Decomposition of Beatty and Complementary Sequences

#A104 INTEGERS 25 (2025) DECOMPOSITION OF BEATTY AND COMPLEMENTARY SEQUENCES Geremias Polanco Department of Mathematics, Smith College, Northampton, Massachusetts Received: 7/7/21, Revised: 1/13/25, Accepted: 10/29/25, Published: 11/25/25 Abstract In this paper, we express the difference of two complementary Beatty sequences as the sum of two other closely related Beatty sequences. In the process, we introduce a new algorithm that generalizes the well-known Minimum Excluded algorithm and provides a method to combinatorially generate any pair of complementary Beatty sequences in a natural way. 1. Introduction Wythoff [31] proved the formula nϕ2−[nϕ]=n, (1) where ϕis the golden ratio. Here, [x] denotes the largest integer not exceeding x. Later, other authors (see, for instance, [15] and [19]) used the formula nα2−[nβ]=tn, with α=2−t+√t2+4 2, and βsatisfying α−1+β−1= 1, to generalize Wythoff’s game. Other generalizations have been made covering Beatty sequences parameterized by limited families of irrational numbers. We obtain a generalization of Identity (1) for all complementary sequences and give a combinatorial interpretation of the corresponding quantity on the right-hand side. For example, "n3 + √3 2#−hn√3i="n√3−1 2#+hn(2 −√3)i+ 1. In particular, we prove Theorems 2, 3 and 4, which together imply that Equation (1) generalizes to any pair of complementary Beatty sequences Aand Bwith irrational slopes β > 2 and αas follows: [nβ]−[nα] = [n(β−2)] + [n(2 −α)] + 1.(2) DOI: 10.5281/zenodo.17711544 INTEGERS: 25 (2025) 2 Moreover, as we shall prove later, the two quantities in the brackets on the right-hand side of the equality count the number of integers that belong to Aand B, respectively, among those integers lying in the half-open interval [an, bn). Incidentally, the title of this paper alludes to Equation (2), its interpretation, and its generalization in Theorems 2 through Theorem 7. We start by setting some useful notation. We will call Aand Bcomplementary sets of natural numbers if A∩B=∅and A∪B=N. For any multiset Xof integers that is bounded below, we number the elements (with multiplicity) of X as x1≤x2≤..., linking in this way indexed lowercase letters (the elements in non-decreasing order) and the uppercase letters (the set). We use set subtraction and addition to denote the term-wise operations: X±Y={xn±yn:n∈N}. A set whose terms are given by bn= [nβ] for some β > 0 is called a Beatty sequence. We remind the reader of Beatty’s Theorem (see, for instance, [28]): Theorem 1 (Beatty-Rayleigh, [8]).The sets Aand Bgiven by an= [nα]and bn= [nβ], respectively, are complementary sets of natural numbers if and only if α is a positive irrational and 1 α+1 β= 1. The rest of the paper is outlined as follows. We end this section by highlighting the importance of the subject. In Section 2 we state the Minimum Excluded (MEX) algorithm, familiar from the theory of impartial games, and generalize it by introducing the novel Minimum Excluded with Skipping (MES) algorithm. The latter will later link to our combinatorial interpretation. In Section 3 we state and prove Theorems 2, 3, and 4, the main results of the paper, and explore further implications of the MES algorithm, such as its combinatorial interpretation. To prove the results of this paper, we use combinatorial arguments, basic analytic techniques, and standard properties of Beatty and Sturmian sequences. For α > 1, the Sturmian sequence with slope 1/α encodes the Beatty sequence with slope α[4, Lemma 9.1.3]. There is also a corresponding relationship between non-homogeneous Beatty and Sturmian sequences [11, Lemma 1]. Sturmian sequences in general and Beatty sequences in particular, are the focus of a growing amount of research, as they play a role in various fields of mathematics [25], biology [17], music [24], computer science [13], and physics (see also [4], [22], [9], [10] and the references therein). The name “Sturmian” was introduced by Hendlund and Morse in their influential work in the 1940s [21]; however, the history of these sequences dates back to 1772 when Bernoulli III worked with what is now known as a non-homogeneous Sturmian sequence (see [4]). In pure mathematics, Sturmian and Beatty sequences have been studied in relation to dynamical systems [21], fixed-point morphisms [28], logarithm of irrational numbers [26], prime numbers [7], algebraic numbers [16], arithmetic functions [2], primitive roots, divisor functions, character sums, INTEGERS: 25 (2025) 3 among others (see [1], [3], [6], [18], [25], [27], [29], and the references therein). More recently, the second and third moments of Beatty sequences and their squares have also been studied [23]. We study differences of Beatty sequences in relation to the well-known MEX algorithm that is widely used in combinatorial game theory, coloring algorithms, and elsewhere (see, for instance, [5], [12], [14], [20], and [30]). 2. Miminum Excluded and Minimum Excluded with Skipping Algorithms In this section, we present the well-known MEX algorithm, introduce the MES algorithm as its generalization, as well as definitions and notations necessary to state and prove Theorem 2. We start with the following definition. Definition 1. Given a set S⊂N, the mex of the set Sis defined as mex(S) := min(Sc∩N), that is, mex(S) is the minimum natural number that is not in S. It is well known that complementary sequences can be generated using the mex rule above (see, for instance, [16]). We call this process the MEX algorithm and define it right below. Throughout the remainder of the paper, if Ais a sequence, we use Anto denote the set An={ak:k≤n}. Definition 2. Given an input sequence H, the Minimum Excluded Algorithm is the following set of algorithmic recursions. •STEP 1: Take a1= 1. •STEP 2: If n≥2, take an= mex{ai, bi|i < n}= mex(An−1∪Bn−1). •STEP 3: If n≥1, take bn=an+hn. •STEP 4: Repeat steps 2 and 3. To our knowledge, the first documented use of the MEX algorithm to generate Beatty sequences combinatorially was by British mathematician Willem A. Wythoff in 1907 [31]. He did so with the purpose of presenting a modification of the combinatorial game Nim. He showed that the winning positions of his game were given by the Beatty pairs (an, bn), defined by the golden ratio. In our context, this corresponds to taking hn=nin the MEX algorithm above. For the purpose of illustration, let us run the first few rounds of the algorithm with hn=n. Applying STEPS 1, 2, and 3, we get a1= 1 and b1=an+n= 1 + 1 = 2, so A1={1} and B1={2}. Applying STEP 4, we first obtain a2= mex(A1∪B1) = 3 and b2=an+n= 3 + 2 = 5, so that A2={1,3}and B2={2,5}. If we follow this process, the first five iterations give Table 1. INTEGERS: 25 (2025) 4 mex(An−1∪Bn−1)bn=an+n An={ak:k≤n}Bn={bk:k≤n} a1= 1 b1= 1 + 1 = 2 A1={1}B1={2} a2= 3 b2= 3 + 2 = 5 A2={1,3}B2={2,5} a3= 4 b3= 4 + 3 = 7 A3={1,3,4}B3={2,5,7} a4= 6 b4= 6 + 4 = 10 A4={1,3,4,6}B4={2,5,7,10} a5= 8 b5= 8 + 5 = 13 A5={1,3,4,6,8}B5={2,5,7, . . .} Table 1: First few iterations of the MEX algorithm Note that the set Anagrees with the first nterms of the sequence Agiven by the golden ratio. Similarly, Bnagrees with the first nterms of its complementary Beatty sequence. We now generalize the MEX algorithm by modifying STEP 3, and call this generalization the Minimum Excluded with Skipping algorithm. To find bn, we choose the minimum excluded element after skipping a number of integers. We base the MES algorithm on the following generalizations of Definition 1. Definition 3. Given a set of integers Aand a non-negative integer k, we define a function denoted by mexk(A) that selects the (k+ 1)st minimum excluded positive integer from the set A. In other words, to find mexk(A), we skip the first kexcluded integers from Aand select the next excluded integer. We now define the MES algorithm. Definition 4. Given a sequence of non-negative integers C={cn}∞ n=1, the Minimun Excluded with Skipping Algorithm is given by the following set of algorithmic recursions. •STEP 1: Set a1= 1, A1={a1}={1}, and B0=∅(the empty set). •STEP 2: For n≥2, set an= mex(An∪Bn). •STEP 3: For n≥1, set bn= mexcn(An∪Bn−1), that is, skip the first cn excluded positive integers and select the next excluded integer from this union. •STEP 4: Repeat steps 2 and 3. Example 1. We use the sequence C={0,1,1,2,3,3,4,4,5,6,6,7,8,8, . . . }to illustrate the MES algorithm. Starting with B0=∅, the first iteration gives a1= 1, and b1= mexc1(A1∪B0) = mex0{1}= 2. So, A1={1}and B1={2}. The second iteration gives a2= mex(A1∪B1) = 3 and b2= mexc2(A2∪B1) = mex1{1,2,3}= 5, as 5 is the second minimum excluded integer after skipping the first excluded integer 4. Continuing this process, subsequent iterations are shown in Table 2. INTEGERS: 25 (2025) 5 anbnAn={ak:k≤n}Bn={bk:k≤n} a1= 1 b1= mex0= 2 A1={1}B1={2} a2= 3 b2= mex1= 5 A2={1,3}B2={2,5} a3= 4 b3= mex1= 7 A3={1,3,4}B3={2,5,7} a4= 6 b4= mex2= 10 A4={1,3,4,6}B4={2,5,7,10} a5= 8 b5= mex3= 13 A5={1,3,4,6,8}B5={2,5,7,10,13} a6= 9 b6= mex3= 15 A6={1,3,4,6,8,9}B6={2,5,7,10,13,15} a7= 11 b7= mex4= 18 A7={1,3,4,6,8,9,11}B7={2,5,7,10,13, . . .} Table 2: MES algorithm: an= mex(An−1∪Bn−1) and bn= mexcn(An∪Bn−1). nth ABC 1) 1 2 0 2) 1 3) 1 4) 2 Table 3: MES Steps. nth ABC 1) 1 2 0 2) 3 5 1 3) 4 7 1 4) 2 5) 3 6) 3 7) 4 8) 4 9) 5 Table 4: MES Steps. nth A B C 1) 1 2 0 2) 3 5 1 3) 4 7 1 4) 6 10 2 5) 8 13 3 6) 9 15 3 7) 11 18 4 8) 12 20 4 9) 14 23 5 10) 16 26 6 11) 17 28 6 12) 19 31 7 Table 5: MES Steps. Remark 1. Because of how it is used in STEP 3 of the MES definition above, we call Cthe skipping sequence of the MES algorithm. After a close look at the resulting Table 2, one can see that the sequence Cused in the MES above can be defined inductively by the rule: a positive integer that has been assigned by the algorithm to Aappears twice in C, otherwise it appears once. In other words, the algorithm can be generated without explicitly listing the sequence C, only using the initial value c0= 0, and the inductive rule just stated. Tables 3, 4, and 5 illustrate how Cis generated inductively in this way. Later we expand on this, showing that the resulting sequences from the MES algorithm (Sequences A and B) are the complementary pair given by the golden ratio. This is Theorem 2 (4a). We need the following definition. Definition 5. The sortjoin of two integer sequences Aand B, denoted by A⋆B, is the union with repetition of the ordered elements of these two sequences. We use A2for the sortjoin of Awith itself, that is, A2=A ⋆ A, and we represent INTEGERS: 25 (2025) 6 the sortjoin of kcopies of Awith itself by Ak. Finally, we use N0to denote the non-negative integers. For instance, if Arepresents the even natural numbers, then A⋆N2 0={0,0,1,1,2,2,2,3,3,4,4,4,5,5, . . .}, because the numbers 2,4,6, . . . repeat three times while the others only repeat twice. We need two more definitions before proving Theorem 2. Part (4) of Theorem 2 mentions the case when a sequence Cis given by the sortjoin C=D ⋆ Nk−1 0in the MES algorithm. In this case, the sequence Cis completely determined by the sequence D. Since Cdefines the MES algorithm (and Ddefines C), it effectively follows that Ddefines the MES algorithm. We record this in the following definition. Definition 6. If there is an increasing sequence of natural numbers Dsuch that the sequence Cof the MES algorithm is given by C=D ⋆ Nk−1 0, we call Dthe defining sequence of the MES algorithm. We also use the following definition. Definition 7. For n≥1, consider the (possibly empty) half-open integer interval (an, bn−1], and let rnbe the cardinality of the intersection of this interval with Bn−1, i.e., rn= #{b|b∈(an, bn−1] and b∈Bn−1}. We define the auxiliary sequence R by R:= {rn}∞ n=1. Remark 2. Consistent with the use of Anand Bn, we use the convention Rn:= {rk:k≤n}. We also note here for future reference that the sequence cnis a counting sequence for some elements of the sequence A. Specifically we have cn= #{a∈A|an< a < bn}. This formula follows because once an integer is skipped when selecting bn, that integer will never be selected by the sequence Bin future steps, thus that element will be picked up by A. This is true whenever cnis non-decreasing. 3. Statement and Proof of the Main Theorems The main results of this paper are Theorems 2, 3, and 4, which together provide a complete picture of the MES algorithm and its relation to decomposing the difference of two Beatty sequences as the sum of two other sequences. Theorem 2, part (a), expresses the difference of two Beatty sequences as the sum of two generalized Beatty sequences, yielding Formula (3). Part (b) of this theorem provides an interpretation of this formula: the MES algorithm. In other words, it states how we can compute anand bnfrom Equation (3), using the MES algorithm. Theorems 3 and 4 state that if the resulting complementary sequences in the MES algorithm, Aand B, are Beatty sequences, then the skipping sequence Chas INTEGERS: 25 (2025) 7 slope β−2, and the counting sequence Rhas slope 2 −α, where αand βare the slopes of the sequences Aand B, respectively. Finally, Theorem 7 states necessary and sufficient conditions for the output sequences Aand Bto be Beatty sequences in the MES algorithm. It also relates their slopes to the slope of the input sequence D, and treats some particular cases of interest. We now state and prove Theorem 2. Theorem 2. Suppose that Aand Bare any pair of complementary sequences. We have the following statements. (a) The sequence formed by the difference A−B, term by term, can always be decomposed in terms of the sum of two other sequences Cand Ras follows: bn−an=cn+rn+ 1,(3) where cncounts the number of integers strictly between anand bnthat belong to the sequence A, and rncounts the number of integers strictly between an and bnthat are in the sequence B. (b) There is a combinatorial interpretation of Formula (3), that is, an algorithm that we call the Minimum Excluded with Skipping algorithm, that takes as input a sequence Dand outputs four sequences A, B and Cand R. If an, bn, cnand rnare the nth term of A, B, C and R, respectively, then Equation (3) holds. Furthermore, the MES algorithm characterizes complementary sequences. Specifically, Aand Bare complementary sequences if and only if there is a non-negative sequence Cthat generates Aand Bthrough the MES algorithm. Proof. Formula (3) of Theorem 2 is evident from the fact that Aand Bare complementary. Thus, the integers in the interval (an, bn) will belong to either Aor B, and hence the stated formula in part (a) must follow. The remainder of part (a) holds by construction as follows. On the one hand, given a non-negative sequence C, the MES can be implemented by skipping cnexcluded integers in the nth step. By construction, this will produce complementary sequences. Conversely, if Cis negative, skipping cannot be performed in the natural sense as it is done in the algorithm. On the other hand, if a pair of complementary sequences is given, one can define rn as in Definition 7. Then cncan be defined as the number of integers in the following complement: [(an, bn)∩Bn]c(in other words, cn= #[(an, bn)∩Bn]c). With this setup, we can implement the algorithm with C={cn}∞ n=1. Then it follows that if R={rn}∞ n=1, then bn−an=cn+rn+ 1. We now develop a series of technical lemmas regarding Sturmian and Beatty sequences, and then use them to prove Theorems 3, 4 and 7. INTEGERS: 25 (2025) 8 Given a Beatty sequence with slope 0 < α < 1, each positive integer will occur in the sequence ktimes or k+ 1 times, for some number k≥1. This is simply a consequence of the size of α. We have the following lemma. Lemma 1. If αbelongs to the interval 1 k+1 < α < 1 kfor ka positive integer, then all non-negative integers appear in the Beatty sequence Cwith slope α, each of them repeating kor k+ 1 times. Both occurrences are infinite. In the language of sortjoin, there exists, in this case, a sequence Bof non-negative integers such that C=B ⋆ Nk−1 0. Proof. It is not hard to see that the inequality 1 k+1 < α < 1 kimplies each nonnegative integer repeats kor k+ 1 times in {[αn]}∞ n=1. In fact, kα < 1 implies that each integer koccurs at least ktimes, and an integer cannot occur k+ 2 times or more. For if [nα] = [(n+ 1)α] = ··· = [(n+k+ 1)α]=k, then 0 = [(n+k+1)α]−[nα]≥[nα] + [(k+ 1)α]−[nα] = [(k+ 1)α]>1. Now, some integers nwill need to be repeated k+1 times. This is so because if all large integers repeat ktimes, then the sequence would be ultimately periodic, and the following equality would need to hold: lim n→∞ [nα] n=1 k, that is, α=1 k, which contradicts the irrationality of α. For the same reasons, all large integers cannot be repeated k+ 1 times. Hence, both kand k+ 1 occur infinitely often. The following definition will be used in the remaining of the paper. Definition 8. For a real number αsuch that 0 <α<1, we define the characteristic function of αas fα(n)=[α(n+ 1)] −[αn]. Clearly, fα(n) = 1 or fα(n) = 0. Since the sum telescopes, we have m X n=1 fα(n)=[α(m+ 1)]. (4) Remark 3. Notice that Equation (4) gives the number of integers n≤mfor which fα(n) = 1. Another definition we will need is the following one. Definition 9. For 1 < β ∈R\Q, define g′ β(n) = (1if n= [kβ],for k∈Z, 0 otherwise. INTEGERS: 25 (2025) 9 It is a well-known fact (see [26] and [4, Lemma 9.1.3]) that for all integers n, g′ β(n)=f1/β(n). Lemma 2. For each irrational number αsuch that 1 k+1 < α < 1 k, there exists a number β:= α 1−α, such that [nα] = tfor k+ 1 different numbers n=m, m + 1, . . . , m +k, if and only if [nβ]=tfor knumbers n=q, q + 1, . . . , q +k−1. In other words, if C= [nα]∞ n=1 and B=α 1−α·n∞ n=1 , then C=D ⋆ Nk 0and B=D ⋆ Nk−1 0for some increasing sequence Dof non-negative integers. Proof. Notice that 1 k+1 < α < 1 kif and only if k < 1 α< k + 1, which is equivalent to k−1<1 α−1< k and 1 α−1 = 1−α α=1 β. Thus, we claim that it is enough to prove that the sequence [nα] equals tfor k+ 1 values of n, if and only if [(t+ 1) 1 α]−[t1 α] = k+ 1. Indeed, if this is true, then [(t+ 1) 1 β]−[t1 β] = [(t+ 1)( 1 α−1)] −[t(1 α−1)] = [(t+ 1) 1 α]−[t1 α]−1=k, it would then follow from this claim that [nβ] repeats ktimes. So we just need to prove the double implication in the claim. For that purpose, from (8), we write f(n):=fα(n) = [(n+ 1)α]−[nα]. And from (3), it follows that f(n) = 1 if and only if there exists a tsuch that n= [t1 α]. Thus, from this last sentence we find that [α(n+m)] = rfor the first k+1 values of mif and only if the following three things happen: first, f(n−1) = 0; second, f(n)=f(n+k+1) = 1; and third, f(n+m) = 0 for all numbers between nand n+k+ 1, i.e., for all msuch that 1 ≤m<k+ 1. For such an n, it follows from (3) that f(n+m) = 0 for 1 ≤m<k+ 1 if and only if there is no jsuch that n+m= [j1 α] for 1 ≤m≤k. Thus, we see that [(t+ 1) 1 α]−[t1 α]> k. Since the difference between two consecutive numbers in the Beatty sequence [t1 α] is either kor k+ 1, we see that [(t+ 1) 1 α]−[t1 α]=k+ 1. Therefore, we have proved that each time the sequence [nα] repeats k+ 1 times, the difference [(t+ 1) 1 α]−[t1 α] equals k+ 1, and thus, the lemma holds. Combining Lemma 1 and Lemma 2, we see that if a Beatty sequence is generated by an αsuch that each non-negative integer repeats kor k+ 1 times, then in the Beatty sequence generated by α 1−α=−1 + 1 1−α, integers repeat k−1 or ktimes. If we iterate this process, we obtain the following lemma. INTEGERS: 25 (2025) 16 Theorem 7. The MES algorithm generalizes the MEX algorithm as follows: Suppose that there is an increasing sequence of positive integers Dsuch that C= D ⋆ Nk−1 0for some k∈N. Then we have the following statements. 1. If k= 2 and Dis the Beatty sequence defined by the golden ratio, then the MES and the MEX algorithms coincide, i.e., they both produce A(given by the golden ratio). In other words, the MES produces the sequence Aif and only if the skipping sequence Ccan be defined as follows: any non-negative integer cappears once or twice in C, and cappears twice in Cif and only if c∈A. 2. Let Dbe the sequence given by α=k−1 + √k2+ 1 k. Then the MES algorithm produces the Beatty sequence Dand its complement. In other words, the MES produces the sequence Dif and only if the skipping sequence Ccan be defined as follows: any non-negative integer cappears k or k−1times in C, and cappears ktimes in Cif and only if c∈D. This coincides with the MEX and the golden ratio when k= 2. Proof. Substitute α=δinto Equation (13), and solve for δin the resulting rational expression to obtain a quadratic irrational with parameter k. The case k= 2 gives the first part of the corollary. Remark 6. As seen in Equations (12) and (13), a given δcan generate infinitely many αs. However, each given αis generated by a unique choice of δand k. Example 2. As an illustration of Theorem 6, consider the number α=187+2√13 113 . The Beatty sequence generated by αis {1,3,5,6,8,10,12,13,15,17,18, . . .}. If we run the MES algorithm with frequency k= 3 and defining sequence given by δ= √13 2, we obtain Table 6. Example 3. Example 1 can be rephrased recursively as follows. The skipping sequence Cof the MES algorithm that generates the sequence Adefined by the golden ratio is given by the condition “if n∈A, then it will repeat twice in C. Otherwise, nwill appear only once in C”. Note that if δ=1+√5 2and k= 2, we obtain the sequence in Example 1, that is, the sequence with the golden ratio. In this case the skipping sequence Cand the output sequence Acoincide with the defining sequence D. Table 6, below, shows the application of the MES algorithm when the frequency kis given by k= 3 and the defining sequence Dis given by δ=√13/2. Note that the output sequences Ais given by α=2−δ+ 2lδ 1+kδ =2−δ+ 6δ 1+6δ=187 + 2√13 113 . INTEGERS: 25 (2025) 17 nth A B C R D 1) 1 2 0 0 1 2) 3 4 0 0 3 3) 5 7 1 0 5 4) 6 9 1 1 7 5) 8 11 1 1 9 6) 10 14 2 1 10 7) 12 16 2 1 12 8) 13 19 3 2 14 9) 15 21 3 2 16 10) 17 23 3 2 18 11) 18 26 4 3 19 12) 20 28 4 3 21 Table 6: Iterations of the MES for k= 2 and δ=√13/2. 4. Concluding Remarks What does the MES algorithm offer that the MEX algorithm does not? Notice that both algorithms allow us to generate any pair of Beatty sequences (and indeed any pair of complementary sequences). The MEX algorithm uses the function bn−an= hnto generate these sequences, and any interesting application requires that we be able to easily describe the function hnindependently of anand bn. To obtain the Beatty sequences generated by αand β, a few cases are straightforward to describe. For example, [31] shows that hn=ngives the case α=1+√5 2. More generally, [15] and [19] show that hn=tn gives the family 2−t+√t2+1 2. The MES algorithm is more advantageous because it provides a general formula for hnas cn+rn+ 1, with explicit formulas for cnand rnin the case of Beatty sequences given in Theorem 2. It also offers the added benefit of a combinatorial interpretation for cnand rn, linking them back to the complementary sequences A and B. We conclude with an open question. Given the many applications of the MEX function or the MEX algorithm as highlighted in Section 1, does the MES algorithm reveal interesting connections that illuminate some aspects of these applications or of complementary sequences in general? Acknowledgements. The author is indebted to his advisor, Kenneth B. Storlarsky, for his guidance and the numerous helpful discussions that made this article possible. He is also deeply grateful to the anonymous referee for a careful reading of the manuscript and significant suggestions that improved the presentation of the ideas in the paper. Also, the author thanks the editor for much appreciated suggestions improving the paper. Finally, he thanks Michel Dekking for some help- INTEGERS: 25 (2025) 18 ful comments incorporated into the initial section of the paper and for his valuable proofreading feedback. This research has been partially supported by the Ministerio de Educaci´on Superior, Ciencia y Tecnolog´ıa (MESCyT) of the Dominican Republic, grants No. FONDOCyT: 2018-2019-1D2-261 and FODOCyT: 2018-2019-1D2-262. References [1] A. G. Abercrombie, Beatty sequences and multiplicative number theory, Acta Arith.70 (1995), 195-207. [2] A. G. Abercrombie, W. Banks and I. Shparlinski, Arithmetic functions on Beatty sequences, Acta Arith.136 (2009), 81-89. [3] J. Allouche and F. Dekking, Generalized Beatty sequences and complementary triples, Mosc. J. Comb. Number Theory .8(2019), 325-341. [4] J. Allouche and J. Shallit, Automatic Sequences: Theory, Applications and Generalizations, Cambridge university press, New York, 2003. 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