
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
If $a = \overline{a_1 a_2 a_3 a_4 a_5 a_6}$, then $P(a) = \overline{a_6 a_1 a_2 a_3 a_4 a_5}$ and $P(P(a)) = \overline{a_5 a_6 a_1 a_2 a_3 a_4}$. Given that $a_5 \neq 0, a_6 \neq 0, a_1 \neq 0$, it follows from $P(P(a)) = a$ that $a_1 = a_5, a_2 = a_6, a_3 = a_1, a_4 = a_2, a_5 = a_3, a_6 = a_4$, i.e., $a_1 = a_3 = a_5 = t$ and $t = 1, 2, \ldots, 9$; $a_2 = a_4 = a_6 = u$ and $u = 1, 2, \ldots, 9$. Thus, the desired $a = \overline{tututu}$, and there are 81 such distinct numbers.

Let $n > 2$ be a prime number and $a = \overline{a_1 a_2 a_3 a_4 \ldots a_{n-3} a_{n-2} a_{n-1} a_n}$. Then

\[
P(a) = \overline{a_n a_1 a_2 a_3 a_4 \ldots a_{n-3} a_{n-2} a_{n-1}} \\
P(P(a)) = \overline{a_{n-1} a_n a_1 a_2 a_3 a_4 \ldots a_{n-3} a_{n-2}}
\]

The property $P(P(a)) = a$ gives the relationships $a_1 = a_{n-1} = a_{n-3} = \ldots = a_1$. For prime $n > 2$, all digits of the number $a$ enter the chain, so they are all equal to each other.

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