Factorions Base 10

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Theorem

The following positive integers are the only factorions base $10$:

$1, 2, 145, 40 \, 585$

This sequence is A014080 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

From examples of factorials:

\(\ds 1\) \(=\) \(\ds 1!\)
\(\ds 2\) \(=\) \(\ds 2!\)
\(\ds 145\) \(=\) \(\ds 1 + 24 + 120\)
\(\ds \) \(=\) \(\ds 1! + 4! + 5!\)
\(\ds 40 \, 585\) \(=\) \(\ds 24 + 1 + 120 + 40 \, 320 + 120\)
\(\ds \) \(=\) \(\ds 4! + 0! + 5! + 8! + 5!\)


A computer search can verify solutions under $2540160 = 9! \times 7$ in seconds.


Let $n$ be a $7$-digit number with $n > 2540160$.

Then the sum of the factorials of its digits is not more than $9! \times 7 = 2540160$.

So $n$ cannot be a factorion base $10$.


Now let $n$ be a $k$-digit number, for $k \ge 8$.

Then the sum of the factorials of its digits is not more than $9! \times k$.

But we have:

\(\ds n\) \(\ge\) \(\ds 10^{k - 1}\)
\(\ds \) \(=\) \(\ds 10^7 \times 10^{k - 8}\)
\(\ds \) \(\ge\) \(\ds 10^7 \times \paren {1 + 9 \paren {k - 8} }\) Bernoulli's Inequality
\(\ds \) \(>\) \(\ds 8 \times 9! \times \paren {9 k - 71}\) $8 \times 9! = 2903040$
\(\ds \) \(=\) \(\ds 9! \paren {72 k - 71 \times 8}\)
\(\ds \) \(>\) \(\ds 9! \times k\) $k \ge 8$
\(\ds \) \(\ge\) \(\ds \text{sum of the factorials of digits of } n\)

So there are no more factorions base $10$.

$\blacksquare$


Historical Note

The fact that $40 \, 585$ is a factorion base $10$ was discovered as late as $1964$ by Leigh Janes.


Sources