Delectable numbers in other bases

So there's a pretty useless number called the delectable number. A delectable number has nine digits, using the numbers 1-9 once in each digit. The first digit of a delectable number must be divisible by one. The first and second digits must be divisible by two, the first through third must be divisible by three, etc. There has only been one delectable number discovered: 381654729. The question I want to ask is whether there exists delectable numbers in bases other than ten. For example, base 16; could there be a 15-digit number such that all digits 1-F are use and the first digit is divisible by 1, the first 2 by 2... the first F by F etc. Anyone have any way of finding these?

• Have you tried googling on the number 381654729? – Barry Cipra Mar 25 '16 at 15:58
• It's the only one in base ten to use all of the numbers from 1-9 exactly once, and I can't find any other numbers in other bases with the same property. – Mathime Mar 25 '16 at 16:03
• You might take a look at the discussion at scienceblogs.com/builtonfacts/2010/08/18/a-conspiracy-of-digits – Barry Cipra Mar 25 '16 at 16:14

Every number must be odd,even,odd,even and so on (every even digit must be divisible by 2) if you want to speed up calculations. I can only find a few up to base 32. Base 2: 1 Base 4: 123, 321 Base 6: 14325, 54321 Base 8: 3254167, 5234761, 5674321 Base 10: 381654729 Base 14: 9C3A5476B812D (A=10, B=11, C=12, D=13)

This is only a partial answer: There are no "delectable" numbers base $b$ if $b$ is odd.

The proof is based on the elementary generalization of casting out $9$'s, namely a number written base $b$ is divisible by $b-1$ if and only if the sum of its base-$b$ digits is divisible by $b-1$. But if $b=2m+1$, then $b-1=2m$ does not divide $1+2+\cdots+(2m)={(2m)(2m+1)\over2}=m(2m+1)$. So even if a number appears delectable for the first $b-2$ divisions, it's guaranteed to go sour at the end.

As for even bases, the first few are fairly easy:

\begin{align} \text{base }2:&\quad1\\ \text{base }4:&\quad123,321\\ \text{base }6:&\quad14325,54321 \end{align}

After that, it gets a little hairy. For base $8$, it's easy to see that delectable numbers must be of the form _2_4_6_ or _6_4_2_, where the _'s are to be filled with the odd digits 1,3,5, and 7, so at worst there are $48$ ($=2\cdot4!$) possibilities to pick through. But naively, at least, it looks like the amount of searching will grow factorially, so even by the time you get to base $16$, you've got something on the order of $7!8!$ numbers to look at -- small enough for a computer to handle, but verging on the unmanagable in the absence of additional ideas.

Remark: I was a little surprised to find that googling on the OP's number 381654729 didn't turn up much that was relevant; to be sure, it gives hits for "polydivisible" numbers, but little that pertains to the question at hand.