to 45360\to 871\to 45361(\to 871)\\ & 540\to 145(\to 145)\end{array}$$

Starting with $69$ produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.

How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?

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数字阶乘链

众所周知，$145$的各位数字的阶乘之和恰好等于本身：

$$1!+4!+5!=1+24+120=145$$

但很少有人注意到，从$169$开始不断地取各位数字的阶乘之和，将会进入循环回到$169$，而且这是所有循环中最长的一个。事实上，只存在三个这样的循环：

$$\begin{array}{rl}& 169\to 363601\to 1454\to 169\\ & 871\to 45361\to 871\\ & 872\to 45362\to 872\end{array}$$

不难证明，从任意数字出发最终都会陷入循环。例如，

$$\begin{array}{rl}& 69\to 363600\to 1454\to 169\to 363601(\to 1454)\\ & 78\to 45360\to 871\to 45361(\to 871)\\ & 540\to 145(\to 145)\end{array}$$

从$69$开始直到进入循环，会得到五个不同的项，而从任意一个小于一百万的数开始，最多能够得到六十个不同的项。

有多少个小于一百万的数满足，从这些数开始不断地取各位数字的阶乘之和，恰好能够得到六十个不同的项？

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