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Can you work out the mathematical rule for this cyclic number sequence?

2, 6, 186, 8, 456, 54, 53, 23, 7, 301, 3, 21, 2, 6,.....

Hint 1

this is a mathematical puzzle

From time to time hints will be updated.

Note: this sequence cannot be found in The On-Line Encyclopedia of Integer Sequences

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The answer appears to be:

The next number in the sequence is derived from the preceding number by the smallest digit cubed, less the smallest digit squared, plus the largest digit.


For a single-digit number $x$ this is simply $x^3 - x^2 + x$. For example $2$:
$$2 \Rightarrow 2^3 - 2^2 + 2 \Rightarrow 8 - 4 + 2 = 6$$
As an example for a three digit number, e.g. $456$:
Smallest digit is $4$, largest is $6$, so:
$$456 \Rightarrow 4^3 - 4^2 + 6 \Rightarrow 64 - 16 + 6 = 54$$

As an example of a two digit number, e.g. $54$:
Smallest digit is $4$, largest is $5$, so:
$$54 \Rightarrow 4^3 - 4^2 + 5 \Rightarrow 64 - 16 + 5 = 53$$

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    $\begingroup$ Pardon me for asking, but how in the ever-loving bull-shattering hello there did you figure that one out? $\endgroup$ – Bass Mar 19 '18 at 5:15
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    $\begingroup$ @Bass, I suppose, by finding the next number from a single-digit number? e.g. 2 to 6, 6 to 186, or 7 to 301. Turned out that the number after $x$ is $x^3 - (x(x-1))$. Then proceed how to generalize it with 2 and 3-digit numbers. (At least this is what I tried, but haven't generalized it. Good work for TheOmegaPostulate!) $\endgroup$ – athin Mar 19 '18 at 6:29
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    $\begingroup$ @Bass, exactly what athin said for the single digit ones. From there I generalised to the two, and then three digits by focusing on "interesting" sequences (e.g. 54 -> 53 -> 23 as there is a change of one followed by a change of 30, and 301 as one digit is a zero). $\endgroup$ – TheOmegaPostulate Mar 19 '18 at 7:12
  • $\begingroup$ Great Job, well done! $\endgroup$ – tom Mar 19 '18 at 7:31

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