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# On some open problems concerning perfect powers

Marco Ripà

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Marco Ripà. On some open problems concerning perfect powers. 2022, 10.48550/arXiv.2205.10163. hal-03725409v2

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Submitted on 13 May 2023

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# On some open problems concerning perfect powers

Marco Ripà

sPIqr Society, World Intelligence Network
Rome, Italy

e-mail: marcokrt1984@yahoo.it

**Abstract:** The starting point of our paper is Kashihara's open problem number 30, concerning the sequence A001292 of the OEIS, asking how many terms are powers of integers. We confirm his last conjecture up to the 100128-th term and provide a general theorem which rules out 4/9 of the candidates. Moreover, we formulate a new, provocative, conjecture involving the OEIS sequence A352991 (which includes all the terms of A001292). Our risky conjecture states that all the perfect powers belonging to the sequence A352991 are perfect squares and they cannot be written as higher order perfect powers if the given term of A352991 is not equal to one. This challenging conjecture has been checked for any integer smaller than 10111121314151617181920212223456789 and no counterexample has been found so far.

**Keywords:** Open problem, Perfect power, Perfect square, Conjecture, Integer sequence.

**2020 Mathematics Subject Classification:** 00A08 (Primary) 11B50, 11-04 (Secondary).

## 1 Introduction

In late 2010, the author of this paper found a recreative open problem by Kenichiro Kashihara (see [1], open problem number 30, p. 25) concerning the sequence A001292 of the On-Line Encyclopedia of Integer Sequences (OEIS) [2]. Kashihara's problem number 30 consists of two independent parts and the author solved the first one quite easily at the time (the complete solution can be found in [4], Section 3.3, pp. 12–15), since it asks to find the probability $0 < p(c) < 1$ that the trailing digit of the generic term of the sequence A001292 is $c \in \{0, 1, 2, \dots, 9\}$ and the formula provided in [4] shows that $p(c) = \frac{11-c}{55}$ for any $c \neq 0$, whereas $p(0) = 0.0\overline{18}$ (e.g., if $c = 7$, then $p(7) = \frac{4}{55} = 0.0\overline{72}$).

In the present paper, we will focus ourselves on the second part of the above mentioned Kashihara's problem number 30, asking how many elements of the sequence A001292 are perfect powers, since Kashihara conjectured that there are none.

Now, bearing in mind that a perfect power of an integer $d > 0$ is a natural number $k \ge 2$ such that $a^k = d$, where also a is a positive integer, we could point out that A001292(1) = 1 can be considered as a solution and argue how this disproves the conjecture, but (from here on) we will disregard this special case and assume that we are looking for a nontrivial counterexample to Kashihara's conjecture.
Lastly, Section 3 is to devoted to introducing a (quite improbable) conjecture concerning
perfect powers belonging to the OEIS sequence A352991 [5, 6].

## 2 The exclusion criterion

In order to be clear on the invoked OEIS sequences, let us introduce a few useful definitions.

**Definition 1.** We define the $m$-th term of the OEIS sequence $A007908$ as
$A007908(m) := 123\dots(m-1)_m$, where $m \in \mathbb{Z}^+$.

**Definition 2.** We define the sequence $A001292$ of the OEIS as the concatenations (sorted in ascending order) of every cyclic permutation of the elements of the sequence $A007908$ (e.g., given $m = 3$, $A001292(A007908(3)) = 123, 231, 312$).

**Definition 3.** We define the OEIS sequence $A352991$ as the concatenation of all the distinct permutations of the first strictly positive $m$ integers, sorted into ascending order (e.g., 12345671089 is a term of the sequence, while 12345670189 does not belong to $A352991$, even if all the digits of the string 123...910 appear once and only once, since "10" is missed).

After having checked the first 100128 terms of the sequence $A001292$ (see Appendix),
exploring any exponent at or above two, we have not found any perfect power so that Kashihara's
conjecture has been verified up to $10^{1235}$ (i.e., the 100129-th term of $A001292$ is the smallest
cyclic permutation of $A007908(448)$ and is greater than $10^{1235}$ by construction). Moreover, we
can prove the following Theorem 1, concerning the sequence $A352991$ which includes every
term of $A001292$.

**Theorem 1.** For any $m > 1$, $A352991(n)$ cannot be a perfect power of an integer if $A352991(n)$ is a permutation of $A007908(m)$ and $m : m \equiv \{2, 3, 5, 6\}(\text{mod } 9)$.

*Proof.* By definition, $A007908(m)$ [3] cannot be a perfect power if $123\dots(m-1)_m$ is divisible by 3 and it is not divisible by $3^2$. Thus, from the well-known divisibility by 3 and 9 criteria, $m : (3 | \sum_{j=1}^m j) \land (3^2 \nmid \sum_{j=1}^m j)$ is a sufficient, but not necessary, condition for letting us disregard any permutation of $123\dots(m-1)_m$ (i.e., given $m$, if a generic permutation of $A007908(m)$ is divisible by 3 and is not congruent to 0 mod 9), then all the permutations of $A007908(m)$ are divisible by 3 once and only once, since the commutativity property holds for addition).

It follows that, for any $n \in \mathbb{Z}^{+}$, $A352991(n)$ cannot be a perfect power if it is a permutation of the string $123\dots(m-1)_m$, where $m$ is such that $A134804(m)$ is divisible by 3. Therefore, the residue modulo 9 of every perfect power belonging to $A352991$ cannot be 2 or 3 or 5 or 6, and this concludes the proof of Theorem 1. $\square$
**Corollary 1.** Kashihara's conjecture is true for the concatenation of any cyclic permutation of $A007908(m)$, where $m : m \equiv \{2, 3, 5, 6\}(\text{mod } 9) \lor m < 448$.

*Proof.* We observe that $A001292$ is a subsequence of $A352991$ [2,5]. By invoking Theorem 1, we can state that every perfect power candidate has to be the concatenation of a (cyclic) permutation of $A007908(m)$, where $m$ is such that $m : m \equiv \{2, 3, 5, 6\}(\text{mod } 9)$. On the other hand, all the remaining terms up to $99\_100\_101\_\dots\_445\_446\_447\_1\_2\_3\_\dots\_96\_97\_98$ have been directly checked (see Appendix for details) and no perfect power has been found.

Therefore, Corollary 1 confirms Kashihara's conjecture for any term of $A001292$ such that $m$ is congruent to $\{2, 3, 5, 6\}(\text{mod } 9)$ or $m \le 447$. $\square$

**Corollary 2.** $\mathbb{Z}_n : A353025(n) \equiv \{2, 3, 4, 5, 6, 7, 8\}(\text{mod } 9)$, and any term of $A001292$ cannot be a perfect power if its digital root is not equal to 0 or 1.

*Proof.* Trivially, $10 \equiv 1(\text{mod } 9)$ and also $(1 + 0) \equiv 1(\text{mod } 9)$ so that any positive integer is congruent modulo 9 to its digital root.

Now, we observe that any terms of $A001292$ belongs to $A353025$.

Since from Theorem 1 it follows that every term of the sequence $A353025$ [6] is a special permutation of $A007908(m)$ which is characterized by $m \equiv \{0, 1, 4, 7, 8\}(\text{mod } 9)$, in order to prove Corollary 2, it is sufficient to note that

$$ \sum_{j=1}^{m} j \equiv \begin{cases} 0(\text{mod } 9) & \text{if } m : m \equiv 0(\text{mod } 9) \\ 1(\text{mod } 9) & \text{if } m : m \equiv 1(\text{mod } 9) \\ 1(\text{mod } 9) & \text{if } m : m \equiv 4(\text{mod } 9) \\ 1(\text{mod } 9) & \text{if } m : m \equiv 7(\text{mod } 9) \\ 0(\text{mod } 9) & \text{if } m : m \equiv 8(\text{mod } 9) \end{cases} . \quad (1) $$

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**Remark 1.** A well-known property of integers is that every perfect power which is congruent modulo 5 to 0 is also necessarily congruent to $\{0, 25, 75\}(\text{mod } 100)$, while if a perfect power is congruent modulo 10 to 6, then its second last digit is odd.

Thus, we are free to combine these additional constraints with Corollary 2 in order to reduce the number of perfect power candidates among the terms of $A352991$.

# 3 Perfect cubes in A353025

In the first half of April 2022, playing with Kashihara's conjecture, a more risky (very likely false but really hard to disprove by brute force) conjecture arose, it is as follows.

**Conjecture 1.** Let $n \in \mathbb{N} - \{0, 1\}$ be given. We (provocatively) conjecture that if $n$ is such that $A352991(n)$ is a perfect power of an integer, then $\mathbb{Z}_k \in \mathbb{N} - \{0, 1, 2\} : A352991(n) = c^k, c \in \mathbb{N}$.
On April 16, 2022, a direct search was performed by the author on the first $10^7$ terms of the sequence and no counterexample has been found (42 perfect squares only). A few days later, Aldo Roberto Pessolano, performed a deeper search running the Mathematica codes published in Appendix, without finding any counterexample and thus confirming Conjecture 1 (at least) up to the smallest permutation of $A007908(22)$ (i.e., for any term of $A352991$ which is strictly greater than 1 and smaller than $10111121314151617181920212223456789)$.

**Remark 2.** If confirmed, Conjecture 1 would imply that all the perfect powers (strictly greater than 1) in $A352991$ are perfect squares and nothing more (no cube, no square of square, and so forth). Nevertheless, under the (arbitrary, but perfectly reasonable) assumption of a standard probability distribution of the cubes in $A352991$ (i.e., we are assuming that $\frac{|n \in \mathbb{N}: A352991(n) \le m \land A352991(n)^{\frac{1}{3}}|}{|n \in \mathbb{N}: A352991(n) \le m|} \cong \frac{|n \in \mathbb{N}: A000578(n) \le m|}{m}$ holds for any sufficiently large $m \in \mathbb{N}$), we would guess the existence of infinitely many counterexamples to Conjecture 1, even if the smallest one is expected to occur in the interval $[10^{58}, 10^{65}]$. On the other hand, the same argument would corroborate Kashihara's conjecture, since the number of perfect powers belonging to $A001292$ cannot probabilistically exceed

$$2 \cdot \sum_{k=2}^{+\infty} \left( \sum_{j=309}^{+\infty} \frac{4 \cdot j \cdot \left( 10^{\frac{4j}{k}} - 10^{\frac{4j-1}{k}} \right)}{9 \cdot 10^{(4 \cdot j - 1)}} \right) \approx \frac{8}{9} \cdot \sum_{j=309}^{+\infty} \frac{j \cdot \left( 10^{\frac{4j}{2}} - 10^{\frac{4j-1}{2}} \right)}{10^{(4 \cdot j - 1)}} \approx 0. \quad (2)$$

**Additional open problems.** Does the sequence $A353025$ have infinitely many perfect squares, infinitely many perfect cubes, infinitely many perfect squares of squares, and so forth? Which is the smallest nontrivial perfect cube (if any) belonging to $A353025$ (we point out that all the terms greater than one and below $(1.01 \cdot 10)^{40}$ have been checked without finding any cube)?

# 4 Conclusion

Kashihara's open problem number 30 has not been completely solved yet. Even if the first part, concerning the probability that the trailing digit of $A001292(n)$ is $c = 1, 2, \dots, 9$, was solved by the author a dozen of years ago [4], the second part and the related conjecture still needs a proof or a nontrivial counterexample (the smallest candidate has 1236 digits). Moreover, in the present paper, we have introduced a wider speculation that allow us to ask to ourselves how to find a term of the OEIS sequence $A353025$ (disregarding $A353025(1)$) which is not a perfect square; a challenging open problem, considering that there is not any perfect cube among the terms of $A352991$ in the interval $(1, 10^{40}]$.

# 5 Appendix

Aldo Roberto Pessolano helped the author of the present paper by verifying Kashihara's conjecture and Conjecture 1 for a very large number of terms. All the provided Mathematica codes run on the M1 processor of his Apple MacBook Air (2020). Kashihara's conjecture has
been currently tested up to the 100128-th term of A001292 and we confirm that it holds for every
element of the set A001292(2), A001292(3), ..., A001292(100128).

The search reached the term 99\_100\_\_\_446\_447\_1\_2\_\_\_97\_98 \approx 9.91 \cdot 10^{1232} \text{ in } 28823 \text{ seconds (about } 8 \text{ hours of calculations) and the code is as follows:

c = True;
p = Table[Prime[q], {q, 1, 565}];
Do[rn = Range[k];
n = ToExpression[StringJoin[ToString[#]&/@rn]]; If[And[Mod[n, 9] != 3, Mod[n, 9] != 6], Do[r = RotateLeft[rn, i - 1]; nk = ToExpression[StringJoin[ToString[#]&/@r]]; If[IntegerQ[nk^ (1/#)], Print[nk, " = ", nk^ (1/#), "^", #]; c = False; Break[] ]&/@p, {i, 1, k}] ]; If[c, Print["1..", k, " checked."], Break[]], {k, 2, 447}]

About our investigation on the perfect powers in A352991, Pessolano has recently
completed the direct check of every term of A352991 which falls in the interval
(1, 987654322120191817161514131211110] (see the code below). As expected, the test has not
returned any perfect power above two.

z = False;
h = 3;
p = Table[Prime[q], {q, 2, 10}];
q[x_, k_, d_, m_] := (y = x^k;
If[DigitCount[y] == d, c = True;
Do[If[Not[StringContainsQ[ToString[x], ToString[i]]], c = False;
Break[], c = True], {i, 10, m}], c = False];
Return[c])
Do[r = Range[k];
n = ToExpression[StringJoin[ToString[#]&/@r]]; If[And[Mod[n, 9] != 3, Mod[n, 9] != 6], d = DigitCount[n]; (s = IntegerPart[(10^ (IntegerLength[n] - 1))^ (1/#)]); f = IntegerPart[(10^ (IntegerLength[n]))^ (1/#)]; Do[If[q[x, #, d, k], Print[x, "^", #, " = ", y]; z = True; Break[]], {x, s, f}])&/@p; g = 2^h; While[g < n, If[q[#, h, d, k], Print[x, "^", h, " = ", y];
z = True;
Break[]&/@{2, 3, 5, 6, 7};
h++;g = 2^h]];
If[z, Break[],
Print["1..", k, " checked."]],
{k, 2, 21}]

On the other hand, the following code returns the complete list of the smallest 42 perfect squares belonging to A352991.

z = 1;
Do[r = Range[k];
n = ToExpression[StringJoin[ToString[#]&/@r]];
If[And[Mod[n, 9] != 3, Mod[n, 9] != 6],
d = DigitCount[n];
s = IntegerPart[Sqrt[10^(IntegerLength[n] - 1)]];
f = IntegerPart[Sqrt[10^(IntegerLength[n])]];
Do[y = x^2;
If[DigitCount[y] == d,
c = True;
Do[
  If[Not[StringContainsQ[ToString[y], ToString[i]]],
    c = False
  ],
  {i, 10, k}];
  If[c, Print[z, " ", y]; z++]
],
{x, s, f}]
],
{k, 2, 10}]

These 42 perfect squares correspond to all the perfect powers in $(1, 10^{16}]$ belonging to A352991, while the next perfect square is 10135681742311129 (we observe that $100676123^2$ is a permutation of 123...16, as suggested by the statement of Theorem 1).

1 13527684

2 34857216

3 65318724

4 73256481

5 81432576

6 139854276
7 152843769
8 157326849
9 215384976
10 245893761
11 254817369
12 326597184
13 361874529
14 375468129
15 382945761
16 385297641
17 412739856
18 523814769
19 529874361
20 537219684
21 549386721
22 587432169
23 589324176
24 597362481
25 615387249
26 627953481
27 653927184
28 672935481
29 697435281
30 735982641
32 743816529
33 842973156
34 847159236
35 923187456
36 14102987536
37 24891057361
38 27911048356
39 28710591364
40 57926381041
41 59710832164
42 75910168324

In the end, our tests have finally confirmed that all the perfect powers which are smaller than $10^{34}$ and that belong to the OEIS sequence A352991 are perfect squares (only). At the present time, Conjecture 1 has been tested for every integer smaller than 10111121314151617181920212223456789, and no counterexample has been found yet.
# Acknowledgments

We sincerely thank Aldo Roberto Pessolano for having helped us very much with the search of perfect powers belonging to $A001292$ and $A352991$, letting us confirm Kashihara's conjecture up to $10^{1235}$ and Conjecture 1 up to $(1.01 \cdot 10)^{34}$.

# References

[1] K. Kashihara. *Comments and topics on Smarandache notions and problems*. Erhus University Press, Arizona, 1996.

[2] R. Muller. Sequence A001292 in the On-line Encyclopedia of Integer Sequences (n.d.). https://oeis.org/A001292. Accessed on May 18, 2022.

[3] R. Muller. Sequence A007908 in the On-line Encyclopedia of Integer Sequences (n.d.). https://oeis.org/A007908. Accessed on May 18, 2022.

[4] M. Ripà. On prime factors in old and new sequences of integers. https://vixra.org/pdf/1101.0092v2.pdf. Accessed on May 17, 2022.

[5] M. Ripà. Sequence A352991 in the On-line Encyclopedia of Integer Sequences (Apr. 16, 2022). https://oeis.org/A352991. Accessed on May 18, 2022.

[6] M. Ripà. Sequence A353025 in the On-line Encyclopedia of Integer Sequences (Apr. 17, 2022). https://oeis.org/A353025. Accessed on May 18, 2022.