These are the only two numbers (besides 1 & 2) that can be written as the sum of the factorials of their digits (in base 10).pic.twitter.com/CHzrPjAhWt
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This sounds like it should have a nice proof. Do share.
Proof is not elegant,, you find an upper bound then test all cases
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Sure, but it was the upper bound I was interested in, it doesn't seen plausible it is on the length of the decimal expansion, as 17, 107, 1007 will all have the same factorial sum. So is it a bound on the sum of the digits? Something else?
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If the number n has d digits, then n must be greater than or equal to 10^(d-1) because n is on the order of 10^d. Additionally, n must be less than or equal to 9!*d (the case when it's all 9's). If you graph these 2 they intersect between d=7 and d=8 so d must be less than 8.
End of conversation
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