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Remember something from seventh grade: \begin{align} & 142857 \\ {}+ {}& 142857 \\ \\ & 285714 \\ {}+{} & 142857 \\ \\ & 428571 \\ {}+{} & 142857 \\ \\ & 571428 \\ {}+{} & 142857 \\ \\ & 714285 \\ {}+{} & 142857 \\ \\ & 857142 \\ {}+{} & 142857 \\ \\ & 999999 \end{align}

So you get all six possible cyclic shifts before you get the repeating $9$s.

So I tried it in base $12$: \begin{align} & 186T35 \\ {}+{} & 186T35 \\ \\ & 35186T \\ {}+{} & 186T35 \\ \\ & 5186T3 \\ {}+{} & 186T35 \\ \\ & 6T3518 \\ {}+{} & 186T35 \\ \\ & 86T351 \\ {}+{} & 186T35 \\ \\ & T35186 \\ {}+{} & 186T35 \\ \\ & EEEEEE \end{align} So at this point you're yawning and saying all this is just what you expected.

But now notice that in base ten, the first digit gets shifted 4 places to the right, then 5, then 2, then 1, then 3. But in base twelve, the last digit gets shifted 4 places to the left, then 5, then 2, then 1, then 3. Why exactly the same pattern in just the opposite order?

Michael Hardy
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    Without a reason to expect any meaningful relationship between base 10 and base 12, why shouldn't we write this off as a coincidence? There are plenty of integers $b$ such that $b$ generates $(\mathbb{Z}/7\mathbb{Z})^\times$, i.e. such that $1/7$ has period $6$ in base $b$, and there are only $120$ cyclic shifts of $6$ elements. The fact that the cyclic shift for $b=12$ is distantly related to the one for $b=10$ - again, with no other reason to think $10$ and $12$ are related in this context - seems unremarkable. – Sal Feb 24 '15 at 06:39
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    Have you tried checking other bases? Maybe it'll give you some clue. – Ruslan Feb 24 '15 at 08:24
  • @Ruslan : In base $8$ I get $0.111111\ldots$. In base $9$ I get $125$, then $251$, then $376$, then $512$, then $637$, then $763$, and finally of course $888$. So $125$ goes through its three shifts and $376$ goes through its three shifts. In base $11$ I also get a three-digit repetend. Clearly the length of the repetend must be a divisor of $6$. And obviously we all knew all this before I posted here. But so far I don't see what you're getting at. How does this shed light on the reversal of order and the reversal of direction of the sequence of shifts? ${}\qquad{}$ – Michael Hardy Feb 24 '15 at 17:53

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I use to represent cyclic numbers... in circle! Well, I do it with $142857$ since it is my (everyone's?) favorite. That way I can decide the starting point around the circle, make one revolution clockwise, and have a multiple of the original number. But to have the correct sequence of growing multiples, one has to follow the correct sequence of starting points. Since there are not repeated digits, it is easy to see that this sequence goes from the smallest digit to the biggest. I always feel amazed by the symmetry of the path of this sequence (depicted in green in the following picture), for which I am not able to give an explanation:

enter image description here

If I repeat this figurate representation with the base-$12$ version of the number, the result is shocking! Ok, the bases are close enough to make the number have the same amount of digits (i.e. $6$), but... not only the starting points path is just as symmetric as before, it is exacly the same just mirrored! This is what you point out in your question, just represented under a figurative point of view. The base-$12$ picture is here below. I don't know what kind of explanation one can expect to find for this phenomenon, but it seems to me that it is somewhat beyond coincidence.

enter image description here

lesath82
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