*Article* **Algebraic Numbers as Product of Powers of Transcendental Numbers**

## **Pavel Trojovský**

Department of Mathematics, Faculty of Science, University of Hradec Králové, 500 03 Hradec Králové, Czech Republic; pavel.trojovsky@uhk.cz; Tel.: +42-049-333-2801

Received: 17 June 2019; Accepted: 4 July 2019; Published: 8 July 2019

**Abstract:** The elementary symmetric functions play a crucial role in the study of zeros of non-zero polynomials in C[*x*], and the problem of finding zeros in Q[*x*] leads to the definition of algebraic and transcendental numbers. Recently, Marques studied the set of algebraic numbers in the form *P*(*T*)*Q*(*T*). In this paper, we generalize this result by showing the existence of algebraic numbers which can be written in the form *<sup>P</sup>*1(*T*)*Q*1(*T*) ··· *Pn*(*T*)*Qn*(*T*) for some transcendental number *T*, where *P*1, ... , *Pn*, *Q*1, ... , *Qn* are prescribed, non-constant polynomials in Q[*x*] (under weak conditions). More generally, our result generalizes results on the arithmetic nature of *z<sup>w</sup>* when *z* and *w* are transcendental.

**Keywords:** Baker's theorem; Gel'fond–Schneider theorem; algebraic number; transcendental number

## **1. Introduction**

The name "transcendental", which comes from the Latin word "transcendˇere", was first used for a mathematical concept by Leibniz in 1682. Transcendental numbers in the modern sense were defined by Leonhard Euler (see [1]).

A complex number *α* is called algebraic if it is a zero of some non-zero polynomial *P* ∈ Q[*x*]. Otherwise, *α* is transcendental. Algebraic numbers form a field, which is denoted by Q. The transcendence of *e* was proved by Charles Hermite [2] in 1872, and two years later Ferdinand von Lindeman [3] extended the method of Hermite's proof to derive that *π* is also transcendental. It should be noted that Lindemann proved the following, much more general statement: The number *eα*, where *α* is any non-zero algebraic number, is always transcendental (see [4]). In 1900, Hilbert raised the question of the arithmetic nature of the power *α <sup>β</sup>* of two algebraic numbers *α* and *β* (it was the seventh problem in his famous list of 23 problems, which he presented at the International Congress of Mathematicians in Paris). The complete solution to this problem was found independently by Gel'fond and Schneider (see [5], p. 9) in 1934. Their results can be formulated as the following theorem (the ideas of the Gel'fond–Schneider proof were used partially in, e.g., [6–8]).

**Theorem 1.** *The Gel'fond–Schneider Theorem: Let α and β be algebraic numbers, with α* = 0 *and α* = 1*, and let β be irrational. Then α <sup>β</sup> is transcendental.*

The Gel'fond–Schneider Theorem classifies the arithmetic nature of *x<sup>y</sup>* when both *x*, *y* are algebraic numbers (because *x<sup>y</sup>* is an algebraic number when *y* is rational). Nevertheless, when at least one of these two numbers is transcendental, anything is possible (see Table 1 below).


**Table 1.** Possible results for the power *x<sup>y</sup>* when *x* or *y* is transcendental.

In all the previous examples we have *x* = *y* (in fact, we used the fact that the logarithm function is the inverse of the exponential function many times). Also, in the cases in which *x* and *y* are both transcendental (in the previous table), these numbers are possibly (though it's not proved) algebraically independent. So, what happens if we consider numbers of the form *x<sup>x</sup>* with *x* transcendental? Is it possible that some of these numbers are algebraic? We remark that the numbers *e<sup>e</sup>* and *π<sup>π</sup>* are expected (but not proved) to be transcendental. In fact, it is easy to use the Gel'fond–Schneider Theorem to prove that every prime number can be written in the form *T<sup>T</sup>* for some transcendental number *T* (for a more general result, see [9]). In this direction, a natural question arises: Given arbitrary, non-constant polynomials *<sup>P</sup>*, *<sup>Q</sup>* <sup>∈</sup> <sup>Q</sup>[*x*], is there always a transcendental number *<sup>T</sup>* such that *<sup>P</sup>*(*T*)*Q*(*T*) is algebraic? Note that *P*(*T*) and *Q*(*T*) are algebraically dependent transcendental numbers (so they do not come from our table). Marques [10] showed that the answer for the previous question is yes. More generally, he proved that for any fixed, non-constant polynomials *P*(*x*), *Q*(*x*) ∈ Q[*x*], the set of algebraic numbers of the form *P*(*T*)*Q*(*T*), with *T* transcendental, is dense in some connected subset of either R or C. A generalization of this result for rational functions with algebraic coefficients was proved by Jensen and Marques [11]. However, the previous results do not apply, e.g., to prove the existence of algebraic numbers which can be written in the form (*T*<sup>2</sup> <sup>+</sup> <sup>1</sup>)*<sup>T</sup>* · *<sup>T</sup>T*2+*T*<sup>+</sup>1, with *<sup>T</sup>* transcendental.

In this paper, we will solve this kind of problem completely by proving a multi-polynomial version of the previous results. The following theorem states our result more precisely.

**Theorem 2.** *Let P*1, ... , *Pn*, *Q*1, ... , *Qn* ∈ Q[*x*] *be non-constant polynomials, such that the leading coefficients of the Qj's have the same sign. Then the set of algebraic numbers of the form <sup>P</sup>*1(*T*)*Q*1(*T*) ··· *Pn*(*T*)*Qn*(*T*)*, with T transcendental, is dense in some open subset of the complex plane. In fact, this dense set can be chosen to be* {*Q*(<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>) : *<sup>Q</sup>* <sup>∈</sup> *<sup>K</sup>*}*, for some dense set <sup>K</sup>* <sup>⊆</sup> <sup>Q</sup>( √−1)\{0}*, <sup>K</sup>* <sup>∩</sup> <sup>Q</sup> <sup>=</sup> <sup>∅</sup>*, and any prime number p* > 2 · ( max 1≤*j*≤*n* {deg *Qj*})!*.*

The proof of the above theorem combines famous classical theorems concerning transcendental numbers (like the Baker's Theorem on linear forms in logarithms and the Gel'fond–Schneider Theorem) and certain purely field-theoretic results. We point out that, in a similar way, we can prove Theorem 2 for rational functions with algebraic coefficients, but we choose to prove this simpler case in order to avoid too many technicalities, which can obscure the essence of the main idea.

#### **2. Proof of Theorem 2**

## *2.1. Auxiliary Results*

Before we proceed to the proof of Theorem 2, we will need the following three lemmas. The first two lemmas come from the work of Baker on linear forms of logarithms of algebraic numbers (see [5], Chapter 2):

**Lemma 1** (*Cf. Theorem 2.4 in [5])***.** *If α*1, *α*2, ... , *α<sup>n</sup> are algebraic numbers other than* 0 *or* 1*, β*1, *β*2, ... , *β<sup>n</sup> are algebraic with* 1, *<sup>β</sup>*1, *<sup>β</sup>*2,..., *<sup>β</sup><sup>n</sup> linearly independent over* <sup>Q</sup>*, then <sup>α</sup>β*<sup>1</sup> <sup>1</sup> *<sup>α</sup>β*<sup>2</sup> <sup>2</sup> ··· *<sup>α</sup>β<sup>n</sup> <sup>n</sup> is transcendental.*

**Lemma 2** (*Cf. Theorem 2.2 in [5]*)**.** *Any non-vanishing linear combination of logarithms of algebraic numbers with algebraic coefficients is transcendental.*

Let F be a family of polynomials. Hereafter, we will denote by RF the set of all the zeros of the polynomials in F. The last of these lemmas is a purely field-theoretical result.

**Lemma 3.** *Let n be any positive integer and let* F *be a family of polynomials in* Q[*x*] *for which there exists a positive integer such that all polynomials in* F *have degree at most* -*. Then for all prime numbers p* > -!*, the following holds:*

$$(1 + \sqrt[n]{2})^n \notin \mathbb{Q}(\mathcal{R}\_{\mathcal{F}}).\tag{1}$$

**Proof of Lemma 3.** Set F = {*F*1, *F*2, ...}, *Kn* = Q(R*F*1···*Fn* ) and *tn* = [*Kn* : Q]. Since *Kn* ⊆ *Kn*+1, then *tn*+<sup>1</sup> = *ntn*, for some positive integer *<sup>n</sup>*. Note that *<sup>n</sup>* = [*Kn*+<sup>1</sup> : *Kn*]=[*Kn*(R*Fn*+<sup>1</sup> ) : *Kn*] ≤ (deg *Fn*+1)! ≤ -!. We claim that (<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>)*<sup>n</sup>* <sup>∈</sup>/ <sup>Q</sup>(RF ) for all integers *<sup>n</sup>* <sup>≥</sup> 1. For the contrary, there exist positive integers *<sup>m</sup>* and *<sup>s</sup>* such that (<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>)*<sup>m</sup>* <sup>∈</sup> *Ks*. Then the degree of (<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>)*<sup>m</sup>* (which is *<sup>p</sup>*) divides *ts*. However, *ts* = *<sup>s</sup>*−<sup>1</sup> ··· -<sup>1</sup>*t*<sup>1</sup> and *p* > -! ≥ max*j*∈[1,*s*−1]{*<sup>j</sup>*, *t*1}, which gives an absurdity. This completes the proof.

With these lemmas in hand, we can proceed to the proof of our main outcome.

## *2.2. The Proof*

In order to simplify our presentation, we use the familiar notation [*a*, *b*] = {*a*, *a* + 1, ... , *b*}, for integers *a* < *b*.

Of course, it is enough to prove our theorem for the case that *P*1, ... , *Pn* are multiplicatively independent. For that, we take an open, simply connected subset Ω of C, such that *Pj*(*x*) ∈ { / 0, 1} for all *x* ∈ Ω and *j* ∈ [1, *n*]. Choosing, for example, the principal branch of the multi-valued logarithm function, the function *f*(*x*) := ∏*<sup>n</sup> <sup>j</sup>*=<sup>1</sup> *Pj*(*x*) *Qj*(*x*) is well defined and analytic in Ω. Moreover, *f*(*x*) is a non-constant function. In fact, if *f* were constant then *f* (*x*) = 0 in Ω and so

$$\sum\_{j=1}^{n} Q\_j'(\mathbf{x}) \log P\_j(\mathbf{x}) + \sum\_{j=1}^{n} \frac{Q\_j(\mathbf{x}) P\_j'(\mathbf{x})}{P\_j(\mathbf{x})} = 0,\tag{2}$$

for all *<sup>x</sup>* <sup>∈</sup> <sup>Ω</sup>. We claim that *<sup>g</sup>*(*x*) :<sup>=</sup> <sup>∑</sup>*<sup>n</sup> <sup>j</sup>*=<sup>1</sup> *Qj*(*x*)*P <sup>j</sup>*(*x*)/*Pj*(*x*) is not the zero function in Ω. In fact, otherwise *G*(*x*) := *P*1(*x*)··· *Pn*(*x*)*g*(*x*) would be the zero polynomial, but the formal polynomial *G* has degree ≤ *<sup>t</sup>* := max*j*∈[1,*n*]{*m*<sup>1</sup> + ··· + *mn* + *tj* − <sup>1</sup>}, where for all *<sup>j</sup>* ∈ [1, *<sup>n</sup>*], *mj* and *tj* are the degree of *Pj* and *Qj*, respectively. Now, if *ti*<sup>1</sup> <sup>=</sup> ··· <sup>=</sup> *tis* <sup>=</sup> max*j*∈[1,*n*]{*tj*}, we get the relation <sup>∑</sup>*<sup>s</sup> <sup>j</sup>*=<sup>1</sup> *mij bij* = 0 (the coefficient of *<sup>x</sup><sup>t</sup>* in *<sup>G</sup>* must be zero), where for all *<sup>j</sup>* ∈ [1, *<sup>n</sup>*], *bj* is the leading coefficient of *Qj*. However ∑*s <sup>j</sup>*=<sup>1</sup> *mij bij* = 0, since *mj* > 0 and *bj* have the same sign. This gives a contradiction. Thus, there exists *<sup>β</sup>* <sup>∈</sup> <sup>Ω</sup> <sup>∩</sup> <sup>Q</sup> such that *<sup>g</sup>*(*β*) <sup>=</sup> 0. Substituting then *<sup>x</sup>* <sup>=</sup> *<sup>β</sup>* in (2), we have that <sup>∑</sup>*<sup>n</sup> <sup>j</sup>*=<sup>1</sup> *Q j* (*β*)log *Pj*(*β*) is a nonzero algebraic number which contradicts Lemma 2. Hence *f* is a non-constant function.

Since *f* is a non-constant analytic function and Ω is an open connected set, *f*(Ω) is an open connected subset of C. Let F be the family of polynomials {*Qi*(*x*) − *<sup>d</sup>* : *<sup>i</sup>* ∈ [1, *<sup>n</sup>*], *<sup>d</sup>* ∈ Q}∪{*x*<sup>2</sup> + <sup>1</sup>}. Clearly, each polynomial in F has degree ≤ 2- := 2 max{deg *Q*1, ... , deg *Qn*}. Thus, the conditions to apply Lemma 3 are fulfilled. Hence, for *p* > 2-!, we have that the set <sup>P</sup> :<sup>=</sup> {*r*(<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>) : *<sup>r</sup>* <sup>∈</sup> Q( √−1)\{0}} forms a dense subset of <sup>C</sup> and no positive integer power of its elements lies in <sup>Q</sup>(RF ). Since *f*(Ω) is open, *f*(Ω) ∩ P is dense in *f*(Ω). Now, it remains to prove that every number in this intersection can be written in the desired form. For that, let *<sup>α</sup>* :<sup>=</sup> *<sup>r</sup>*(<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>) <sup>∈</sup> *<sup>f</sup>*(Ω) ∩ P, then

$$\mathfrak{a} = f(T) = \prod\_{j=1}^{n} P\_j(T)^{Q\_j(T)} \, , \tag{3}$$

where *T* ∈ Ω. Therefore, it is enough to prove that *T* is a transcendental number. To get a contradiction, suppose the contrary; i.e., that *T* is algebraic. Then *P*1(*T*), ... , *Pn*(*T*), *Q*1(*T*), ... , *Qn*(*T*) are also algebraic numbers. By the choice of Ω, Lemma 1 ensures the existence of a nontrivial Q-relation among 1, *Q*1(*T*),..., *Qn*(*T*) (this implies, in particular, that the degree of *T* is at most -). Without loss of generality we can assume that *anQn*(*T*) = *a*<sup>0</sup> + ∑*n*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> *ajQj*(*T*), where *aj* is an integer, with *an* > 0. Therefore, identity (3) becomes

$$\alpha^{a\_n} = P\_n(T)^{a\_0} \left( P\_1(T)^{a\_n} P\_n(T)^{a\_1} \right)^{Q\_1(T)} \cdots \left( P\_{n-1}(T)^{a\_n} P\_n(T)^{a\_{n-1}} \right)^{Q\_{n-1}(T)}.$$

Note that *αanPn*(*T*)−*a*<sup>0</sup> is an algebraic number and *Pj*(*T*)*anPn*(*T*) *aj* <sup>=</sup> 0 for *<sup>j</sup>* <sup>∈</sup> [1, *<sup>n</sup>* <sup>−</sup> <sup>1</sup>]. We claim that *Pj*(*T*)*anPn*(*T*) *aj* <sup>=</sup> 1 for some *<sup>j</sup>* <sup>∈</sup> [1, *<sup>n</sup>* <sup>−</sup> <sup>1</sup>]. In fact, otherwise we would have *<sup>α</sup>an* <sup>=</sup> *Pn*(*T*)*a*<sup>0</sup> <sup>∈</sup> <sup>Q</sup>(*T*) and so (<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>)*an* <sup>∈</sup> <sup>Q</sup>(*T*, √−1) has degree at most 2-. However, this gives an absurdity since the degree of (<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>)*an* is *<sup>p</sup>* <sup>&</sup>gt; <sup>2</sup>-!. Thus, sometimes *Pj*(*T*)*anPn*(*T*) *aj* is an algebraic number different from 0 and 1, so we can apply Lemma <sup>1</sup> again to get a <sup>Z</sup>-relation *bn*−<sup>1</sup>*Qn*−1(*T*) = *<sup>b</sup>*<sup>0</sup> <sup>+</sup> <sup>∑</sup>*n*−<sup>2</sup> *<sup>j</sup>*=<sup>1</sup> *bjQj*(*T*), where *bj* is an integer, with *bn*−<sup>1</sup> > 0. Analogously, one can iterate this process *n* − 1 times to conclude that

$$\alpha^{q} = A(T) \left( P\_1(T)^{\varepsilon\_1} \cdots P\_n(T)^{\varepsilon\_n} \right)^{Q\_1(T)},\tag{4}$$

where *<sup>A</sup>*(*T*) <sup>∈</sup> <sup>Q</sup>(*P*1(*T*), ... , *Pn*(*T*)) and *<sup>q</sup>*, *cj*'s <sup>∈</sup> <sup>Z</sup>, with *<sup>q</sup>* <sup>&</sup>gt; 0. If <sup>∏</sup>*<sup>n</sup> <sup>j</sup>*=<sup>1</sup> *Pj*(*T*) *cj* = 1, we would arrive at the same absurdity as before since <sup>Q</sup>(*P*1(*T*), ... , *Pn*(*T*)) <sup>⊆</sup> <sup>Q</sup>(*T*). Thus <sup>∏</sup>*<sup>n</sup> <sup>j</sup>*=<sup>1</sup> *Pj*(*T*) *cj* <sup>∈</sup> Q\{0, 1}, so by the Gel'fond–Schneider Theorem we deduce that *Q*1(*T*) is a rational number, say *<sup>r</sup>*/*s*, with some integers *<sup>r</sup>* and *<sup>s</sup>*, *<sup>s</sup>* <sup>&</sup>gt; 0. Hence, *<sup>T</sup>* belongs to <sup>R</sup>*Q*1(*x*)−*r*/*<sup>s</sup>* ⊆ RF . But then *<sup>α</sup>qs* <sup>=</sup> *<sup>A</sup>*(*T*)*sP*1(*T*)*rc*<sup>1</sup> ··· *Pn*(*T*)*rcn* (see (4)) and thus (<sup>1</sup> <sup>+</sup> <sup>√</sup>*<sup>p</sup>* <sup>2</sup>)*qs* <sup>∈</sup> <sup>Q</sup>(RF ), contradicting the choice of *<sup>p</sup>* in Lemma 3. In conclusion, *T* must be transcendental, and this completes the proof.

#### **3. Conclusions**

In this paper, we use analytic (complex analysis), algebraic (Galois' extensions and symmetry) and transcendental tools (Baker's theory) to prove, in particular, the existence of infinitely many algebraic numbers of the form *<sup>P</sup>*1(*T*)*Q*1(*T*) ··· *Pn*(*T*)*Qn*(*T*), where *<sup>T</sup>* is a transcendental number and *P*1, ... , *Pn*, *Q*1, ... , *Qn* are previously fixed rational polynomials (under some weak technical conditions).

**Funding:** The author was supported by Project of Excelence PrF UHK, University of Hradec Králové, Czech Republic 01/2019.

**Acknowledgments:** The author thanks anonymous referees for their careful corrections and their valuable comments that helped to improve the paper's quality and readability.

**Conflicts of Interest:** The author declares no conflict of interest.

#### **References**


c 2019 by the author. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

## *Article* **On Some Formulas for Kaprekar Constants**

## **Atsushi Yamagami \* and Y ¯uki Matsui**

Department of Information Systems Science, Soka University, Tokyo 192-8577, Japan **\*** Correspondence: yamagami@soka.ac.jp

Received: 3 June 2019 ; Accepted: 2 July 2019; Published: 5 July 2019

**Abstract:** Let *b* ≥ 2 and *n* ≥ 2 be integers. For a *b*-adic *n*-digit integer *x*, let *A* (resp. *B*) be the *b*-adic *n*-digit integer obtained by rearranging the numbers of all digits of *x* in descending (resp. ascending) order. Then, we define the *Kaprekar transformation T*(*b*,*n*)(*x*) := *A* − *B*. If *T*(*b*,*n*)(*x*) = *x*, then *x* is called a *b*-*adic n*-*digit Kaprekar constant*. Moreover, we say that a *b*-adic *n*-digit Kaprekar constant *x* is *regular* when the numbers of all digits of *x* are distinct. In this article, we obtain some formulas for regular and non-regular Kaprekar constants, respectively. As an application of these formulas, we then see that for any integer *b* ≥ 2, the number of *b*-adic odd-digit regular Kaprekar constants is greater than or equal to the number of all non-trivial divisors of *b*. Kaprekar constants have the symmetric property that they are fixed points for recursive number theoretical functions *T*(*b*,*n*).

**Keywords:** Kaprekar constants; Kaprekar transformation; fixed points for recursive functions

**MSC:** 2010: 11A99; 11P99


## **1. Introduction**

Let Z be the set of all rational integers. In this article, the symbol [*α*] with any rational number *α* stands for the greatest integer that is less than or equal to *α*.

For integers *b* ≥ 2 and *n* ≥ 2, we denote by Z(*b*, *n*) the set of all *b*-adic *n*-digit integers, i.e.,

$$\begin{aligned} \mathbb{Z}(b,n) &= \{ \mathbf{x} \in \mathbb{Z} \mid 0 \le \mathbf{x} \le b^n - 1 \} \\ &= \{ a\_{n-1} b^{n-1} + \dots + a\_1 b + a\_0 \mid 0 \le a\_0, a\_1, \dots, a\_{n-1} \le b - 1 \} .\end{aligned}$$

For any:

$$\mathbf{x} = a\_{n-1}b^{n-1} + \dots + a\_1b + a\_0 \in \mathbb{Z}(b, n)$$

with 0 ≤ *a*0, *a*1,..., *an*−<sup>1</sup> ≤ *b* − 1, we denote the *b*-adic expression of *x* by:

$$\mathbf{x} = (a\_{n-1} \cdot \cdots \cdot a\_1 a\_0)\_b \dots$$

In the case where *b* = 10, we omit the subscript as:

$$\mathbf{x} = a\_{n-1} \cdot \cdots \cdot a\_1 a\_0$$

as usual if any confusion occurs with the product of *a*0, *a*1,..., *an*−1.

**Definition 1.** *Let cn*−<sup>1</sup> ≥···≥ *c*<sup>1</sup> ≥ *c*<sup>0</sup> *be the rearrangement of the numbers a*0, *a*1, ... , *an*−<sup>1</sup> *of all digits of x* ∈ Z(*b*, *n*) *in descending order. We define the Kaprekar transformation as:*

$$T\_{(b,n)}: \mathbb{Z}(b,n) \to \mathbb{Z}(b,n); \quad \mathbf{x} \mapsto (\mathfrak{c}\_{n-1} \cdot \cdots \cdot \mathfrak{c}\_1 \mathfrak{c}\_0)\_b - (\mathfrak{c}\_0 \mathfrak{c}\_1 \cdot \cdots \cdot \mathfrak{c}\_{n-1})\_{b-1}$$

**Definition 2.** (1) *For any x* ∈ Z(*b*, *n*)*, we say that x is a b-adic n-digit Kaprekar constant if T*(*b*,*n*)(*x*) = *x.*

(2) *We see immediately that zero is a b-adic n-digit Kaprekar constant for any b* ≥ 2 *and n* ≥ 2*, which we call the trivial Kaprekar constant. Then, we denote by ν*(*b*, *n*) *the number of all b-adic n-digit non-trivial Kaprekar constants. By Ref. [1] (Proposition 1.3), we see that:*

$$\nu(b, n) \le\_{b-1+\left[\frac{n}{2}\right]} \mathbb{C}\_{\left[\frac{n}{2}\right]} - 1,$$

*where we put:*

$$\,\_r\mathbb{C}\_s := \frac{r!}{s!(r-s)!} = \frac{r(r-1)\cdots(r-s+1)}{s\cdots 1}$$

*for any integers r* > *s* > 0*.*

(3) *We say that a b-adic n-digit non-trivial Kaprekar constant x* = (*an*−<sup>1</sup> ··· *a*1*a*0)*<sup>b</sup> is regular when ai* = *aj for any i* = *j. We denote by ν*reg(*b*, *n*) *(resp. ν*non*-*reg(*b*, *n*)*) the number of all b-adic n-digit regular (resp. non-regular) Kaprekar constants. By the definition, we see immediately that:*

$$\nu(b, n) = \nu\_{\text{reg}}(b, n) + \nu\_{\text{ron}\cdot \text{reg}}(b, n)$$

*and if b* < *n, then ν*reg(*b*, *n*) = 0 *and ν*(*b*, *n*) = *ν*non*-*reg(*b*, *n*)*.*

**Example 1.** *Kaprekar [2,3], who was the initiator of this research, discovered that ν*(10, 4) = 1*, and the only non-trivial* 10*-adic four-digit Kaprekar constant is:* 6174*.*


**Example 2.** *Here is the list of all b-adic n-digit non-trivial Kaprekar constants for* 2 ≤ *b* ≤ 15 *and* 2 ≤ *n* ≤ 7*. Note that, in the list below, we omit the subscript b. Further, the symbol* − *means that ν*(*b*, *n*) = 0*, and non-trivial Kaprekar constants with the symbol* ∗ *are regular.*

*Then, we obtain the following list of the numbers ν* = *ν*(*b*, *n*)*, ν*<sup>r</sup> = *ν*reg(*b*, *n*) *and ν*nr = *ν*non*-*reg(*b*, *n*)*:*


Now, we have the following:

**Questions:** (1) Are there any formulas for *ν*(*b*, *n*), *ν*reg(*b*, *n*) and *ν*non-reg(*b*, *n*) in terms of *b* and *n*?

(2) Are there any formulas for *b*-adic *n*-digit regular or non-regular Kaprekar constants in terms of *b* and *n*?

**Known results:** There are some known results that answer some parts of the questions above as follows:

(1) In the case where *n* = 2, by the results on the two-digit Kaprekar transformation given by Young [4] (cf. [1], Theorem 3.1), we see that for any integer *b* ≥ 2, there exists a *b*-adic two-digit non-trivial Kaprekar constant if and only if *b* + 1 is divisible by three.

Since there is no two-digit non-regular Kaprekar constant by definition, we see immediately that for any integer *b* ≥ 2,

$$
\nu\_{\text{norm-reg}}(b, \mathcal{Q}) = 0 \quad \text{and} \quad \nu(b, \mathcal{Q}) = \nu\_{\text{reg}}(b, \mathcal{Q}) .
$$

In this article, we shall prove in Theorem 4(1) and Corollary 3(1) that any two-digit regular Kaprekar constant is of the form:

$$(m(2m+1))\_{3m+2}$$

with any integers *m* ≥ 0 and:

$$\nu\_{\text{reg}}(b,2) = \begin{cases} 1 & \text{if } 3 \mid (b+1), \\ 0 & \text{otherwise}. \end{cases}$$

(2) In the case where *n* = 3, Eldridge and Sagong [5] proved that any three-digit non-trivial Kaprekar constant is of the form:

$$(m(2m+1)(m+1))\_{2m+2}$$

with any integers *m* ≥ 0 and that for any integer *b* ≥ 2,

$$\nu(b,3) = \begin{cases} 1 & \text{if } b \text{ is even,} \\ 0 & \text{if } b \text{ is odd.} \end{cases}$$

In particular, we see immediately that:

$$\nu\_{\text{reg}}(b,3) = \begin{cases} 1 & \text{if } b \ge 4 \text{ is even,} \\ 0 & \text{if } b = 2 \text{ or } b \ge 3 \text{ is odd.} \end{cases}$$

and:

$$\nu\_{\text{non-reg}}(b, \mathfrak{z}) = \begin{cases} 1 & \text{if } b = 2, \\ 0 & \text{if } b \ge 3. \end{cases}$$

(3) In the case where *n* = 4, Hasse and Prichett [6] obtained a formula:

$$((3m+3)m(4m+3)(2m+2))\_{5m+5}$$

for (5*m* + 5)-adic four-digit non-trivial Kaprekar constants with any integer *m* ≥ 0. This implies that if *b* ≥ 5 and 5 | *b*, then *ν*reg(*b*, 4) ≥ 1.

In this article, we shall prove in Theorem 4(2) and Corollary 3(2) that any four-digit regular Kaprekar constant is equal to (3021)<sup>4</sup> or given by the above formula obtained by Hasse and Prichett with *m* ≥ 1 and that for any integer *b* ≥ 2,

$$\nu\_{\text{reg}}(b,4) = \begin{cases} 1 & \text{if } b = 4 \text{ or, } b \ge 10 \text{ and } 5 \mid b, \\ 0 & \text{otherwise.} \end{cases}$$

(4) In the case where *n* = 5, Prichett [7] obtained a formula:

$$((2m+2)m(3m+2)(2m+1)(m+1))\_{3m+3}$$

for (3*m* + 3)-adic five-digit non-trivial Kaprekar constants with any integers *m* ≥ 0. This implies that if *b* ≥ 6 and 3 | *b*, then *ν*reg(*b*, 5) ≥ 1.

In this article, we shall prove in Theorem 3(1) and Corollary 3(3) that any five-digit regular Kaprekar constant is given by the above formula obtained by Prichett with *m* ≥ 1 and that for any integer *b* ≥ 2,

$$\nu\_{\text{reg}}(b,5) = \begin{cases} 1 & \text{if } b \ge 6 \text{ and } 3 \mid b, \\ 0 & \text{otherwise.} \end{cases}$$

(5) In the case where *b* = 2, the first author [1] showed that for any *n* ≥ 2, all two-adic *n*-digit non-trivial Kaprekar constants are of the form:

$$\underbrace{1\cdots1}\_{(1\cdots1}0\underbrace{1\cdots1}\_{0\cdots1}0\cdots01\_{1\cdots01}1\_{2\cdots1}$$

with all integers 1 ≤ *k* ≤ *n* 2 and *<sup>ν</sup>*(2, *<sup>n</sup>*) = *<sup>n</sup>* 2 . In particular, we see immediately that:

$$\nu\_{\text{reg}}(2,n) = \begin{cases} 1 & \text{if } n = 2, \\ 0 & \text{if } n \ge 3 \end{cases}$$

and:

$$\nu\_{\text{non-reg}}(2,n) = \begin{cases} 0 & \text{if } n = 2, \\ \left\lceil \frac{n}{2} \right\rceil & \text{if } n \ge 3. \end{cases}$$

(6) In the case where *b* = 3, the authors [8] showed that for any *n* ≥ 2, all three-adic *n*-digit non-trivial Kaprekar constants are of the form:

$$\underbrace{\cdots^{k}}\_{(2\cdots 2}\underbrace{1\cdots 1}\_{1\cdots 1}0\underbrace{2\cdots 2}\_{2\cdots 2}\overbrace{1\cdots 1}^{\ell-k}\overbrace{0\cdots 0}^{k-1}\_{1\cdots 0}1\_{3\cdots 1}$$

with all pairs (*k*, -) of integers satisfying 0 < *k* < and *n* = 3-− *k*, and:

$$\nu(3,n) = \left[\frac{1}{6}\left(n - \frac{1+3(-1)^n}{2}\right)\right].$$

In particular, we see immediately that:

$$\nu\_{\text{reg}}(\mathfrak{J}, \mathfrak{n}) = 0, \quad \nu\_{\text{norm-reg}}(\mathfrak{J}, \mathfrak{n}) = \nu(\mathfrak{J}, \mathfrak{n}).$$

We have the impression that the behavior of the values of *ν*(*b*, *n*), *ν*reg(*b*, *n*) and *ν*non-reg(*b*, *n*) in the list in Example 2 is not only complicated, but also suggestive of some general rules. It seems that it is very hard to obtain general results without observing any case-by-case results. The aim of this article is to see formulas for *b*-adic *n*-digit regular and non-regular Kaprekar constants and to study the properties of *ν*reg(*b*, *n*) and *ν*non-reg(*b*, *n*) towards answers to the questions above.

Firstly, we see formulas for Kaprekar constants in the following:

**Theorem 1.** *Let m* ≥ 0 *and n* ≥ 2 *be any integers. We put:*

$$b(m,n) = \begin{cases} 3m+2 & \text{if } n=2, \\ 2\frac{n-4}{2}(4m+3)+m+2 & \text{if } n \text{ is even and } n \ge 4, \\ \frac{n+1}{2}(m+1) & \text{if } n \text{ is odd.} \end{cases}$$

(1) *We assume that n is even and define the b*(*m*, *n*)*-adic n-digit integer:*

$$K(m,n) = \begin{cases} (m(2m+1))\_{h(m,2)} & \text{if } n=2, \\ ((3m+3)m(4m+3)(2m+2))\_{h(m,4)} & \text{if } n=4, \end{cases}$$

$$K(m,n) = \begin{cases} ((3m+3)m(4m+3)(2m+2))\_{b(m,4)} & \text{if } n=4, \\ (a\_{n-1}a\_{n-2}\cdots a\_{\frac{n}{2}+1}a\_{\frac{n}{2}}a\_{\frac{n}{2}-1}\cdots a\_{\frac{n}{j}}\cdots a\_1a\_0)\_{b(m,n)} & \text{if } n\ge 6, \end{cases}$$

*where we put:*

$$\begin{aligned} a\_{n-1} &= 2^{\frac{n-4}{2}}(4m+3) - m, \\ a\_i &= (2^{\frac{n-4}{2}} - 2^{n-i-2})(4m+3) + m + 1 \qquad \text{for } n-2 \ge i \ge \frac{n}{2} + 1, \\ a\_{\frac{n}{2}} &= m, \\ a\_j &= 2^{j-1}(4m+3) \qquad \text{for } \frac{n}{2} - 1 \ge j \ge 1, \\ a\_0 &= 2m+2. \end{aligned}$$

*Then, K*(*m*, *n*) *is a non-trivial Kaprekar constant, which is regular if and only if n* = 2 *or m* ≥ 1*.* (2) *We assume that n is odd and define the b*(*m*, *n*)*-adic n-digit integer:*

$$L(m,n) = \begin{cases} (m(2m+1)(m+1))\_{b(m,3)} & \text{if } n=3, \\ (b\_{n-1}\cdot\cdot b\_i\cdot\cdots\cdot b\_{\frac{n+3}{2}}b\_{\frac{n+1}{2}}b\_{\frac{n-1}{2}}b\_{\frac{n-3}{2}}\cdot\cdots\cdot b\_j\cdot\cdots\cdot b\_1b\_0)\_{b(m,n)} & \text{if } n\ge 5, \end{cases}$$

*where we put:*

$$\begin{aligned} b\_i &= \left(i - \frac{n-1}{2}\right)(m+1) & \text{for } n-1 \ge i \ge \frac{n+3}{2}, \\ b\_{\frac{n+1}{2}} &= m, \\ b\_{\frac{n-1}{2}} &= \frac{n+1}{2}(m+1) - 1, \\ b\_j &= (j+1)\left(m+1\right) - 1\left(=b\_{\frac{n+1}{2}+j} - 1\right) & \text{for } \frac{n-3}{2} \ge j \ge 1, \\ b\_0 &= m+1. \end{aligned}$$

*Then, L*(*m*, *n*) *is a non-trivial Kaprekar constant, which is regular if and only if m* ≥ 1*.*

**Remark 1.** (1) *We can see that for any integer n* ≥ 2*, the sequence:*

$$b(n) := \{ b(m, n) \mid m = 0, 1, 2, \dots \}$$

*consisting of bases defined in Theorem 1 is the arithmetic progression with the common difference:*

$$\begin{cases} 3 & \text{if } n = 2, \\ 2^{\frac{n}{2}} + 1 & \text{if } n \text{ is even and } n \ge 4, \\ \frac{n+1}{2} & \text{if } n \text{ is odd} \end{cases}$$

*and the first term:*

$$\begin{cases} 2 & \text{if } n = 2, \\ 3 \times 2^{\frac{n-4}{2}} + 2 & \text{if } n \text{ is even and } n \ge 4, \\ \frac{n+1}{2} & \text{if } n \text{ is odd.} \end{cases}$$

(2) *As we have seen in the known results above, the regular Kaprekar constants K*(*m*, 4), *L*(*m*, 3)*, and L*(*m*, 5) *have already been obtained by Hasse and Prichett [6], Eldridge and Sagong [5], and Prichett [7], respectively.*

**Definition 3.** (1) *We call the double series:*

$$\begin{aligned} K &:= \{ K(m, n) \mid m = 1, 2, 3, \dots, n = 2, 4, 6, \dots \}, \\ L &:= \{ L(m, n) \mid m = 1, 2, 3, \dots, n = 3, 5, 7, \dots \} \end{aligned}$$

*the systems of regular Kaprekar constants.*

(2) *Let n* ≥ 2 *be any integer. We call the sequence:*

$$\begin{aligned} K(n) &:= \{ K(m, n) \mid m = 1, 2, 3, \dots \} \quad \text{with even } n \text{ or } \\ L(n) &:= \{ L(m, n) \mid m = 1, 2, 3, \dots \} \quad \text{with odd } n \end{aligned}$$

*the progression of n-digit regular Kaprekar constants with arithmetic progression b*(*n*) \ {*b*(0, *n*)} *of bases.*

*By Theorem 1, we see that the formulas for the numbers an*−1, ... , *a*<sup>0</sup> *(resp. bn*−1, ... , *b*0*) of digits of members in K*(*n*) *(resp. L*(*n*)*) are given by polynomials in m of degree one. This implies that they can be regarded as arithmetic progressions indexed by m* = 1, 2, 3, ...*, as well as the arithmetic progression b*(*n*) \ {*b*(0, *n*)} *of bases.*

(3) *Let m* ≥ 1 *be any integer. We call the sequences:*

$$\begin{aligned} K[m] &:= \{ K(m, n) \mid n = 2, 4, 6, \dots \} \\ \text{(resp. } L[m] &:= \{ L(m, n) \mid n = 3, 5, 7, \dots \} \end{aligned}$$

*the m-th chain of regular Kaprekar constants in the system K (resp. L) with ascending even (resp. odd) digits.*

**Example 3.** (1) *Here are examples of some members K*(*m*, *n*) *in the progressions K*(*n*) *and the chains K*[*m*] *of regular Kaprekar constants with* 1 ≤ *m* ≤ 5 *and n* = 2, 4, 6*.*


(2) *Here are examples of some members L*(*m*, *n*) *in the progressions L*(*n*) *and the chains L*[*m*] *of regular Kaprekar constants with* 1 ≤ *m* ≤ 5 *and n* = 3, 5, 7*.*


**Remark 2.** *By the cases where n* = 4 *and n* = 6 *in the lists in Examples 2 and 3, we see that the progressions K*(*n*) *and L*(*n*) *of n-digit regular Kaprekar constants may not consist of all n-digit regular Kaprekar constants in general. Actually, for any n* ≥ 2*, it seems that it is very hard to obtain formulas for all n-digit regular Kaprekar constants. In Section 2, we obtain some partial results on them with specified n.*

As a corollary of Theorem 1, we immediately obtain some results on the positivity of the numbers *ν*reg(*b*, *n*) of all *b*-adic *n*-digit regular Kaprekar constants as in the following:

**Corollary 1.** (1) *Let n* ≥ 2 *and b* ≥ 2 *be any integers. If n and b satisfy one of the following conditions:*

$$\begin{aligned} \text{(i) } n &= 2 \text{ and } b = 3m + 2 \text{ with } m \ge 1, \\ \text{(ii) } n &\text{ is even, } n \ge 4 \text{ and } b = 2^{\frac{n-4}{2}}(4m+3) + m + 2 \text{ with } m \ge 1, \\ \text{(iii) } n &\text{ is odd and } b = \frac{n+1}{2}(m+1) \text{ with } m \ge 1, \end{aligned}$$

*then:*

$$
v\_{\text{reg}}(b, n) \ge 1.
$$

(2) *If an integer b* ≥ 4 *is not a prime number, then for any non-trivial divisor d of b,*

$$\nu\_{\text{reg}}(b, 2d - 1) \ge 1.$$

*Therefore, the number of all b-adic odd-digit regular Kaprekar constants is greater than or equal to the number of all non-trivial divisors of b.*

Secondly, we obtain formulas for non-regular Kaprekar constants by means of double series of regular Kaprekar constants obtained in Theorem 1 in the following:

**Theorem 2.** *Let the notation be as in Theorem 1.*

(1) *We assume that m* ≡ 1 (mod 3) *and n* ≡ 0 (mod 4)*, and put:*

$$
\beta\_{m,n} = \frac{b(m,n) - 1}{3}.
$$

*For any integer r* ≥ 2*, we denote by K*(*m*, *n*,*r*) *the b*(*m*, *n*)*-adic* (*n* + 2*r*)*-digit integer:*

$$\left( (3m+3) \overbrace{\beta\_{m,4} \cdot \cdots \cdot \beta\_{m,4}}^{r} m(4m+3) \overbrace{(2\beta\_{m,4}) \cdot \cdots \cdot (2\beta\_{m,4})}^{r} (2m+2) \right)\_{b(m,4)}$$

*in the case where n* = 4*, and:*

$$\left(a\_{n-1}\cdot\cdot a\_{\frac{n}{2}+1}\widetilde{\beta\_{m,n}\cdot\cdot\cdot\beta\_{m,n}}a\_{\frac{n}{2}}a\_{\frac{n}{2}-1}\widetilde{(2\beta\_{m,n})\cdot\cdot\cdot(2\beta\_{m,n})}a\_{\frac{n}{2}-2}\cdot\cdot\cdot a\_0\right)\_{b(m,n)}$$

*in the case where n* ≥ 8*. Then, K*(*m*, *n*,*r*) *is a non-regular Kaprekar constant.*

(2) *We assume that m* = 1*, n* ≡ 3 (mod 6) *and n* ≥ 9*. For any integer r* ≥ 2*, we denote by L*(1, *n*,*r*) *the b*(1, *n*)(= *n* + 1)*-adic* (*n* + 2*r*)*-digit integer:*

$$\left( b\_{n-1} \cdots b\_{\frac{2n}{3}} \overbrace{\frac{n}{3} \cdots \frac{n}{3}}^{r} b\_{\frac{2n}{3}} \cdots b\_{\frac{n}{3}} \overbrace{\frac{2n}{3} \cdots \frac{2n}{3}}^{r} b\_{\frac{n}{3}-1} \cdots b\_0 \right)\_{b(1,n)}$$

$$\left( \left( \mathrel{\phantom{\phantom{\phantom{\pi}}}}\_{\left(b(1,n),n\right)} \left( n \cdots \frac{2n+3}{3} \underbrace{\overbrace{\frac{2n}{3} \cdots \frac{2n}{3}}^{r} \overbrace{\frac{2n}{3}}^{r} \cdots \overbrace{\frac{n+3}{3}}^{r+3} \overbrace{\frac{n}{3} \cdots \frac{n}{3}}^{r} \overbrace{\frac{n}{3} \cdots \frac{n}{3}}^{r} \dots 1 \right) \right) \right)\_{b(1,n)}$$

*Then, L*(1, *n*,*r*) *is a non-regular Kaprekar constant.*

**Example 4.** (1) *Here is an example of the non-regular constant K*(*m*, *n*,*r*) *obtained in Theorem 2*(1) *in the case where m* = 4*, n* = 8*, and r* = 2*.*


(2) *Here is an example of the non-regular constant L*(1, *n*,*r*) *obtained in Theorem 2*(2) *in the case where n* = 9 *and r* = 4*.*


As a corollary of Theorem 2, we immediately obtain the following result on the positivity of the numbers *ν*reg(*b*, *n*) of all *b*-adic *n*-digit non-regular Kaprekar constants:

**Corollary 2.** *For any integers m* ≥ 1 *and n* ≥ 4 *satisfying:*

*m* ≡ 1 (mod 3) *and n* ≡ 0 (mod 4)

*or:*

$$m = 1, \ n \equiv 3 \pmod{6} \text{ and } n \ge 9,$$

*and for any integer r* ≥ 2*, we see that:*

$$\nu\_{\text{non-reg}}(b(m,n),n+2r) \ge 1.$$

In Section 1, we shall prove Theorems 1 and 2 and Corollaries 1 and 2. In Section 2.1, we shall obtain some formulas for all *n*-digit regular Kaprekar constants in Theorem 3 for *n* = 5, 7, 9, 11 and Theorem 4 for *n* = 2, 4, 6, 8. Moreover, we shall see some conditional results on formulas for *n*-digit regular Kaprekar constants in Proposition 1 for *n* = 13, 15, 17. Then, we shall see in Section 2.2 some observations on the values of *ν*reg(*b*, *n*). We think that this article is fit for the Special Issue "Number Theory and Symmetry," since Kaprekar constants have the symmetric property that they are fixed points for recursive number theoretical functions *T*(*b*,*n*).

## **2. Proofs of Theorems and Corollaries in the Introduction**

In this section, we prove Theorem 1 and Corollary 1 on regular Kaprekar constants and Theorem 2 and Corollary 2 on non-regular Kaprekar constants, respectively.

#### *2.1. A Proof of Theorem 1*

(1) Let the notation be as in Part (1) of Theorem 1. Here, we omit proving the Parts (i)–(iii), since they can be checked by direct calculations.

(iv) In the case where *n* ≥ 8 is even, let:

$$\mathcal{K}(m,n) = (a\_{n-1}a\_{n-2}\cdots a\_{i}\cdots \cdot a\_{\frac{n}{2}+1}a\_{\frac{n}{2}}a\_{\frac{n}{2}-1}\cdots \cdot a\_{j}\cdots \cdot a\_{1}a\_{0})\_{b(m,n)}$$

be the *b*(*m*, *n*)-adic *n*-digit integer defined in the assertion of Theorem 1(1). Let *cn*−<sup>1</sup> ≥···≥ *c*<sup>1</sup> ≥ *c*<sup>0</sup> be the rearrangement of the numbers *a*0, ... , *an*−<sup>1</sup> of all digits of *K*(*m*, *n*) in descending order. Then, the relation between *a*0,..., *an*−<sup>1</sup> and *c*0,..., *cn*−<sup>1</sup> is given as in the following:

**Lemma 1.** *In the situation above, we see that:*

$$\begin{array}{ll} \mathcal{c}\_{n-1} = a\_{\frac{n}{2}-1} \; \mathcal{c}\_{n-2} = a\_{n-1}, \\\\ \mathcal{c}\_{i} = a\_{i+1} \; \mathcal{c}\_{n-i-1} = a\_{n-i-2} \quad \text{for } n-3 \ge i \ge \frac{n}{2}, \\\\ \mathcal{c}\_{1} = a\_{0} \; \mathcal{c}\_{0} = a\_{\frac{n}{2}}. \end{array}$$

**Proof.** Since for any *<sup>n</sup>* <sup>−</sup> <sup>3</sup> <sup>≥</sup> *<sup>i</sup>* <sup>≥</sup> *<sup>n</sup>* 2 ,

$$\begin{aligned} a\_{i+1} &= (2^{\frac{n-4}{2}} - 2^{n-i-3})(4m+3) + m+1, \\ a\_{n-i-2} &= 2^{n-i-3}(4m+3), \end{aligned}$$

we see easily that:

$$a\_{n-2} > a\_{n-3} > \cdots > a\_{\frac{n}{2}+1}$$

and:

$$a\_{\frac{\Phi}{2}-2} > a\_{\frac{\Phi}{2}-3} > \cdots > a\_1.$$

Moreover,

$$\begin{aligned} a\_{\frac{n}{2}-1} &= 2^{\frac{n-4}{2}}(4m+3) \\ &\ge 2^{\frac{n-4}{2}}(4m+3) - m = a\_{n-1} \\ &> 2^{\frac{n-4}{2}}(4m+3) - (3m+2) = a\_{n-2}, \\ a\_{\frac{n}{2}+1} - a\_{\frac{n}{2}-2} &= (2^{\frac{n-4}{2}} - 2^{\frac{n}{2}-3})(4m+3) + (m+1) - 2^{\frac{n}{2}-3}(4m+3) \\ &= m+1 > 0 \end{aligned}$$

and:

$$a\_1 = 4m + 3 > a\_0 = 2m + 2 > a\_{\frac{m}{2}} = m.$$

Therefore, the lemma is proven.

We put:

$$T\_{\left(b(m,n),n\right)}\left(K(m,n)\right) = \left(a'\_{n-1} \cdot \cdots \cdot a'\_1 a'\_0\right)\_{b\left(m,n\right)}$$

with integers 0 ≤ *a* <sup>0</sup>, *a* <sup>1</sup>, ... , *a <sup>n</sup>*−<sup>1</sup> <sup>≤</sup> *<sup>b</sup>*(*m*, *<sup>n</sup>*) <sup>−</sup> 1. By Ref. [1] (Theorem 1.1 (6)) and Lemma 1, we then see that:

*a <sup>n</sup>*−<sup>1</sup> <sup>=</sup> *cn*−<sup>1</sup> <sup>−</sup> *<sup>c</sup>*<sup>0</sup> <sup>=</sup> *<sup>a</sup> <sup>n</sup>* <sup>2</sup> <sup>−</sup><sup>1</sup> − *a <sup>n</sup>* <sup>2</sup> = <sup>2</sup> *n*−4 <sup>2</sup> (4*m* + 3) − *m* = *an*−1, *a <sup>n</sup>*−<sup>2</sup> = *cn*−<sup>2</sup> − *<sup>c</sup>*<sup>1</sup> = *an*−<sup>1</sup> − *<sup>a</sup>*<sup>0</sup> = <sup>2</sup> *n*−4 <sup>2</sup> (4*m* + 3) − *m* − (2*m* + 2) = (2 *n*−4 <sup>2</sup> − 1)(4*m* + 3) + *m* + 1 = *an*−2, *a n* 2 = *c <sup>n</sup>* <sup>2</sup> − *c <sup>n</sup>* <sup>2</sup> <sup>−</sup><sup>1</sup> − <sup>1</sup> = *<sup>a</sup> <sup>n</sup>* <sup>2</sup> <sup>+</sup><sup>1</sup> − *a <sup>n</sup>* <sup>2</sup> <sup>−</sup><sup>2</sup> − 1 = (2 *n*−4 <sup>2</sup> − 2 *n* <sup>2</sup> <sup>−</sup>3)(4*<sup>m</sup>* <sup>+</sup> <sup>3</sup>)+(*<sup>m</sup>* <sup>+</sup> <sup>1</sup>) <sup>−</sup> <sup>2</sup> *n* <sup>2</sup> <sup>−</sup>3(4*<sup>m</sup>* <sup>+</sup> <sup>3</sup>) <sup>−</sup> <sup>1</sup> = *m* = *a <sup>n</sup>* 2 , *a n* <sup>2</sup> <sup>−</sup><sup>1</sup> <sup>=</sup> *<sup>b</sup>*(*m*, *<sup>n</sup>*) <sup>−</sup> <sup>1</sup> <sup>−</sup> (*<sup>c</sup> <sup>n</sup>* <sup>2</sup> − *c <sup>n</sup>* <sup>2</sup> <sup>−</sup>1) = 2 *n*−4 <sup>2</sup> (4*m* + 3) + *m* + 2 − 1 − (*m* + 1) = 2 *n*−4 <sup>2</sup> (4*m* + 3) = *a <sup>n</sup>* <sup>2</sup> <sup>−</sup>1, *a* <sup>1</sup> = *b*(*m*, *n*) − 1 − (*cn*−<sup>2</sup> − *c*1) = 2 *n*−4 <sup>2</sup> (4*m* + 3) + *m* + 2 − 1 − ((2 *n*−4 <sup>2</sup> − 1)(4*m* + 3) + *m* + 1) = 4*m* + 3 = *a*1, *a* <sup>0</sup> = *b*(*m*, *n*) − (*cn*−<sup>1</sup> − *c*0) = 2 *n*−4 <sup>2</sup> (4*m* + 3) + *m* + 2 − (2 *n*−4 <sup>2</sup> (4*m* + 3) − *m*) = 2*m* + 2 = *a*0.

Moreover, we see that for any *<sup>n</sup>* <sup>−</sup> <sup>3</sup> <sup>≥</sup> *<sup>i</sup>* <sup>≥</sup> *<sup>n</sup>* <sup>2</sup> <sup>+</sup> 1,

$$\begin{aligned} a'\_i &= c\_i - c\_{n-i-1} = a\_{i+1} - a\_{n-i-2} \\ &= (2^{\frac{n-4}{2}} - 2^{n-i-3})(4m+3) + m + 1 - 2^{n-i-3}(4m+3) \\ &= (2^{\frac{n-4}{2}} - 2^{n-i-2})(4m+3) + m + 1 = a\_{i\prime} \\ a'\_{n-i-1} &= b(m,n) - 1 - a'\_i \\ &= 2^{\frac{n-4}{2}}(4m+3) + m + 2 - 1 - ((2^{\frac{n-4}{2}} - 2^{n-i-2})(4m+3) + m + 1) \\ &= 2^{n-i-2}(4m+3) = a\_{n-i-1}. \end{aligned}$$

Therefore, we see that:

$$\begin{aligned} T\_{(b(m,n),n)}(K(m,n)) &= (a\_{n-1}' \cdots a\_1' a\_0')\_{b(m,n)} \\ &= (a\_{n-1} \cdots a\_1 a\_0)\_{b(m,n)} \\ &= K(m,n), \end{aligned}$$

i.e., *K*(*m*, *n*) is a non-trivial Kaprekar constant, which is regular if and only if *m* ≥ 1, which implies that *a <sup>n</sup>* <sup>2</sup> <sup>−</sup><sup>1</sup> = *an*−1.

(2) Let the notation be as in Part (2) of Theorem 1.

As we have seen in the known results (2) and (4) in the Introduction, the cases where *n* = 3 and *n* = 5 have already been proven by Eldridge and Sagong [5] and Prichett [7], respectively. Therefore, it suffices to prove Part (2) in the case where *n* ≥ 7.

For any odd integer *n* ≥ 7, let:

$$L(m,n) = (b\_{n-1} \cdot \cdots \cdot b\_i \cdot \cdots \cdot b\_{\frac{n+3}{2}} b\_{\frac{n+1}{2}} b\_{\frac{n-1}{2}} b\_{\frac{n-3}{2}} \cdots \cdot b\_j \cdot \cdots \cdot b\_1 b\_0)\_{\mathbb{b}(m,n)}$$

be the *b*(*m*, *n*)-adic *n*-digit integer defined in the assertion of Theorem 1(2). Let *cn*−<sup>1</sup> ≥···≥ *c*<sup>1</sup> ≥ *c*<sup>0</sup> be the rearrangement of the numbers *b*0, ... , *bn*−<sup>1</sup> of all digits of *L*(*m*, *n*) in descending order. Then, the relation between *b*0, *b*1,..., *bn*−<sup>1</sup> and *c*0, *c*1,..., *cn*−<sup>1</sup> is given as in the following:

**Lemma 2.** *In the situation above, we see that:*

$$\begin{aligned} c\_{n-1} &= b\_{\frac{n-1}{2}}, \\ c\_{2i-1} &= b\_{\frac{n-1}{2} + i'} \ c\_{2i-2} = b\_{i-1} \quad \text{for } \frac{n-1}{2} \ge i \ge 2, \\ c\_1 &= b\_{0'} \ c\_0 = b\_{\frac{n+1}{2}}. \end{aligned}$$

**Proof.** By the definition of the numbers of all digits of *L*(*m*, *n*) in Theorem 1(2), we see immediately that:

$$\begin{aligned} c\_{n-1} &= \frac{n+1}{2}(m+1) - 1 = b\_{\frac{n-1}{2}}, \\ c\_{2i-1} &= i(m+1) = b\_{\frac{n-1}{2} + i'}, \\ c\_1 &= m+1 = b\_{0'} \ c\_0 = m = b\_{\frac{n+1}{2}}. \end{aligned}$$

Therefore, the lemma is proven.

Then, we can prove Part (2) in the case where *n* ≥ 7 by the same argument as in the proof of Theorem 1(1)(iv). Therefore, we omit the details of the calculations here.

## *2.2. A Proof of Corollary 1*

(1) In Cases (i) and (ii), we have the *b*(*m*, *n*)-adic *n*-digit regular Kaprekar constant *K*(*m*, *n*) by Theorem 1 (1). On the other hand, in Case (iii), we have the *b*(*m*, *n*)-adic *n*-digit regular Kaprekar constant *L*(*m*, *n*) by Theorem 1(2). Therefore, we see that for any integers *b* ≥ 2 and *n* ≥ 2 satisfying Condition (i), (ii), or (iii),

$$\vee\_{\text{reg}}(b, n) \ge 1,$$

and Part (1) is proven.

(2) For any integer *b* ≥ 4 that is not a prime number, let *d* be any non-trivial divisor of *b*, i.e., *d* is a divisor of *b* satisfying 1 < *d* < *b*. We put:

$$m\_d = \frac{b}{d} - 1, \quad n\_d = 2d - 1.$$

Since *md* ≥ 1 is an integer and *nd* ≥ 3 is an odd integer satisfying *b*(*md*, *nd*) = *b*, by Theorem 1(2), we have the *b*-adic *nd*-digit regular Kaprekar constant *L*(*md*, *nd*). Therefore, we see that:

$$\nu\_{\text{reg}}(b, n\_d) \ge 1.$$

Moreover, since *nd* = *nd* for any non-trivial divisors *d* = *d* of *b*, we see that *L*(*md*, *nd*) = *L*(*md* , *nd* ). Therefore, the number of all *b*-adic odd-digit regular Kaprekar constants is greater than or equal to the number of all non-trivial divisors of *b*, and Part (2) is proven.

*2.3. A Proof of Theorem 2*

(1) We assume that:

```
m ≡ 1 (mod 3) and n ≡ 0 (mod 4).
```
(a) In the case where *n* = 4, *b*(*m*, 4) = 5*m* + 5, and:

$$
\beta\_{m,4} = \frac{b(m,4) - 1}{3} = \frac{5m + 4}{3}
$$

which is an integer, since the assumption *m* ≡ 1 (mod 3) implies that:

$$b(m, 4) \equiv 2m - 1 \equiv 1 \pmod{3}.$$

Then, for any *r* ≥ 2, the *b*(*m*, 4)-adic (2*r* + 4)-digit integer obtained by rearranging of the numbers of all digits of *K*(*m*, 4,*r*) in descending order is:

$$\left( (4m+3)\overbrace{(2\beta\_{m,4})\cdot\cdot(2\beta\_{m,4})}^{r}(3m+3)(2m+2)\overbrace{\beta\_{m,4}\cdot\cdot\cdot\beta\_{m,4}}^{r}m \right)\_{b\left(m,4\right)} \dots$$

By Ref. [1] (Theorem 1.1 (6)) and the case where *n* = 4 in Theorem 1(1), we then see that:

$$T\_{(b(m,4),2r+4)}(K(m,4,r)) = K(m,4,r)\_{r'} $$

since *b*(*m*, 4) − 1 − *βm*,4 = 2*βm*,4. Therefore, *K*(*m*, 4,*r*) is a non-regular Kaprekar constant. (b) In the case where *n* ≥ 8, *b*(*m*, *n*) = 2 *n*−4 <sup>2</sup> (4*m* + 3) + *m* + 2, and

$$\beta\_{m,m} = \frac{b(m,n) - 1}{3} = \frac{1}{3} \left( 2^{\frac{n-4}{2}} (4m + 3) + m + 1 \right)$$

which is an integer, since *<sup>n</sup>* <sup>≡</sup> <sup>0</sup> (mod 4) implies that *<sup>n</sup>*−<sup>4</sup> <sup>2</sup> is even and *m* ≡ 1 (mod 3) implies that:

$$b(m,n) \equiv (-1)^{\frac{n-4}{2}}m + m - 1 \equiv 1 \pmod{3}.$$

Let the notation be as in Theorem 1(1). Since, *n* ≥ 8, we see that:

$$\begin{split} a\_{\frac{n}{2}-2} - \beta\_{m,n} &= 2^{\frac{n}{2}-3}(4m+3) - \frac{1}{3} \left( 2^{\frac{n-4}{2}}(4m+3) + m + 1 \right) \\ &= \left( \frac{2^{\frac{n}{2}}}{6} - \frac{1}{3} \right) m + \frac{2^{\frac{n}{2}}}{8} - \frac{1}{3} > 0, \\ \beta\_{m,n} - a\_{\frac{n}{2}-3} &= \frac{1}{3} \left( 2^{\frac{n-4}{2}}(4m+3) + m + 1 \right) - 2^{\frac{n}{2}-4}(4m+3) \\ &= \left( \frac{2^{\frac{n}{2}}}{12} + \frac{1}{3} \right) m + \frac{2^{\frac{n}{2}}}{16} + \frac{1}{3} > 0, \\ a\_{\frac{n}{2}+2} - 2\beta\_{m,n} &= \left( 2^{\frac{n}{2}-2} - 2^{\frac{n}{2}-4} \right)(4m+3) + m + 1 \\ & \qquad - \frac{2}{3} \left( 2^{\frac{n-4}{2}}(4m+3) + m + 1 \right) \end{split}$$

$$\begin{aligned} &= \left(\frac{2^{\frac{\pi}{2}}}{12} + \frac{1}{3}\right)m + \frac{2^{\frac{\pi}{2}}}{16} + \frac{1}{3} > 0, \\ 2\beta\_{m,n} - a\_{\frac{\pi}{2} + 1} &= \frac{2}{3}\left(2^{\frac{n-4}{2}}(4m+3) + m + 1\right) \\ &\qquad - \left(\left(2^{\frac{\pi}{2} - 2} - 2^{\frac{\pi}{2} - 3}\right)(4m+3) + m + 1\right) \\ &= \left(\frac{2^{\frac{\pi}{2}}}{6} - \frac{1}{3}\right)m + \frac{2^{\frac{\pi}{2}}}{8} - \frac{1}{3} > 0. \end{aligned}$$

By Ref. [1] (Theorem 1.1 (6)) and Lemma 1, we then see that:

$$T\_{(b(m,n),n+2r)}(K(m,n,r)) = K(m,n,r)\_{\ast}$$

since *b*(*m*, *n*) − 1 − *βm*,*<sup>n</sup>* = 2*βm*,*n*. Therefore, *K*(*m*, *n*,*r*) is a *b*(*m*, *n*)-adic (*n* + 2*r*)-digit non-regular Kaprekar constant for any *r* ≥ 2.

By (a) and (b) above, Part (1) of Theorem 2 is proven.

(2) We assume that:

 $m = 1$ ,  $n \equiv 3$  (mod 6) and  $n \ge 9$ .

Let the notation be as in Theorem 1 (2). By the definition in *loc. cit.*, the *b*(1, *n*)(= *n* + 1)-adic *n*-digit integer obtained by rearranging of the numbers of all digits *b*0, *b*1, ... , *bn*−<sup>1</sup> of *L*(1, *n*) in descending order is:

$$(n \ (n-1) \cdot \cdot \cdot \cdot \cdot \, \, \, \, \, \, 1)\_{b(1\mu)}$$

given by all integers from 1–*n*. By Ref. [1] (Theorem 1.1 (8)) and Theorem 1 (2), we then see that:

$$\begin{aligned} &T\_{(b(1,n),n+2r)}(L(1,n,r)) \\ &= T\_{(b(1,n),n)}\left(n\cdot\cdot\frac{2n+3}{3}\cdot\frac{2n}{3}\cdot\cdot\frac{2n}{3}\cdot\frac{2n}{3}\cdot\cdot\frac{n+3}{3}\cdot\frac{n}{3}\cdot\frac{n}{3}\cdot\frac{n}{3}\cdot\cdots\right), \\ &= L(1,n,r), \end{aligned}$$

since *n* ≥ 9 and *b*(1, *n*) − 1 − -2*n* <sup>3</sup> <sup>−</sup> *<sup>n</sup>* 3 <sup>=</sup> <sup>2</sup>*<sup>n</sup>* <sup>3</sup> . Therefore, *<sup>L</sup>*(1, *<sup>n</sup>*,*r*) is a *<sup>b</sup>*(1, *<sup>n</sup>*)-adic (*<sup>n</sup>* <sup>+</sup> <sup>2</sup>*r*)-digit non-regular Kaprekar constant for any *r* ≥ 2, and Part (2) of Theorem 2 is proven.

**Remark 3.** *Although we omit the proof here, we can see that for any integer m* ≥ 2 *and odd integer n* ≥ 3*, it is impossible to construct any b*(*m*, *n*)*-adic* (*n* + 2*r*)*-digit non-regular Kaprekar constant by adding βm*,*n's and* (2*βm*,*n*)*'s to the b*(*m*, *n*)*-adic expression of the b*(*m*, *n*)*-adic n-digit regular Kaprekar constant L*(*m*, *n*)*, as well as in Part* (1) *of Theorem 2.*

## *2.4. A Proof of Corollary 2*

We assume that:

$$m \equiv 1 \pmod{3} \text{ and } n \equiv 0 \pmod{4}$$

(resp.

$$m = 1, \ n \equiv 3 \pmod{6} \text{ and } n \ge 9).$$

By Theorem 2, for any integer *r* ≥ 2, we then have the *b*(*m*, *n*)-adic (*n* + 2*r*)-digit non-regular Kaprekar constant *K*(*m*, *n*,*r*) (resp. *L*(1, *n*,*r*)). Therefore, we see that:

$$\nu\_{\text{non-reg}}(b, n+2r) \ge 1,$$

and Corollary 2 is proven.

## **3. On** *n***-Digit Regular Kaprekar Constants with Specified** *n*

## *3.1. Some Formulas for All n-Digit Regular Kaprekar Constants with Specified n*

Let *K*(*n*) and *L*(*n*) be the progressions of *n*-digit regular Kaprekar constants defined in Definition 3(2) for even and odd positive integers *n*, respectively. On the other hand, it seems that it is very hard to obtain formulas for *all n*-digit regular Kaprekar constants. In this subsection, we shall obtain partial results on such formulas by case-by-case arguments.

Firstly, we shall see formulas for all *n*-digit regular Kaprekar constants in the cases where *n* = 5, 7, 9, 11 in Theorem 3. Note that, in the case where *n* = 3, Eldridge and Sagong [5] already proved that a three-digit integer *x* is a regular Kaprekar constant if and only if *x* ∈ *L*(3), i.e., *x* is of the form:

$$(m(2m+1)(m+1))\_{2m+2}$$

with *m* ≥ 1.

Although one can obtain a similar result for each odd integer *n* ≥ 13, the authors would not like to do tedious calculations for solving simultaneous equations obtained by the uniqueness of *b*-adic expressions of any positive integer for any integer *b* ≥ 2.

**Theorem 3.** (1) *A five-digit integer x is a regular Kaprekar constant if and only if x* ∈ *L*(5)*, i.e., x is of the form:*

$$((2m+2)m(3m+2)(2m+1)(m+1))\_{3m+3}$$

*with m* ≥ 1*.*

(2) *A seven-digit integer x is a regular Kaprekar constant if and only if x* ∈ *L*(7)*, i.e., x is of the form:*

$$((\mathfrak{Im} + \mathfrak{Z})(2m + 2)m(4m + \mathfrak{Z})(\mathfrak{Im} + \mathfrak{Z})(2m + 1)(m + 1))\_{4m + 4}$$

*with m* ≥ 1*.*

(3) *For any integer b* ≥ 2*, a b-adic nine-digit integer x is a regular Kaprekar constant if and only if x is of the form:*

$$((b-m-1)(b-2m-2)(b-3m-3)m(b-1)(b-m-2)(3m+2)(2m+1)(m+1))\_{b-1}$$

*where the base b is in the range* 5*m* + 4 < *b* < 6*m* + 5 *with m* ≥ 1*.*

*In particular, when b* = 5*m* + 5*, x is a member of L*(9)*.*

(4) *An* 11*-digit integer x is a regular Kaprekar constant if and only if x* ∈ *L*(11)*, i.e., x is of the form:*

$$\begin{aligned} ((5m+5)(4m+4)(3m+3)(2m+2)m(6m+5) \\ (5m+4)(4m+3)(3m+2)(2m+1)(m+1))\_{6m+6} \end{aligned}$$

*with m* ≥ 1*.*

**Proof.** By Theorem 1, it suffices to show that any regular Kaprekar constant in each case is of the form stated in the assertion. In the following, let *b* ≥ 2 be any integer.

(1) For any *b*-adic five-digit regular Kaprekar constant *x*, we denote by (*c*4*c*3*c*2*c*1*c*0)*<sup>b</sup>* with:

$$b - 1 \ge c\_4 > c\_3 > c\_2 > c\_1 > c\_0 \ge 0$$

the rearrangement in descending order of the numbers of all digits of *x*. By Ref. [1] (Theorem 1.1 (7)),

$$\begin{aligned} \mathbf{x} &= T\_{(b,5)}((c\_4 c\_3 c\_2 c\_1 c\_0)\_b) \\ &= ((c\_4 - c\_0)(c\_3 - c\_1 - 1)(b - 1)(b - 1 - (c\_3 - c\_1))(b - (c\_4 - c\_0)))\_b .\end{aligned}$$

We see the following magnitude relations among the numbers of all digits of *x*:

$$\begin{aligned} b - 1 &\ge c\_4 - c\_0 > c\_3 - c\_1 - 1, \\ b - 1 &> b - 1 - (c\_3 - c\_1) > b - (c\_4 - c\_0). \end{aligned}$$

Then, we obtain the following:

**Lemma 3.**

$$\begin{aligned} b - 1 &= c\_{4\prime} & c\_4 - c\_0 &= c\_3 & b - 1 - (c\_3 - c\_1) &= c\_{2\prime}, \\ b - (c\_4 - c\_0) &= c\_1 & \text{and} & c\_3 - c\_1 - 1 &= c\_0. \end{aligned}$$

**Proof.** Since *c*<sup>4</sup> is the maximum number among all digits of *x*,

$$b - 1 = c\_{4\cdot}$$

This implies that:

$$c\_4 - c\_0 = b - 1 - c\_0 \quad \text{and} \quad b - (c\_4 - c\_0) = c\_0 + 1.$$

Since *c*<sup>1</sup> is the second smallest number among all digits of *x*, we then see that:

$$b - (c\_4 - c\_0) = c\_1.$$

This implies that:

$$\infty - c\_1 - 1 = \alpha$$

by the two inequalities above. Moreover, we see that:

$$\begin{aligned} b - 1 - (c\_3 - c\_1) &= b - 2 - c\_0 \\ &< b - 1 - c\_0 = c\_4 - c\_0 \end{aligned}$$

which implies that:

$$c\_4 - c\_0 = c\_3 \quad \text{and} \quad b - 1 - (c\_3 - c\_1) = c\_2$$

as desired.

We then see that the following equality holds:

$$\begin{aligned} & ( (c\_4 - c\_0)(c\_3 - c\_1 - 1)(b - 1)(b - 1 - (c\_3 - c\_1))(b - (c\_4 - c\_0)) )\_b \\ & = (c\_3 c\_0 c\_4 c\_2 c\_1)\_b. \end{aligned}$$

This implies that *b* = 3*c*<sup>0</sup> + 3 and:

$$c\_4 = 3c\_0 + 2, \ c\_3 = 2c\_0 + 2, \ c\_2 = 2c\_0 + 1, \ c\_1 = c\_0 + 1.$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\infty = ((2m+2)m(3m+2)(2m+1)(m+1))\_{3m+3}.$$

If *m* = 0, then we see a contradiction that *x* = (20211)<sup>3</sup> is not regular. Therefore, *m* ≥ 1, and Part (1) is proven.

(2) For any *b*-adic seven-digit regular Kaprekar constant *x*, we denote by (*c*6*c*5*c*4*c*3*c*2*c*1*c*0)*<sup>b</sup>* with:

$$b - 1 \ge c\_6 > c\_5 > c\_4 > c\_3 > c\_2 > c\_1 > c\_0 \ge 0$$

the rearrangement in descending order of the numbers of all digits of *x*. By the same argument as in the proof of Part (1), we then see that one of the following two equalities holds:

((*c*<sup>6</sup> − *c*0)(*c*<sup>5</sup> − *c*1)(*c*<sup>4</sup> − *c*<sup>2</sup> − 1)(*b* − 1)(*b* − 1 − (*c*<sup>4</sup> − *c*2)) (*b* − 1 − (*c*<sup>5</sup> − *c*1))(*b* − (*c*<sup>6</sup> − *c*0)))*<sup>b</sup>* = 2 (*c*5*c*2*c*0*c*6*c*4*c*3*c*1)*<sup>b</sup>* ··· (i) (*c*5*c*3*c*0*c*6*c*4*c*2*c*1)*<sup>b</sup>* ··· (ii)

The equality (i) implies a contradiction that *<sup>c</sup>*<sup>2</sup> <sup>=</sup> <sup>−</sup> <sup>1</sup> 2 . The equality (ii) implies that *b* = 4*c*<sup>0</sup> + 4 and:

$$\begin{aligned} c\mathfrak{c} &= 4\mathfrak{c}\_0 + 3, & c\mathfrak{c} &= 3\mathfrak{c}\_0 + 3, & c\_4 &= 3\mathfrak{c}\_0 + 2, \\ c\mathfrak{c}\_3 &= 2\mathfrak{c}\_0 + 2, & c\_2 &= 2\mathfrak{c}\_0 + 1, & c\_1 &= \mathfrak{c}\_0 + 1. \end{aligned}$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\mathbf{x} = ((3m+3)(2m+2)m(4m+3)(3m+2)(2m+1)(m+1))\_{4m+4}.$$

If *m* = 0, then we see a contradiction that *x* = (3203211)<sup>4</sup> is not regular. Therefore, *m* ≥ 1, and Part (2) is proven.

(3) For any *b*-adic nine-digit regular Kaprekar constant *x*, we denote by (*c*8*c*7*c*6*c*5*c*4*c*3*c*2*c*1*c*0)*<sup>b</sup>* with:

$$b - 1 \ge c\_8 > c\_7 > c\_6 > c\_5 > c\_4 > c\_3 > c\_2 > c\_1 > c\_0 \ge 0$$

the rearrangement in descending order of the numbers of all digits of *x*. By the same argument as in the proof of Part (1), we then see that one of the following six equalities holds:

$$= \begin{cases} (c\_8 - c\_0)(c\_7 - c\_1)(c\_6 - c\_2)(c\_5 - c\_3 - 1)(b - 1)(b - 1 - (c\_5 - c\_3)) \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \begin{cases} (c\_7 c\_6 - c\_1)(c\_6 - c\_2)(c\_5 - c\_3 - 1)(b - 1)(b - 1 - (c\_5 - c\_3)) \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \end{cases} & \text{if } \quad \begin{cases} (c\_7 c\_5 c\_4 c\_0 c\_8 c\_6 c\_3 c\_2 c\_1)\_b & \text{\$ (-c\_7 - c\_1) \choose (b - (c\_8 - c\_0)) \choose b \choose c\_5} \\ (c\_7 c\_5 c\_3 c\_0 c\_8 c\_6 c\_4 c\_2 c\_1)\_b & \text{\$ \cdots \ \$ (i) \$ \$ (c\_7 - c\_3) \choose (b - 1) \choose b \choose c\_5 \text{\$ (i) \$ (c\_7 - c\_3) \choose b \choose c\_5} \\ (c\_7 c\_4 c\_3 c\_0 c\_8 c\_6 c\_5 c\_2 c\_1)\_b & \text{\$ \cdots \ \$ (i) \$ (c\_7 - 1) \choose b \choose c\_5 \text{\$ (i) \$ (c\_7 - c\_3) \choose b \choose c\_5} \\ (c\_7 c\_4 c\_2 c\_0 c\_8 c\_6 c\_5 c\_3 c\_1)\_b & \text{\$ \cdots \ \$ (v) \$ (c\_7 - 1) \choose b \choose c\_5} \\ (c\_7 c\_3 c\_2 c\_0 c\_8 c\_6 c\_5 c\_4 c\_1)\_b & \text{\$ \cdots \ \$ (v) \$ (\$ )} \\ \end{cases}$$

The equalities (i) and (v) imply a contradiction that *c*<sup>4</sup> = *c*3. The equalities (iii), (iv), and (vi) imply a contradiction that *c*<sup>5</sup> = *c*4. The equality (ii) implies that *b* = *c*<sup>3</sup> + 3*c*<sup>0</sup> + 3 and:

$$\begin{aligned} c\_8 &= c\_3 + 3c\_0 + 2, & c\_7 &= c\_3 + 2c\_0 + 2, & c\_6 &= c\_3 + 2c\_0 + 1, \\ c\_8 &= c\_3 + c\_0 + 1, & c\_4 &= 3c\_0 + 2, & c\_2 &= 2c\_0 + 1, & c\_1 &= c\_0 + 1. \end{aligned}$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that *x* is equal to:

$$((b-m-1)(b-2m-2)(b-3m-3)m(b-1)(b-m-2)(3m+2)(2m+1)(m+1))\_{b-m}$$

where the base *b* is in the range 5*m* + 4 < *b* < 6*m* + 5, since:

$$c\_4 = 3m + 2 > c\_3 = b - 3m - 3 > c\_2 = 2m + 1.$$

If *m* = 0, then we see a contradiction that *b* is in the range 4 < *b* < 5. Therefore, *m* ≥ 1, and Part (3) is proven.

(4) For any *b*-adic 11-digit regular Kaprekar constant *x*, we denote by (*c*10*c*9*c*8*c*7*c*6*c*5*c*4*c*3*c*2*c*1*c*0)*<sup>b</sup>* with:

$$b - 1 \ge c\_{10} > c\_9 > c\_8 > c\_7 > c\_6 > c\_5 > c\_4 > c\_3 > c\_2 > c\_1 > c\_0 \ge 0$$

the rearrangement in descending order of the numbers of all digits of *x*. By the same argument as in the proof of Part (1), we then see that one of the following twenty equalities holds:

((*c*<sup>10</sup> − *c*0)(*c*<sup>9</sup> − *c*1)(*c*<sup>8</sup> − *c*2)(*c*<sup>7</sup> − *c*3)(*c*<sup>6</sup> − *c*<sup>4</sup> − 1)(*b* − 1)(*b* − 1 − (*c*<sup>6</sup> − *c*4)) (*b* − 1 − (*c*<sup>7</sup> − *c*3))(*b* − 1 − (*c*<sup>8</sup> − *c*2))(*b* − 1 − (*c*<sup>9</sup> − *c*1))(*b* − (*c*<sup>10</sup> − *c*0)))*<sup>b</sup>* = ⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ (*c*9*c*7*c*6*c*5*c*0*c*10*c*8*c*4*c*3*c*2*c*1)*<sup>b</sup>* ··· (i) (*c*9*c*7*c*6*c*4*c*0*c*10*c*8*c*5*c*3*c*2*c*1)*<sup>b</sup>* ··· (ii) (*c*9*c*7*c*6*c*3*c*0*c*10*c*8*c*5*c*4*c*2*c*1)*<sup>b</sup>* ··· (iii) (*c*9*c*7*c*6*c*2*c*0*c*10*c*8*c*5*c*4*c*3*c*1)*<sup>b</sup>* ··· (iv) (*c*9*c*7*c*5*c*4*c*0*c*10*c*8*c*6*c*3*c*2*c*1)*<sup>b</sup>* ··· (v) (*c*9*c*7*c*5*c*3*c*0*c*10*c*8*c*6*c*4*c*2*c*1)*<sup>b</sup>* ··· (vi) (*c*9*c*7*c*5*c*2*c*0*c*10*c*8*c*6*c*4*c*3*c*1)*<sup>b</sup>* ··· (vii) (*c*9*c*7*c*4*c*3*c*0*c*10*c*8*c*6*c*5*c*2*c*1)*<sup>b</sup>* ··· (viii) (*c*9*c*7*c*4*c*2*c*0*c*10*c*8*c*6*c*5*c*3*c*1)*<sup>b</sup>* ··· (ix) (*c*9*c*7*c*3*c*2*c*0*c*10*c*8*c*6*c*5*c*4*c*1)*<sup>b</sup>* ··· (x) (*c*9*c*4*c*3*c*2*c*0*c*10*c*8*c*7*c*6*c*5*c*1)*<sup>b</sup>* ··· (xi) (*c*9*c*5*c*3*c*2*c*0*c*10*c*8*c*7*c*6*c*4*c*1)*<sup>b</sup>* ··· (xii) (*c*9*c*5*c*4*c*2*c*0*c*10*c*8*c*7*c*6*c*3*c*1)*<sup>b</sup>* ··· (xiii) (*c*9*c*5*c*4*c*3*c*0*c*10*c*8*c*7*c*6*c*2*c*1)*<sup>b</sup>* ··· (xiv) (*c*9*c*6*c*3*c*2*c*0*c*10*c*8*c*7*c*5*c*4*c*1)*<sup>b</sup>* ··· (xv) (*c*9*c*6*c*4*c*2*c*0*c*10*c*8*c*7*c*5*c*3*c*1)*<sup>b</sup>* ··· (xvi) (*c*9*c*6*c*4*c*3*c*0*c*10*c*8*c*7*c*5*c*2*c*1)*<sup>b</sup>* ··· (xvii) (*c*9*c*6*c*5*c*2*c*0*c*10*c*8*c*7*c*4*c*3*c*1)*<sup>b</sup>* ··· (xviii) (*c*9*c*6*c*5*c*3*c*0*c*10*c*8*c*7*c*4*c*2*c*1)*<sup>b</sup>* ··· (xix) (*c*9*c*6*c*5*c*4*c*0*c*10*c*8*c*7*c*3*c*2*c*1)*<sup>b</sup>* ··· (xx)

The equality (i) implies a contradiction that *c*<sup>5</sup> ≤ *c*4.

The equalities (ii), (x), (xii), and (xiii) imply a contradiction that *c*<sup>10</sup> = *c*9.

The equalities (iii), (iv), (vii), (xi), and (xviii) imply a contradiction that *c*<sup>7</sup> < *c*6.

The equality (v) implies a contradiction that *c*<sup>6</sup> < *c*5.

The equalities (viii) and (xvi) imply a contradiction that *c*<sup>7</sup> = *c*6.

The equality (ix) implies that:

$$c\_7 = c\_2 + 2c\_0 + 1, \quad c\_6 = 4c\_0 + 2, \quad c\_3 = 2c\_0 + 1, 2$$

which yields a contradiction that *c*<sup>2</sup> > 2*c*<sup>0</sup> + 1 > *c*2.

The equality (xiv) implies a contradiction that *c*<sup>6</sup> = *c*5. The equality (xv) implies a contradiction that *c*<sup>8</sup> = *c*7.

The equality (xvii) implies that:

$$c\_7 = 2c\_3, \quad c\_6 = 3c\_3 - 2c\_0 - 1, \quad c\_2 = 2c\_0 + 1,$$

which implies a contradiction that *c*<sup>3</sup> > 2*c*<sup>0</sup> + 1 > *c*3.

The equality (xix) implies a contradiction that *c*<sup>7</sup> = 4*c*<sup>0</sup> + 8 3 . The equality (xx) implies a contradiction that *c*<sup>8</sup> < *c*7. The equality (vi) implies that *b* = 6*c*<sup>0</sup> + 6 and:

$$\begin{aligned} \mathbf{c}\_{10} &= \mathbf{6c}\_0 + 5, \ c\_9 = 5\mathbf{c}\_0 + 5, \ c\_8 = 5\mathbf{c}\_0 + 4, \ c\_7 = 4\mathbf{c}\_0 + 4, \ c\_6 = 4\mathbf{c}\_0 + 3, \\ \mathbf{c}\_5 &= 3\mathbf{c}\_0 + 3, \ c\_4 = 3\mathbf{c}\_0 + 2, \ c\_3 = 2\mathbf{c}\_0 + 2, \ c\_2 = 2\mathbf{c}\_0 + 1, \ c\_1 = \mathbf{c} + 1. \end{aligned}$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\begin{aligned} \infty &= ((5m+5)(4m+4)(3m+3)(3m+2)m(6m+5) \\ &= (5m+4)(4m+3)(3m+2)(2m+1)(m+1) \end{aligned}$$

If *m* = 0, then we see a contradiction that *x* = (54320543211)<sup>6</sup> is not regular. Therefore, *m* ≥ 1, and Part (4) is proven.

Secondly, we see formulas for all *n*-digit regular Kaprekar constants in the cases where *n* = 2, 4, 6, 8 in Theorem 4. Although one can obtain a similar result for each even integer *n* ≥ 10, the authors would not like to do tedious calculations for solving simultaneous equations obtained by the uniqueness of *b*-adic expressions of any positive integer for any integer *b* ≥ 2.

Note that we shall need more calculations of solving simultaneous equations in the proof for even cases in Theorem 4 than odd cases in Theorem 3, because, in the case where *n* ≥ 2 is even, the Kaprekar transformation *T*(*b*,*n*) may not necessarily give us the maximum number *b* − 1 among the numbers of all digits.

**Theorem 4.** (1) *A two-digit integer x is a regular Kaprekar constant if and only if x* ∈ *K*(2) ∪ {(01)2}*, i.e, x is of the form:*

$$(m(2m+1))\_{3m+2}$$

*with m* ≥ 0*.*

(2) *A four-digit integer x is a regular Kaprekar constant if and only if x* = (3021)<sup>4</sup> *or x* ∈ *K*(4)*, i.e., x is of the form:*

$$((\mathfrak{Z}m+\mathfrak{Z})m(4m+\mathfrak{Z})(2m+\mathfrak{Z}))\_{5m+5}$$

*with m* ≥ 1*.*

(3) *A six-digit integer x is a regular Kaprekar constant if and only if x is equal to:*

$$\begin{aligned} &(530421)\_6 \\ &((9m+6)(5m+3)(3m+1)(2m+7)(10m+6)(6m+4))\_{15m+10m} \\ &((5m+4)(3m+2)m(6m+4)(4m+3)(2m+2))\_{7m+6} ∨ \\ &((7m+6)(5m+4)m(8m+6)(4m+1)(2m+2))\_{9m+8} \left(\in K(6)\right) \end{aligned}$$

*with m* ≥ 1*.*

(4) *An eight-digit integer x is a regular Kaprekar constant if and only if x is equal to:*

$$\begin{aligned} (97508421)\_{10} \quad &(75306421)\_{8} \\ ((11m+7)(7m+4)(5m+3)(3m+1)(14m+8) \\ &(12m+7)(10m+6)(6m+4))\_{17m+11} \\ ((15m+9)(9m+5)(7m+4)(3m+1)(18m+10) \\ &(14m+8)(12m+7)(6m+4))\_{21m+13} \\ ((13m+10)(11m+8)(7m+5)m(14m+10) \\ &(8m+6)(4m+3)(2m+2))\_{15m+12} \quad ∨ \\ ((15m+12)(13m+10)(9m+7)m(16m+12) \\ &(8m+6)(4m+3)(2m+2))\_{17m+14} \left( \in K(8) \right) \end{aligned}$$

*with m* ≥ 1*.*

**Proof.** (1) For any *b*-adic two-digit regular Kaprekar constant *x*, we denote by *x* = (*c*1*c*0)*<sup>b</sup>* with *b* − 1 ≥ *c*<sup>1</sup> > *c*<sup>0</sup> ≥ 0 the rearrangement in descending order of numbers of all digits of *x*. By Ref. [1] (Theorem 1.1 (2)),

$$\mathfrak{a} = T\_{(b,2)}((c\_1c\_0)\_b) = ((c\_1 - c\_0 - 1)(b - (c\_1 - c\_0)))\_b.$$

We then see that one of the following two equalities holds:

$$((c\_1 - c\_0 - 1)(b - (c\_1 - c\_0)))\_b = \begin{cases} (c\_1 c\_0)\_b & \cdots \text{ (i)}\\ (c\_0 c\_1)\_b & \cdots \text{ (ii)} \end{cases}$$

The equality (i) implies a contradiction that *c*<sup>0</sup> = −1. The equality (ii) implies that:

$$c\_1 = \frac{2b - 1}{3} \quad \text{and} \quad c\_0 = \frac{b - 2}{3}.$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$b = 3m + 2 \quad \text{and} \quad c\_1 = 2m + 1$$

as desired.

(2) For any *b*-adic four-digit regular Kaprekar constant *x*, we denote by (*c*3*c*2*c*1*c*0)*<sup>b</sup>* with *b* − 1 ≥ *c*<sup>3</sup> > *c*<sup>2</sup> > *c*<sup>1</sup> > *c*<sup>0</sup> ≥ 0 the rearrangement in descending order of the numbers of all digits of *x*. By Ref. [1] (Theorem 1.1 (6)),

$$\begin{aligned} \mathbf{x} &= T\_{(b,4)}((c\_3 c\_2 c\_1 c\_0)\_b) \\ &= ((c\_3 - c\_0)(c\_2 - c\_1 - 1)(b - 1 - (c\_2 - c\_1)))(b - (c\_3 - c\_0)))\_b .\end{aligned}$$

Since:

$$c\_3 - c\_0 > c\_2 - c\_1 - 1 \quad \text{and} \quad b - 1 - (c\_2 - c\_1) > b - (c\_3 - c\_0).$$

we see that one of the following six equalities holds:

$$= \begin{cases} (c\_3 - c\_0)(c\_2 - c\_1 - 1)(b - 1 - (c\_2 - c\_1))(b - (c\_3 - c\_0)))\_b \\ \qquad \qquad \qquad \cdot \cdot \text{ (i)}\\ (c\_3 c\_1 c\_2 c\_0)\_b & \text{  $\cdot$  (i)}\\ (c\_3 c\_0 c\_2 c\_1)\_b & \text{  $\cdot$  (i)}\\ (c\_1 c\_0 c\_3 c\_2)\_b & \text{  $\cdot$  (i)}\\ (c\_2 c\_0 c\_3 c\_1)\_b & \text{  $\cdot$  (i)}\\ (c\_2 c\_0 c\_3 c\_0)\_b & \text{  $\cdot$  (i)}\\ (c\_2 c\_1 c\_3 c\_0)\_b & \text{  $\cdot$  (i)} \end{cases}$$

The equalities (i), (ii), and (vi) imply a contradiction that *c*<sup>3</sup> = *b*. The equality (iii) implies that *x* = (3021)4. The equality (iv) implies a contradiction that *c*<sup>3</sup> < *c*2. The equality (v) implies that *b* = 5*c*<sup>0</sup> + 5 and:

$$c\_3 = 4c\_0 + 3, \quad c\_2 = 3c\_0 + 3, \quad c\_1 = 2c\_0 + 2.$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\mathbf{x} = ((3m+3)m(4m+3)(2m+2))\_{5m+5}\mathbf{x}$$

If *m* = 0, then we see a contradiction that *x* = (3032)<sup>5</sup> is not regular. Therefore, *m* ≥ 1, and Part (2) is proven.

(3) For any *b*-adic six-digit regular Kaprekar constant *x*, we denote by (*c*5*c*4*c*3*c*2*c*1*c*0)*<sup>b</sup>* with:

$$b - 1 \ge c\_5 > c\_4 > c\_3 > c\_2 > c\_1 > c\_0 \ge 0$$

the rearrangement in descending order of the numbers of all digits of *x*. By Ref. [1] (Theorem 1.1 (6)),

$$\begin{aligned} \mathbf{x} &= T\_{(b,6)}((c\_5c\_4c\_3c\_2c\_1c\_0)\_b) \\ &= ((c\_5 - c\_0)(c\_4 - c\_1)(c\_3 - c\_2 - 1)(b - 1 - (c\_3 - c\_2)) \\ & & (b - 1 - (c\_4 - c\_1))(b - (c\_5 - c\_0)))\_b. \end{aligned}$$

Since *c*<sup>5</sup> − *c*<sup>0</sup> > *c*<sup>4</sup> − *c*<sup>1</sup> > *c*<sup>3</sup> − *c*<sup>2</sup> − 1 and:

$$b - 1 - (c\_3 - c\_2) > b - 1 - (c\_4 - c\_1) > b - (c\_5 - c\_0)\_2$$

we see that *c*<sup>3</sup> − *c*<sup>2</sup> − 1 = *c*<sup>0</sup> or *b* − (*c*<sup>5</sup> − *c*0) = *c*0. The equality *b* − (*c*<sup>5</sup> − *c*0) = *c*<sup>0</sup> implies a contradiction that *b* = *c*5, and the equality *c*<sup>4</sup> − *c*<sup>1</sup> = *c*<sup>4</sup> implies a contradiction that *c*<sup>1</sup> = 0 > *c*0. Therefore, we see that one of the following nine equalities holds:

$$= \begin{pmatrix} (c\_5 - c\_0)(c\_4 - c\_1)(c\_3 - c\_2 - 1)(b - 1 - (c\_3 - c\_2)) \\ \\ (b - 1 - (c\_4 - c\_1))(b - (c\_5 - c\_0)))\_b \\ \\ (c\_5 c\_2 c\_0 c\_4 c\_3 c\_1)\_b & \cdots & (\text{i}) \\ (c\_5 c\_2 c\_0 c\_4 c\_3 c\_1)\_b & \cdots & (\text{ii}) \\ (c\_2 c\_1 c\_0 c\_3 c\_2 c\_3)\_b & \cdots & (\text{iii}) \\ (c\_3 c\_1 c\_0 c\_5 c\_4 c\_3)\_b & \cdots & (\text{iv}) \\ (c\_3 c\_2 c\_0 c\_5 c\_4 c\_1)\_b & \cdots & (\text{v}) \\ (c\_4 c\_1 c\_0 c\_5 c\_3 c\_2)\_b & \cdots & (\text{vi}) \\ (c\_4 c\_2 c\_0 c\_5 c\_3 c\_1)\_b & \cdots & (\text{vii}) \\ (c\_4 c\_3 c\_0 c\_5 c\_2 c\_1)\_b & \cdots & (\text{vii}) \\ (c\_4 c\_3 c\_0 c\_5 c\_2 c\_1)\_b & \cdots & (\text{vi}) \\ \end{pmatrix}$$

The equality (i) implies that *x* = (530421)6.

The equality (ii) and (iii) imply a contradiction that *c*<sup>2</sup> = *c*1.

The equality (iv) implies that *c*<sup>2</sup> = *c*<sup>0</sup> + 1, which contradicts the condition that *c*<sup>2</sup> > *c*<sup>1</sup> > *c*0. The equality (vi) implies a contradiction that *c*<sup>2</sup> = *c*0.

The equality (vii) implies a contradiction that *x* = (420432)<sup>6</sup> is not regular. The equality (v) implies that *b* = 5*c*<sup>0</sup> + 5 and:

$$\begin{aligned} c\_5 &= 4c\_0 + 3, \quad c\_4 = \frac{10c\_0 + 8}{3}, \quad c\_3 = 3c\_0 + 3, \\ c\_2 &= 2c\_0 + 2, \quad c\_1 = \frac{5c\_0 + 4}{3}. \end{aligned}$$

Putting *c*<sup>0</sup> = 3*m* + 1 with *m* ≥ 0, we then see that:

$$\mathbf{x} = ((9m+6)(5m+3)(3m+1)(12m+7)(10m+6)(6m+4))\_{15m+10\cdot 1}$$

If *m* = 0, then we see a contradiction that *x* = (631764)<sup>10</sup> is not regular. Therefore, *m* ≥ 1. The equality (viii) implies that *b* = 7*c*<sup>0</sup> + 6 and:

$$\begin{aligned} c\_5 &= 6c\_0 + 4, & c\_4 &= 5c\_0 + 4, & c\_3 &= 4c\_0 + 3, \\ c\_2 &= 3c\_0 + 2, & c\_1 &= 2c\_0 + 2. \end{aligned}$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\infty = ((5m+4)(3m+2)m(6m+4)(4m+3)(2m+2))\tau\_{m+6}.$$

If *m* = 0, then we see a contradiction that *x* = (420432)<sup>6</sup> is not regular. Therefore, *m* ≥ 1. The equality (ix) implies that *b* = 9*c*<sup>0</sup> + 8 and:

$$\begin{aligned} c\mathfrak{s} &= 8\mathfrak{c}\_0 + 6, & c\_4 &= 7\mathfrak{c}\_0 + 6, & c\_3 &= 5\mathfrak{c}\_0 + 4, \\ c\_2 &= 4\mathfrak{c}\_0 + 3, & c\_1 &= 2\mathfrak{c}\_0 + 2. \end{aligned}$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\mathbf{x} = ((7m+6)(5m+4)m(8m+6)(4m+3)(2m+2))\_{9m+8-4}$$

If *m* = 0, then we see a contradiction that *x* = (640632)<sup>8</sup> is not regular. Therefore, *m* ≥ 1, and Part (3) is proven.

(4) For any *b*-adic eight-digit regular Kaprekar constant *x*, we denote by (*c*7*c*6*c*5*c*4*c*3*c*2*c*1*c*0)*<sup>b</sup>* with:

$$b - 1 \ge c\_7 > c\_6 > c\_5 > c\_4 > c\_3 > c\_2 > c\_1 > c\_0 \ge 0$$

the rearrangement in descending order of the numbers of all digits of *x*. By Ref. [1] (Theorem 1.1 (6)),

$$\begin{aligned} \mathbf{x} &= T\_{\left(b, \mathbf{8}\right)} \left( \left( c\gamma c\_6 c\_5 c\_4 c\_3 c\_2 c\_1 c\_0 \right)\_b \right) \\ &= \left( \left( c\gamma - c\_0 \right) \left( c\_6 - c\_1 \right) \left( c\_5 - c\_2 \right) \left( c\_4 - c\_3 - 1 \right) \left( b - 1 - \left( c\_4 - c\_3 \right) \right) \right) \\ &\quad \left( b - 1 - \left( c\gamma - c\_2 \right) \right) \left( b - 1 - \left( c\_6 - c\_1 \right) \right) \left( b - \left( c\gamma - c\_0 \right) \right) \right)\_b. \end{aligned}$$

Since *c*<sup>7</sup> − *c*<sup>0</sup> > *c*<sup>6</sup> − *c*<sup>1</sup> > *c*<sup>5</sup> − *c*<sup>2</sup> > *c*<sup>4</sup> − *c*<sup>3</sup> − 1 and:

$$b - 1 - (c\_4 - c\_3) > b - 1 - (c\_5 - c\_2) > b - 1 - (c\_6 - c\_1) > b - (c\_7 - c\_0)\_2$$

we see that *c*<sup>4</sup> − *c*<sup>3</sup> − 1 = *c*<sup>0</sup> or *b* − (*c*<sup>7</sup> − *c*0) = *c*0. The equality *b* − (*c*<sup>7</sup> − *c*0) = *c*<sup>0</sup> implies a contradiction that *b* = *c*7, and the equality *c*<sup>6</sup> − *c*<sup>1</sup> = *c*<sup>6</sup> implies a contradiction that *c*<sup>1</sup> = 0 > *c*0. Therefore, we see that one of the following thirty equalities holds:

((*c*<sup>7</sup> − *c*0)(*c*<sup>6</sup> − *c*1)(*c*<sup>5</sup> − *c*2)(*c*<sup>4</sup> − *c*<sup>3</sup> − 1)(*b* − 1 − (*c*<sup>4</sup> − *c*3)) (*b* − 1 − (*c*<sup>5</sup> − *c*2))(*b* − 1 − (*c*<sup>6</sup> − *c*1))(*b* − (*c*<sup>7</sup> − *c*0)))*<sup>b</sup>* = ⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ (*c*7*c*5*c*4*c*0*c*6*c*3*c*2*c*1)*<sup>b</sup>* ··· (i) (*c*7*c*5*c*3*c*0*c*6*c*4*c*2*c*1)*<sup>b</sup>* ··· (ii) (*c*7*c*5*c*2*c*0*c*6*c*4*c*3*c*1)*<sup>b</sup>* ··· (iii) (*c*7*c*5*c*1*c*0*c*6*c*4*c*3*c*2)*<sup>b</sup>* ··· (iv) (*c*7*c*4*c*3*c*0*c*6*c*5*c*2*c*1)*<sup>b</sup>* ··· (v) (*c*7*c*4*c*2*c*0*c*6*c*5*c*3*c*1)*<sup>b</sup>* ··· (vi) (*c*7*c*4*c*1*c*0*c*6*c*5*c*3*c*2)*<sup>b</sup>* ··· (vii) (*c*7*c*3*c*2*c*0*c*6*c*5*c*4*c*1)*<sup>b</sup>* ··· (viii) (*c*7*c*3*c*1*c*0*c*6*c*5*c*4*c*2)*<sup>b</sup>* ··· (ix) (*c*7*c*2*c*1*c*0*c*6*c*5*c*4*c*3)*<sup>b</sup>* ··· (x) (*c*3*c*2*c*1*c*0*c*7*c*6*c*5*c*4)*<sup>b</sup>* ··· (xi) (*c*4*c*2*c*1*c*0*c*7*c*6*c*5*c*3)*<sup>b</sup>* ··· (xii) (*c*4*c*3*c*1*c*0*c*7*c*6*c*5*c*2)*<sup>b</sup>* ··· (xiii) (*c*4*c*3*c*2*c*0*c*7*c*6*c*5*c*1)*<sup>b</sup>* ··· (xiv) (*c*5*c*2*c*1*c*0*c*7*c*6*c*4*c*3)*<sup>b</sup>* ··· (xv) (*c*5*c*3*c*1*c*0*c*7*c*6*c*4*c*2)*<sup>b</sup>* ··· (xvi) (*c*5*c*3*c*2*c*0*c*7*c*6*c*4*c*1)*<sup>b</sup>* ··· (xvii) (*c*5*c*4*c*1*c*0*c*7*c*6*c*3*c*2)*<sup>b</sup>* ··· (xviii) (*c*5*c*4*c*2*c*0*c*7*c*6*c*3*c*1)*<sup>b</sup>* ··· (xix) (*c*5*c*4*c*3*c*0*c*7*c*6*c*2*c*1)*<sup>b</sup>* ··· (xx) (*c*6*c*2*c*1*c*0*c*7*c*5*c*4*c*3)*<sup>b</sup>* ··· (xxi) (*c*6*c*3*c*1*c*0*c*7*c*5*c*4*c*2)*<sup>b</sup>* ··· (xxii) (*c*6*c*3*c*2*c*0*c*7*c*5*c*4*c*1)*<sup>b</sup>* ··· (xxiii) (*c*6*c*4*c*1*c*0*c*7*c*5*c*3*c*2)*<sup>b</sup>* ··· (xxiv) (*c*6*c*4*c*2*c*0*c*7*c*5*c*3*c*1)*<sup>b</sup>* ··· (xxv) (*c*6*c*4*c*3*c*0*c*7*c*5*c*2*c*1)*<sup>b</sup>* ··· (xxvi) (*c*6*c*5*c*1*c*0*c*7*c*4*c*3*c*2)*<sup>b</sup>* ··· (xxvii) (*c*6*c*5*c*2*c*0*c*7*c*4*c*3*c*1)*<sup>b</sup>* ··· (xxviii) (*c*6*c*5*c*3*c*0*c*7*c*4*c*2*c*1)*<sup>b</sup>* ··· (xxix) (*c*6*c*5*c*4*c*0*c*7*c*3*c*2*c*1)*<sup>b</sup>* ··· (xxx)

The equality (i) implies that *x* = (97508421)10. The equality (ii) implies that *x* = (75306421)8. The equality (iii) implies a contradiction that *c*<sup>6</sup> = *c*4. The equality (iv) implies a contradiction that *c*<sup>5</sup> = *c*3. The equalities (v), (x), (xv), and (xxi) imply a contradiction that *c*<sup>6</sup> = *c*5. The equality (vi) implies a contradiction that *<sup>c</sup>*<sup>2</sup> <sup>=</sup> <sup>5</sup> 3 . The equality (vii) implies a contradiction that *c*<sup>7</sup> < *c*6.

The equalities (viii) and (ix) imply a contradiction that *c*<sup>3</sup> = *c*1. The equalities (xi), (xii), (xiii), and (xiv) imply a contradiction that *c*<sup>2</sup> = *c*1. The equality (xvii) implies a contradiction that *c*<sup>1</sup> = *c*<sup>0</sup> = −2. The equality (xviii) implies a contradiction that *b* = 5*c*<sup>0</sup> + 14 3 . The equality (xix) implies a contradiction that *<sup>b</sup>* <sup>=</sup> <sup>2</sup>*c*<sup>2</sup> <sup>−</sup> <sup>2</sup> 3 . The equality (xx) implies a contradiction that *c*<sup>7</sup> = *c*5. The equality (xxii) implies a contradiction that 4 > *c*<sup>1</sup> > 3. The equality (xxiv) implies a contradiction that *b* = 2*c*<sup>1</sup> + 7 3 . The equality (xxv) implies a contradiction that *c*<sup>5</sup> = 6*c*<sup>1</sup> + 14 3 . The equality (xxvi) implies a contradiction that *c*<sup>4</sup> = *c*1. The equality (xxvii) implies a contradiction that *c*<sup>0</sup> = −1. The equality (xxviii) implies a contradiction that *c*<sup>7</sup> = *c*4. The equality (xvi) implies that *<sup>b</sup>* <sup>=</sup> <sup>17</sup>*c*<sup>0</sup> <sup>+</sup> <sup>16</sup> <sup>3</sup> and:

$$\begin{aligned} c\_7 &= \frac{14c\_0 + 10}{3}, \quad c\_6 = 4c\_0 + 3, \quad c\_5 = \frac{11c\_0 + 10}{3}, \quad c\_4 = \frac{10c\_0 + 8}{3}, \\\ c\_3 &= \frac{7c\_0 + 5}{3}, \quad c\_2 = 2c\_0 + 2, \quad c\_1 = \frac{5c\_0 + 4}{3}. \end{aligned}$$

Putting *c*<sup>0</sup> = 3*m* + 1 with *m* ≥ 0, we then see that:

$$\begin{aligned} \mathbf{x} &= ((11m+7)(7m+4)(5m+3)(3m+1) \\ &= (4m+8)(12m+7)(10m+6)(6m+4) \end{aligned}$$

If *m* = 0, then we see a contradiction that *x* = (74318764)<sup>11</sup> is not regular. Therefore, *m* ≥ 1. The equality (xxiii) implies that *b* = 7*c*<sup>0</sup> + 6 and:

$$\begin{aligned} c\gamma &= 6c\_0 + 4, \quad c\_6 = 5c\_0 + 4, \quad c\_5 = \frac{14c\_0 + 10}{3}, \quad c\_4 = 4c\_0 + 3, \\ c\_3 &= 3c\_0 + 2, \quad c\_2 = \frac{7c\_0 + 5}{3}, \quad c\_1 = 2c\_0 + 2. \end{aligned}$$

Putting *c*<sup>0</sup> = 3*m* + 1 with *m* ≥ 0, we then see that:

$$\begin{aligned} \infty &= ((15m+9)(9m+5)(7m+4)(3m+1) \\ &= (18m+10)(14m+8)(12m+7)(6m+4))\_{21m+13} \end{aligned}$$

If *m* = 0, then we see a contradiction that *x* = (9541(10)874)<sup>13</sup> is not regular. Therefore, *m* ≥ 1. The equality (xxix) implies that *b* = 15*c*<sup>0</sup> + 12 and:

$$\begin{aligned} c\_7 &= 14c\_0 + 10, & c\_6 &= 13c\_0 + 10, & c\_5 &= 11c\_0 + 8, & c\_4 &= 8c\_0 + 6, \\ c\_3 &= 7c\_0 + 5, & c\_2 &= 4c\_0 + 3, & c\_1 &= 2c\_0 + 2. \end{aligned}$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\begin{aligned} x &= ((13m+10)(11m+8)(7m+5)m \\ &(14m+10)(8m+6)(4m+3)(2m+2))\_{15m+12}. \end{aligned}$$

If *m* = 0, then we see a contradiction that *x* = ((10)850(10)632)<sup>12</sup> is not regular. Therefore, *m* ≥ 1.

The equality (xxx) implies that *b* = 17*c*<sup>0</sup> + 14 and:

$$\begin{aligned} \mathbf{c}\_7 &= 16\mathbf{c}\_0 + 12, & \mathbf{c}\_6 &= 15\mathbf{c}\_0 + 12, & \mathbf{c}\_5 &= 13\mathbf{c}\_0 + 10, & \mathbf{c}\_4 &= 9\mathbf{c}\_0 + 7, \\ \mathbf{c}\_3 &= 8\mathbf{c}\_0 + 6, & \mathbf{c}\_2 &= 4\mathbf{c}\_0 + 3, & \mathbf{c}\_1 &= 2\mathbf{c}\_0 + 2. \end{aligned}$$

Putting *m* = *c*<sup>0</sup> ≥ 0, we then see that:

$$\begin{aligned} x &= ((15m+12)(13m+10)(9m+7)m \\ &(16m+12)(8m+6)(4m+3)(2m+2))\_{17m+14}. \end{aligned}$$

If *m* = 0, then we see a contradiction that *x* = ((12)(10)70(12)632)<sup>14</sup> is not regular. Therefore, *m* ≥ 1, and Part (4) is proven.

We shall also obtain some conditional results on formulas for *n*-digit regular Kaprekar constants in the following proposition for which we omit the proof because one can prove them by the same arguments as in the proof of Theorem 3:

**Proposition 1.** *Let the notation be as in Theorem 3. For any integer b* ≥ 2*, we see the following:*

$$(1) \text{ A } b \text{-adic 13-digit integer } \mathbf{x} = (a\_{12} \cdot \dots \cdot a\_0)\_b \text{ with } 0 \le a\_0, \dots, a\_{12} \le b - 1 \text{ satisfying the condition:} $$

$$a\_{11} > a\_4 > a\_{10} > a\_3 > a\_9 > a\_2 > a\_8 > a\_{11}$$

*is a regular Kaprekar constant if and only if x* ∈ *L*(13) *with b* ∈ *b*(13)*, i.e., x is of the form:*

$$\begin{aligned} &((6m+6)(5m+5)(4m+4)(3m+3)(2m+2)m \\ &(7m+6)(6m+5)(5m+4)(4m+3)(3m+2)(2m+1)(m+1))\_{7m+7} \end{aligned}$$

*with m* ≥ 1*.*

(2) *A b-adic* 15*-digit integer x* = (*a*<sup>14</sup> ··· *a*0)*<sup>b</sup> with* 0 ≤ *a*0,..., *a*<sup>14</sup> ≤ *b* − 1 *satisfying the condition:*

$$a\_{13} > a\_5 > a\_{12} > a\_4 > a\_{11} > a\_3 > a\_{10} > a\_2 > a\_3 > a\_1$$

*is a regular Kaprekar constant if and only if x is of the form:*

$$\begin{aligned} \left( (b - m\_1 - 1)(b - 2m\_1 - 2)(b - 3m\_1 - 3)(b - 2m\_1 - m\_2 - 2) \right) \\ (b - 3m\_1 - m\_2 - 3)m\_2 m\_1 (b - 1)(b - m\_1 - 2)(b - m\_2 - 1) \\ (3m\_1 + m\_2 + 2)(2m\_1 + m\_2 + 1)(3m\_1 + 2)(2m\_1 + 1)(m\_1 + 1) \end{aligned}$$

*where m*<sup>1</sup> ≥ 1*, m*<sup>2</sup> *is in the range:*

$$2m\_1 + 1 < m\_2 < 3m\_1 + 2$$

*and b is in the range:*

$$6m\_1 + m\_2 + 5 < b < 5m\_1 + 2m\_2 + 4.1$$

(3) *A b-adic* 17*-digit integer x* = (*a*<sup>16</sup> ··· *a*0)*<sup>b</sup> with* 0 ≤ *a*0,..., *a*<sup>16</sup> ≤ *b* − 1 *satisfying the condition:*

$$a\_{15} > a\_6 > a\_{14} > a\_5 > a\_{13} > a\_4 > a\_{12} > a\_3 > a\_{11} > a\_2 > a\_{10} > a\_{13}$$

*is a regular Kaprekar constant if and only if x is of the form:*

$$\begin{aligned} &\left((b-m-1)(b-2m-2)(b-3m-3) \\ &\quad \left(\frac{3b-7m-7}{4}\right)\left(\frac{3b-11m-11}{4}\right)\left(\frac{b-3m-2}{2}\right)\left(\frac{b-m-1}{4}\right) \\ &\quad m(b-1)(b-m-2) \\ &\quad \left(\frac{3b+m-3}{4}\right)\left(\frac{b+3m+1}{2}\right)\left(\frac{b+11m+7}{4}\right)\left(\frac{b+7m+3}{4}\right) \\ &\quad (3m+2)(2m+1)(m+1))\_{b\prime} \end{aligned}$$

*where b satisfies the conditions:*

$$9m + 7 < b < 11m + 9 \quad \text{and} \quad b \equiv m + 1 \pmod{4}$$

*with m* ≥ 1*.*

## *3.2. Some Observations on ν*reg(*b*, *n*) *with Specified n*

As a corollary to Theorems 3 and 4, we can make some observations on the numbers *ν*reg(*b*, *n*) of all *b*-adic *n*-digit regular Kaprekar constants for *n* = 2, 4, 5, 6, 7, 8, 9, 11 as in the following:

**Corollary 3.** *Let b* ≥ 2 *be any integer. Then, we see the following:*

$$\begin{aligned} (1) \quad & \nu\_{\text{reg}}(b,2) = \begin{cases} 1 & \text{if } 3 \mid (b+1), \\ 0 & \text{otherwise.} \end{cases} \\ (2) \quad & \nu\_{\text{reg}}(b,4) = \begin{cases} 1 & \text{if } b=4 \text{ or, } b \ge 10 \text{ and } 5 \mid b, \\ 0 & \text{else.} \end{cases} \end{aligned}$$

$$(\pi)^{-\text{reag}}\begin{pmatrix} 0 & \text{otherwise.} \end{pmatrix}$$

$$\mathbf{(3)} \quad \nu\_{\text{reg}}(b, \mathbf{5}) = \begin{cases} 1 & \text{if } b \ge \mathbf{6} \text{ and } \mathbf{3} \mid b, \\ 0 & \text{otherwise.} \end{cases}$$

$$(4)\qquad \nu\_{\text{reg}}(b,6) = \begin{cases} 2 & \text{if } b \in (A\_1 \cap A\_2) \cup (A\_2 \cap A\_3), \\ 1 & \text{otherwise,} \\ 0 & \text{if } b \neq 6 \text{ and } b \notin A\_1 \cup A\_2 \cup A\_3. \end{cases}$$

*where the sets A*1*, A*2*, and A*<sup>3</sup> *are defined as:*

$$\begin{aligned} A\_1 &= \{ b \in \mathbb{Z} \mid b \ge 25 \text{ and } b \equiv 10 \pmod{15} \}, \\ A\_2 &= \{ b \in \mathbb{Z} \mid b \ge 13 \text{ and } b \equiv 6 \pmod{7} \}, \\ A\_3 &= \{ b \in \mathbb{Z} \mid b \ge 17 \text{ and } b \equiv 8 \pmod{9} \}. \end{aligned}$$

$$\begin{aligned} (5) \quad &\nu\_{\text{reg}}(b,7) = \begin{cases} 1 & \text{if } b \ge 8 \text{ and } 4 \mid b, \\ 0 & \text{otherwise.} \end{cases} \\ (6) \quad &\nu\_{\text{reg}}(b,8) = \begin{cases} 2 & \text{if } b \in (B\_1 \cap B\_2) \cup (B\_1 \cap B\_3) \cup (B\_2 \cap B\_3) \cup (B\_3 \cap B\_4), \\ 1 & \text{otherwise.} \end{cases} \\ (9) \quad &\text{if } b \ne 8, 10 \text{ and } b \not\not\le B\_1 \cup B\_2 \cup B\_3 \cup B\_4. \end{aligned}$$

*where the sets B*1*, B*2*, B*3*, and B*<sup>4</sup> *are defined as:*

$$\begin{aligned} B\_1 &= \{ b \in \mathbb{Z} \mid b \ge 28 \text{ and } b \equiv 11 \pmod{17} \}, \\ B\_2 &= \{ b \in \mathbb{Z} \mid b \ge 34 \text{ and } b \equiv 13 \pmod{21} \}, \\ B\_3 &= \{ b \in \mathbb{Z} \mid b \ge 27 \text{ and } b \equiv 12 \pmod{15} \}, \\ B\_4 &= \{ b \in \mathbb{Z} \mid b \ge 31 \text{ and } b \equiv 14 \text{ (mod 17)} \}. \\\\ (7) \qquad \nu\_{\text{reg}}(b, 9) &= \begin{cases} \left\lceil \frac{b}{30} \right\rceil + 1 & \text{if } b \equiv 10, 15, 16, 20, 21, 22, 25, 26, 27, 28 \text{ (mod 30)}, \\ & \qquad \left\lceil \frac{b}{30} \right\rceil & \text{otherwise.} \end{cases} \end{aligned}$$

(8) *<sup>ν</sup>*reg(*b*, 11) = <sup>2</sup> 1 if *b* ≥ 12 and 6 | *b*, 0 otherwise.

**Remark 4.** (1) *The intersections of the sets A*1*, A*2*, and A*<sup>3</sup> *in Corollary 3* (4) *are the following:*

$$\begin{aligned} A\_1 \cap A\_2 &= \{ b \in \mathbb{Z} \mid b \ge 55 \text{ and } b \equiv 55 \pmod{105} \}, \\ A\_2 \cap A\_3 &= \{ b \in \mathbb{Z} \mid b \ge 62 \text{ and } b \equiv 62 \pmod{63} \}, \\ A\_1 \cap A\_3 &= \bigotimes. \end{aligned}$$

(2) *The intersections of the sets B*1*, B*2*, B*3*, and B*<sup>4</sup> *in Corollary 3* (6) *are the following:*

$$\begin{aligned} B\_1 \cap B\_2 &= \{ b \in \mathbb{Z} \mid b \ge 181 \text{ and } b \equiv 181 \text{ (mod 357)} \}, \\ B\_1 \cap B\_3 &= \{ b \in \mathbb{Z} \mid b \ge 147 \text{ and } b \equiv 147 \text{ (mod 255)} \}, \\ B\_2 \cap B\_4 &= \{ b \in \mathbb{Z} \mid b \ge 286 \text{ and } b \equiv 286 \text{ (mod 357)} \}, \\ B\_3 \cap B\_4 &= \{ b \in \mathbb{Z} \mid b \ge 255 \text{ and } b \equiv 255 \text{ (mod 255)} \}, \\ B\_1 \cap B\_4 &= B\_2 \cap B\_3 = \mathcal{O}. \end{aligned}$$

**Remark 5.** *We can see that Corollary 3*(1)*–*(5) *matches the values of ν*<sup>r</sup> *in the list in Example 2.*

**Proof.** We see immediately that Parts (1)–(6) and (8) are implied by the respective formulas obtained in Theorem 3(1), (2), (4) and Theorem 4 for the respective digits *n*, since these formulas give distinct *n*-digit regular Kaprekar constants for distinct positive integers *m*, and we see that:

$$A\_1 \cap A\_3 = B\_1 \cap B\_4 = B\_2 \cap B\_3 = \bigcirc$$

as mentioned in Remark 4.

Now, we prove Part (7) for the case where *n* = 9. Since the formula obtained in Theorem 3(3) gives distinct *b*-adic nine-digit regular Kaprekar constants for distinct pairs (*b*, *m*) of suitable integers *b* and *m*, we see that:

$$\nu\_{\text{reg}}(b,9) = \sharp \left\{ m \in \mathbb{Z} \: \middle| \: m \ge 1, \frac{b-5}{6} < m < \frac{b-4}{5} \right\},$$

where the symbol stands for the number of all elements in the set.

For any integer *b* ≥ 0, we then see that:

$$v\_{\text{reg}}(b,9) = \begin{cases} b' & \text{if } 30b' + 2 \le b \le 30b' + 9, \\ b' + 1 & \text{if } b = 30b' + 10, \\ b' & \text{if } 30b' + 11 \le b \le 30b' + 14, \\ b' + 1 & \text{if } 30b' + 15 \le b \le 30b' + 16, \\ b' & \text{if } 30b' + 17 \le b \le 30b' + 19, \\ b' + 1 & \text{if } 30b' + 20 \le b \le 30b' + 22, \\ b' & \text{if } 30b' + 23 \le b \le 30b' + 24, \\ b' + 1 & \text{if } 30b' + 25 \le b \le 30b' + 28, \\ b' & \text{if } b = 30b' + 29, \\ b' + 1 & \text{if } 30b' + 30 \le b \le 30b' + 31. \end{cases}$$

Therefore, Part (7) is proven.

Moreover, as a corollary to Proposition 1, we can obtain lower bounds for *ν*reg(*b*, *n*) with *n* = 13, 15, 17 as in the following:

**Corollary 4.** *Let b* ≥ 2 *be any integer. Then, we have the following estimations:*

"

$$(1)\qquad \nu\_{\text{reg}}(b,13) \ge 1 \text{ if } b \ge 14 \text{ and } 7 \mid b.$$

$$\begin{aligned} (2) \quad &\nu\_{\text{reg}}(b, 15) \ge \sum\_{\frac{k-2}{9} \le m \le \frac{k-8}{8}} (b - 8m - 7) + \sum\_{\frac{k-5}{11} \le m \le \frac{k-8}{9}} \left( m - \left\lceil \frac{b - 9m}{2} \right\rceil + 3 \right), \\ &\text{the sums stands for positive integers.} \\ (3) \quad &\nu\_{\text{reg}}(b, 17) \ge \# \left\{ k \in \mathbb{Z} \, \middle| \, k \ge 2, b \equiv k \pmod{4}, 0 \le \frac{b - 9k}{4} \le \left\lceil \frac{k}{2} \right\rceil - 1 \right\}. \end{aligned}$$

**Proof.** (1) We see immediately that Part (1) is implied by the conditional formula obtained in Proposition 1(1), since the formula gives distinct (7*m* + 7)-adic 13-digit regular Kaprekar constants for distinct positive integers *m*.

(2) Since the conditional formula obtained in Proposition 1(2) gives distinct *b*-adic 15-digit regular Kaprekar constants for distinct triples (*b*, *m*1, *m*2) of suitable integers *b*, *m*1, and *m*2, we see that:

$$\begin{aligned} \nu\_{\text{reg}}(b, 15) \ge \sharp \{ (m\_1, m\_2) \in \mathbb{Z} \times \mathbb{Z} \mid m\_1 \ge 1, \, 2m\_1 + 1 < m\_2 < 3m\_1 + 2, \\ 6m\_1 + m\_2 + \dots + b &< 5m\_1 + 2m\_2 + 4 \}. \end{aligned}$$

For any integer *m*<sup>1</sup> ≥ 1, the list of *m*<sup>2</sup> and *b* satisfying the conditions:

$$2m\_1 + 1 < m\_2 < 3m\_1 + 2, \quad 6m\_1 + m\_2 + 5 < b < 5m\_1 + 2m\_2 + 4$$

is the following:


Since the number of *b*'s appearing in the list above is equal to:

$$\begin{cases} (b+1) - (8m\_1 + 8) & \text{if } 8m\_1 + 8 \le b \le 9m\_1 + 7, \\\\ (m\_1 - 1) - \left[ \frac{b - (9m\_1 + 8)}{2} \right] & \text{if } 9m\_1 + 8 \le b \le 11m\_1 + 5, \end{cases}$$

the right-hand side in the inequality above is equal to:

$$\sum\_{\frac{b-\mathcal{T}}{\mathfrak{P}} \le m \le \frac{b-\mathcal{S}}{\mathfrak{S}}} (b - 8m - \mathcal{T}) + \sum\_{\frac{b-\mathcal{S}}{\mathfrak{T}\mathcal{T}} \le m \le \frac{b-\mathcal{S}}{\mathfrak{P}}} \left( m - \left\lceil \frac{b-9m}{2} \right\rceil + 3 \right) \mathcal{T}$$

where the symbol *m* in the sums stands for positive integers. Therefore, Part (2) is proven.

(3) Since the conditional formula obtained in Proposition 1(3) gives distinct *b*-adic 17-digit regular Kaprekar constants for distinct pairs (*b*, *m*) of suitable integers *b* and *m*, we see that:

$$\nu\_{\text{reg}}(b, 17) \ge \sharp \{ m \in \mathbb{Z} \mid m \ge 1, 9m + 7 < b < 11m + 9, b \equiv m + 1 \pmod{4} \}.$$

For any integer *m* ≥ 1, the first term and the final term in the range 9*m* + 7 < *b* < 11*m* + 9 of the arithmetic progression with the common difference of four, which are congruent to *m* + 1 modulo four, are 9*m* + 9 and (9*m* + 9) + 4 -7*m* + 1 2 8 − 1 , respectively. Putting *k* = *m* + 1, we then see that:

$$\begin{aligned} &\sharp\{m \in \mathbb{Z} \mid m \ge 1, \, 9m + 7 < b < 11m + 9, \, b \equiv m + 1 \pmod{4}\}, \\ &= \sharp\left\{k \in \mathbb{Z} \mid k \ge 2, \, b \equiv k \pmod{4}, \, 0 \le \frac{b - 9k}{4} \le \left\lceil \frac{k}{2} \right\rceil - 1 \right\}, \end{aligned}$$

and Part (3) is proven.

**Author Contributions:** Conceptualization, A.Y.; investigation, A.Y. and Y.M.; writing, original draft, A.Y.

**Funding:** This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

**Acknowledgments:** The A.Y. is very grateful to the Y.M., who was one of his students at Soka University, for giving some interesting talks about formulas for Kaprekar constants in seminars held in 2017 at Soka University.

**Conflicts of Interest:** The authors declare no conflict of interest.

**Errata of [1]:** Since the reference [1] is very important to readers of this article, we would like to describe the errata of [1] here:

p. 263, -. 32, *N*(*b*, 2) and -(*b*, 2) → *N*(*b*, 5) and -(*b*, 5) p. 266, -.7, 14, 16, 18, 19, 20, 21, 23, 24: (*c*0)<sup>2</sup> → (*c*0)*<sup>b</sup>* p. 266, -.16: ((*c* − 1)(*b* − *c*))<sup>2</sup> → ((*c* − 1)(*b* − *c*))*<sup>b</sup>* p. 266, -.14, 19: ((*δ*1(*c*) − 1)(*b* − *δ*1(*c*)))<sup>2</sup> → ((*δ*1(*c*) − 1)(*b* − *δ*1(*c*)))*<sup>b</sup>* p. 266, -.21: (*c* − 1)(*b* − *c*))<sup>2</sup> → ((*c* − 1)(*b* − *c*))*<sup>b</sup>* p. 266, -.24: ((*δv*2(*b*+1)−*v*2+1(*c*) − <sup>1</sup>)(*<sup>b</sup>* − *<sup>δ</sup>v*2(*b*+1)−*v*2+1(*c*)))<sup>2</sup> → ((*δv*2(*b*+1)−*v*2(*c*)+1(*c*) − <sup>1</sup>)(*<sup>b</sup>* − *<sup>δ</sup>v*2(*b*+1)−*v*2(*c*)+1(*c*)))*<sup>b</sup>* p. 267, -.2, 3: (*c*0)<sup>2</sup> → (*c*0)*<sup>b</sup>* p. 269, -.11: *n* ≥ 7 and → *n* ≥ 7; *n* is odd and p. 269, -.12: *c <sup>n</sup>* <sup>2</sup> <sup>−</sup><sup>2</sup> → *c <sup>n</sup>*−<sup>1</sup> <sup>2</sup> −2 p.280, -.16: Delete the sentence "A.L. Ludington, A bound on Kaprekar constants, J. Reine Angew. Math. 310 (1979) 196–203."

## **References**


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