Self-Factorial Number

Question

Self-factorial number is the number where its digits' factorials summation is equal to the number itself. But there are only a few amount of them. For example;

$1=1!$

$2=2!$

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

So what is the fourth self-factorial number? How about fifth?

Answer 1

A self-factorial number can have no more than 7 digits. Proof: Suppose it has $n$ digits. Then the sum of the factorials of the digits is at most $$ 9! \cdot n = 362880n $$ whereas the number itself is at least $$ 10^{n-1}. $$ For $n \ge 8$, $10^{n-1} > 362880n$ so these two cannot be equal.

So it remains that we check all numbers with up to $7$ digits. A brute-force search checking $1$ to $9999999$ finds that the only solutions are

1, 2, 145, 40585.

Answer 2

Its

40585. They are factorians-A014080. And no there is no fifth. Reference

Answer 3

I wrote a Java program to find a solution by brute-force. I found

40585

as a fourth number.

The fifth number should have a ton of digits. I'm still searching.

EDIT: it looks like there is not a fifth one.