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Inspired from the question Is there any perfect square in the sequence $12,123,1234,12345,...$?, there is no perfect square other than $1$ in the sequence of Smarandache numbers. But I wonder if are there any perfect powers other than $1$ in the sequence of Smarandache numbers?

Using Pari GP, I checked up to the 20th term of Smarandache numbers to see if are there any perfect powers in the sequence, but I wasn't succesful.

Are there any perfect powers other than $1$ in the sequence of Smarandache numbers?

Thirdy Yabata
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    Perhaps you didn't read that question and its comments and answers carefully. We do not know there are no perfect squares there other than $1$, it's just extremely unlikely that there are any, Since cubes are rarer than squares, it is even more unlikely that there are perfect cubes other than $1$ in the sequence. But again, there is no proof. – Robert Israel Jul 17 '23 at 01:41
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    Indeed, is there any perfect power other than $1$ in the sequence? And if not, why? – Juan Moreno Jul 23 '23 at 22:36
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    Since cubes are rarer than squares, it is even more unlikely that there are perfect cubes other than 1 in the sequence. It is not for me a plausible argument; for example, ${n^2+2}_{n\ge1}$ is never a square but can contains a cube. Respect to the first objection, I agree completely. (Why not to ask about a $n-$power in the Scharamandache numbers..........? – Piquito Aug 10 '23 at 17:58
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    The is no perfect power in the first $10^6$ terms of this series (except for $1$), adapting the heuristics from the linked question now results in a very low probability that there exists one. – jorisperrenet Aug 16 '23 at 22:21
  • We have

    $$a_n = a_{n-1}\cdot 10^{\lfloor log_{10}(10n) \rfloor} + n$$

    Suppose $a_{n-1} = k^2$.

    $$ \begin{align} a_n &= k^2\cdot 10^{\lfloor log_{10}(10n) \rfloor} + n \ a_n - n &\equiv k^2 \pmod{9} \end{align} $$

    We have

    $$(a_n - n \pmod{9})+1 = \text{dr}_{10}(a_n - n)$$

    where $\text{dr}_{b}(x)$ is the Digital Root of $x$ in base $b$.

    The squares modulo $9$ are $0, 1, 4, 7$. Therefore, the digital root of $a_n - n$ must be one of $1, 2, 5, 8$ for $a_{n-1} = k^2$.

     – vvg Aug 17 '23 at 05:50
  • PARI/GP has the command ispower() which also works well and quickly for numbers with thousands of digits. – Peter Aug 17 '23 at 15:33
  • What kind of answer are you looking for ? A proof is almost surely out of reach and that there is no such perfect power ($1$ is usually also not considered to be a perfect power, PARI/GP also does not do that) is extremely likely. I do not think you will get anything better. – Peter Aug 17 '23 at 15:41

2 Answers2

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Here is what I discovered so far about the existence of perfect powers of degree $> 2$ among the terms of the OEIS sequence $A352991$ (which includes all the terms of the sequence $A001292$): https://arxiv.org/pdf/2205.10163.pdf

I also agree with the old Kashihara's conjecture that the sequence $A001292$ does not contain any nontrivial perfect power.

Marco Ripà
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A perfect power $x$ must not contain a prime divisor $p$ such that $p^2$ divides $x$. So some of the non-perfect squares are $4k+2,25k+\{5;10;15;20\}$ and $9k+\{3;6\}$

With the first two cases, we can conclude $a_n$ isn't a perfect power if $n=4k+2$ or $n=25k+\{5;10;15;20\}$. With the third one, by caculating the sum of $a_n$'s digits through $n$ we can say $a_n$ isn't a perfect power if $n=9k+\{2;3;5;6\}$

Le_Square
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