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Question: Does there exist $a,n \in \mathbb{Z}^+$, where $n \geq 2$, such that $$a, an, an^2,an^3,\ldots,an^5$$ are all palindromes in base 10?

We see that $a=1$ and $n=11$ give rise to $$1, 11, 121, 1331, 14641$$ which is basically the first few rows of Pascal's triangle. There are infinitely many other examples of length 5, which we can generate by generalising the Pascal's triangle construction, giving $$11, 1111, 112211, 11333311, 1144664411$$ $$111, 111111, 111222111, 111333333111, 111444666444111$$ and so on.

Some other examples are:

$$147741, 13444431, 1223443221, 111333333111, 10131333313101$$

$$1478741, 134565431, 12245454221, 1114336334111, 101404606404101$$

However, I checked for $a,n \leq 10^7+1$, and didn't find any of length greater than 5.

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  • $\begingroup$ The reason it gets unlikely is that the fifth row of Pascal's triangle contains numbers with more than one digit. $\endgroup$ Jan 13, 2018 at 22:51

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