Number which is Sum of Subfactorials of Digits

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Theorem

The only integer which is the sum of the subfactorials of its digits is $148 \, 349$:

$148 \, 349 = \mathop !1 + \mathop !4 + \mathop !8 + \mathop !3 \mathop + \mathop !4 \mathop + \mathop !9$


Proof

We have:

\(\ds 148 \, 349\) \(=\) \(\ds 0 + 9 + 14 \, 833 + 2 + 9 + 133 \, 496\)
\(\ds \) \(=\) \(\ds \mathop !1 + \mathop !4 + \mathop !8 + \mathop !3 \mathop + \mathop !4 \mathop + \mathop !9\)


A computer search can verify solutions under $10^6$ (that is, with no more than $6$ digits) in seconds.


Let $n$ be a $k$-digit number, for $k \ge 7$.

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

But we have:

\(\ds n\) \(\ge\) \(\ds 10^{k - 1}\)
\(\ds \) \(=\) \(\ds 10^6 \times 10^{k - 7}\)
\(\ds \) \(\ge\) \(\ds 10^6 \times \paren {1 + 9 \paren {k - 7} }\) Bernoulli's Inequality
\(\ds \) \(>\) \(\ds 7 \times \mathop !9 \times \paren {9 k - 62}\) $7 \times \mathop !9 = 934472$
\(\ds \) \(=\) \(\ds \mathop !9 \paren {63 k - 62 \times 7}\)
\(\ds \) \(\ge\) \(\ds \mathop !9 \times k\) $k \ge 7$
\(\ds \) \(\ge\) \(\ds \text{sum of the subfactorials of digits of } n\)

So no more numbers have this property.

$\blacksquare$


Historical Note

This result was apparently obtained by R.S. Dougherty, but details are hard to come by.


Sources