COUNTING SUBWORD OCCURRENCES IN BASE-b EXPANSIONS. Julien Leroy 1 Department of Mathematics, University of Liège, Liège, Belgium

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1 #A13 INEGERS 18A (2018) COUNING SUBWORD OCCURRENCES IN BASE-b EXPANSIONS Julien Leroy 1 Department of Mathematics, University of Liège, Liège, Belgium JLeroy@uliegebe Michel Rigo Department of Mathematics, University of Liège, Liège, Belgium MRigo@uliegebe Manon Stipulanti 2 Department of Mathematics, University of Liège, Liège, Belgium MStipulanti@uliegebe Received: 4/12/17, Revised: 11/20/17, Accepted: 1/9/18, Published: 3/16/18 Abstract We consider the sequence (S b (n)) n 0 counting the number of distinct (scattered) subwords occurring in the base-b expansion of the non-negative integers By using a convenient tree structure, we provide recurrence relations for (S b (n)) n 0 leading to the b-regularity of the latter sequence hen we deduce the asymptotics of the summatory function of the sequence (S b (n)) n 0 JeffShallit sinfluenceincombinatoricsonwords cannot be underestimated We are therefore very happy to contribute to this special issue dedicated to his birthday his paper contains small bits of the recurrent topics he has been working on 1 Introduction A finite word is a finite sequence of letters belonging to a finite set called the alphabet Let u, v be two finite words We say that v is a scattered subword of u and we write v u, if v is a subsequence of u All along the paper, we let b denote an integer greater than 1 We let rep b (n) denotethe(greedy)base-b expansion of n N \{0} starting with a non-ero digit We set rep b (0) to be the empty word denoted by ε 1 J Leroy is an FNRS post-doctoral fellow 2 M Stipulanti is supported by a FRIA grant 1E03016

2 INEGERS: 18A (2018) 2 We let L b = {1,,b 1}{0,,b 1} {ε} be the set of base-b expansions of the non-negative integers For all w {0,,b 1},wealsodefineval b (w) tobethevalueofw in base b, ie,ifw = w n w 0 with w i {0,,b 1} for all i, thenval b (w) = n i=0 w ib i In this paper, we also make extensive use of the genealogical order defined as follows If u, v {0,,b 1} are two words, we say u is less than v in the genealogical order, andwewriteu<v, if either u < v, or if u = v and there exist words p, q, r {0,,b 1} and letters a, a {0,,b 1} with u = paq, v = pa r and a<a Byu v, wemean that either u<v,oru = v Definition 1 For n 0, we define the sequence (S b (n)) n 0 by setting S b (n) := # {v L b v rep b (n)} (1) We also consider the summatory function (A b (n)) n 0 of the sequence (S b (n)) n 0 defined by A b (0) = 0 and for all n 1, n 1 A b (n) := S b (j) he quantity A b (n) canbethoughtofasthetotalnumberofbase-b expansions occurring as scattered subwords in the base-b expansion of integers less than n (the same subword is counted k times if it occurs in the base-b expansion of k distinct integers) Example 1 If b =3,thenthefirstfewtermsofthesequence 3 (S 3 (n)) n 0 A are 1, 2, 2, 3, 3, 4, 3, 4, 3, 4, 5, 6, 5, 4, 6, 7, 7, 6, 4, 6, 5, 7, 6, 7, 5, 6, 4, 5, 7, 8, 8, 7, 10, For instance, the subwords of the word 121 are ε, 1, 2, 11, 12, 21, 121 hus, we have S 3 (val 3 (121)) = S 3 (16) = 7 he first few terms of (A 3 (n)) n 0 A are 0, 1, 3, 5, 8, 11, 15, 18, 22, 25, 29, 34, 40, 45, 49, 55, Motivated by generaliations of the Pascal triangle [12], we dealt with the case b = 2, considering the sequences (S 2 (n)) n 0 [13] and (A 2 (n)) n 0 [14] Firstly, using recurrence relations, we showed that the sequence (S 2 (n)) n 0 A is 2-regular in the sense of Allouche and Shallit [1] We also conjectured six recurrence relations for (S 3 (n)) n 0 depending on the position of n between two consecutive powers of 3 Using the heuristic from [3] suggesting recurrence relations, 3 Some of the sequences of this paper are uploaded in [16]

3 INEGERS: 18A (2018) 3 the sequence (S 3 (n)) n 0 was expected to be 3-regular In Section 2 of this paper, we show that for all b 2, the recurrence relations satisfied by (S b (n)) n 0 reduce to three forms; see Proposition 1 In particular, this proves the conjecture stated in [13] for b =3 hen,insection3,wededucetheb-regularity of (S b (n)) n 0 ;see heorem 1 Moreover we obtain a linear representation of the sequence with b b matrices We also show that (S b (n)) n 0 is palindromic over [(b 1)b l,b l+1 ] Secondly, we studied the behavior of (A 2 (n)) n 0 A282720, exhibiting a continuous periodic function H of period 1 such that A 2 (n) = 3 log 2 (n) H(log 2 (n)) for all n o this aim, from the 2-regularity of (S 2 (n)) n 0, we derived recurrence relations for (A 2 (n)) n 0 involving powers of 3, leading to a particular decomposition of A 2 (n) Sustained by computer experiments, we also conjectured that A b (nb) = (2b 1)A b (n) In Section 4 of this paper, generaliing the previous approach, we obtain specific recurrence relations for (A b (n)) n 0 involving powers of 2b 1(Proposition6)andprovingtheaboveconjectureaboutA b (nb) (Corollary 1) Using these relations, we consider the so-called (2b 1)-decompositions of A b (n) and exhibit a continuous periodic function H b of period 1 such that A b (n) =(2b 1) log b (n) H b (log b (n)) for all n (heorem 2) 2 General Recurrence Relations in Base b he aim of this section is to prove the following result exhibiting recurrence relations satisfied by the sequence (S b (n)) n 0 his result is useful to prove that the summatory function of the latter sequence also satisfies recurrence relations;see Section 4 Proposition 1 he sequence (S b (n)) n 0 satisfies S b (0) = 1, S b (1) = = S b (b 1) = 2, and, for all x, y {1,,b 1} with x y, alll 1 and all r {0,,b l 1 1}, S b (xb l + r) = S b (xb l 1 + r)+s b (r); (2) S b (xb l + xb l 1 + r) = 2S b (xb l 1 + r) S b (r); (3) S b (xb l + yb l 1 + r) = S b (xb l 1 + r)+2s b (yb l 1 + r) 2S b (r) (4) Most of the results are proved by induction and the base case usually takes into account the values of S b (n) for0 n<b 2 hese values are easily obtained from Definition 1 and summaried in able 1 For the sake of completeness, we recall the definition of a particularly useful tool called the trie of subwords to prove Proposition 1 his tool is also useful to prove the b-regularity of the sequence (S b (n)) n 0 ;seesection3

4 INEGERS: 18A (2018) 4 rep b (n) ε x x0 xx xy x00 x0x x0y S b (n) rep b (n) xx0 xxx xxy xy0 xyx xyy xy S b (n) able 1: he first few values of S b (n) for 0 n < b 3, with pairwise distinct x, y, {1,,b 1} Definition 2 Let w be a finite word over {0,,b 1} he language of its subwords is factorial, ie, if xy is a subword of w, theny is also a subword of w hus we may associate with w, thetrie 4 of its subwords he root is ε and if u and ua are two subwords of w with a {0,,b 1}, thenus a child of u We let (w) denote the subtree in which we only consider the children 1,,b 1ofthe root ε and their successors, if they exist Remark 1 he number of nodes on level l 0 in (w) countsthenumberof subwords of length l in L b occurring in w In particular, the number of nodes of the trie (rep b (n)) is exactly S b (n) for all n 0 Definition 3 For each non-empty word w L b, we consider a factoriation of w into maximal blocks of consecutively distinct letters (ie, +1 for all i) ofthe form w = a n1 1 anm M, with n l 1 for all l For each l {0,,M 1}, weconsiderthesubtree l of (w) whoserootisthenodea n1 1 an l l a l+1 Forconvenience,weset M to be an empty tree with no node Roughly speaking, we have a root of a new subtree l for each new alternation of digits in w For each l {0,,M 1}, we also let # l denote the number of nodes of the tree l Note that for k i 2, one could possibly have a k = Foreachl {0,,M 1}, we let Alph(l) denote the set of letters occurring in a l+1 a M henforeach letter a Alph(l), we let j(a, l) denote the smallest index in {l +1,,M} such that a j(a,l) = a Example 2 In this example, we set b =3andw = L 3 Using the previous notation, we have M =4,a 1 =2,a 2 =0,a 3 =1anda 4 = 2 For instance, Alph(0) = {0, 1, 2}, Alph(2)={1, 2} and j(0, 0) = 2, j(1, 0) = 3, j(2, 0) = 1 and j(2, 1) = 4 he following result describes the structure of the tree (w) It directly follows from the definition 4 his tree is also called prefix tree or radix tree All successors of a node have a common prefix and the root is the empty word

5 INEGERS: 18A (2018) 5 Proposition 2 ([13, Proposition 27]) Let w be a finite word in L b Withtheabove notation about M and the subtrees l,thetree (w) has the following properties 1 Every letter a Alph(0) \{0} is a child of the the node of label ε his node has thus #(Alph(0) \{0}) children Each child s the root of a tree isomorphic j(a,0) 1 2 For each l {0,,M 1} and each i {0,,n l+1 1} with (l, i) (0, 0), the node of label x = a n1 1 an l l ai l+1 has #(Alph(l)) children that are xa for a Alph(l) Each child xa with a a l+1 is the root of a tree isomorphic to j(a,l) 1 Example 3 Let us continue Example 2 he tree ( ) is depicted in Figure 1 We use three different colors to represent the letters 0, 1, 2 he tree 0 Figure 1: he trie ( ) (resp, 1 ; resp, 2 ; resp, 3 ) is the subtree of (w) with root 2 (resp, 2 2 0; resp, ; resp, ) hese subtrees are represented in Figure 1 using dashed lines he tree 3 is limited to a single node since the number of nodes of M 1 is n M,whichisequalto1inthisexample Using tries of subwords, we prove the following five lemmas heir proofs are essentially the same, so we only prove two of them Lemma 1 For each letter x {1,,b 1} and each word u {0,,b 1},we have # {v L b v x00u} =2 # {v L b v x0u} # {v L b v xu}

6 INEGERS: 18A (2018) 6 x 0,i i x 0 x,i i x R 1 a 1 x 0,i i x 0 a 2 0 R 2 (a) he tree (x0u) (b) he tree (xu) (c) he tree (x00u) Figure 2: Schematic structure of the trees (x0u), (xu) and (x00u) Proof Recall that from Remark 1, we need to prove that # (x00u) =2# (x0u) # (xu) Assume first that u is of the form u =0 n, n 0 he tree (xu) is linear and has n+2 nodes, (x0u) hasn+3 nodes and (x00u) hasn+4 nodes he formula holds Now suppose that u contains letters other than 0 We let a 1,,a m denote all the pairwise distinct letters of u different from 0 hey are implicitly ordered with respect to their first appearance in u If x {a 1,,a m }, we let i x {1,,m} denote the index such that x = x For all i {1,,m}, we let u i denote the prefix of u that ends with the first occurrence of the letter in u, and we let denote the subtree of (xu) withrootxu i First, observe that the subtree of (xu) withrootx is equal to the subtree of (x0u) withrootx0 and also to the subtree of (x00u) withrootx00 Secondly, for all i {1,,m}, the subtree of (x0u) withrootx is Similarly, (x00u) containstwocopiesof :thesubtreesofrootx and x0 Finally, for all i {1,,m} with i i x,thesubtreeof(x0u) withroot is and the subtree of (x00u) withroot is

7 INEGERS: 18A (2018) 7 he situation is depicted in Figure 2 where we put a unique edge for several indices when necessary, eg, the edge labeled by stands for m edges labeled by a 1,,a m he claimed formula holds since 2 2+# +2 =3+# +3 i i x i i x # +#x 1+# + # +2#x i i x # Lemma 2 For each letter x {1,,b 1} and each word u {0,,b 1},we have # {v L b v xx0u} =#{v L b v x0u} +#{v L b v xu} Proof he proof is similar to the proof of Lemma 1 Lemma 3 For all letters x, y {1,,b 1} and each word u {0,,b 1}, we have # {v L b v x0yu} =#{v L b v xyu} +#{v L b v yu} Proof he proof is similar to the proof of Lemma 1 Observe that one needs to divide the proof into two cases according to whether x is equal to y or not As a first case, also consider u = y n with n 0 instead of u =0 n with n 0 Lemma 4 For all letters x, y {1,,b 1} and each word u {0,,b 1}, we have # {v L b v xxyu} =2 # {v L b v xyu} # {v L b v yu} Proof he proof is similar to the proof of Lemma 3 he next lemma having a slightly more technical proof, we present it Lemma 5 For all letters x, y {1,,b 1} with x y, {0,,b 1} and each word u {0,,b 1},wehave # {v L b v xyu} = #{v L b v xu} +2 # {v L b v yu} 2 # {v L b v rep b (val b (u))}

8 INEGERS: 18A (2018) 8 Proof Let x, y {1,,b 1} with x y, {0,,b 1}, and let u {0,,b 1} Our reasoning is again based on the structure of the associated trees he proof is divided into two cases depending on whether =0ornot As a first case, suppose that 0 hen,observethatrep b (val b (u)) = u Now assume that u is of the form u = n, n 0 If x and y, thetree (u) is linear and has n +2 nodes, (xu) and (yu) have2(n +2) nodes and (xyu) has 4(n + 2) nodes and the claimed formula holds If x and y =, thetree (u) is linear and has n +2nodes, (xu) has2(n +2)nodes, (yu) hasn +3 nodes and (xyu) has2(n + 3) nodes and the claimed formula holds If x = and y, thetree (u) is linear and has n +2nodes, (xu) hasn +3nodes, (yu) has2(n +2)nodesand (xyu) has3(n +2)+1 nodes andthe claimed formula holds Now suppose that u contains other letters than We let a 1,,a m denote all the pairwise distinct letters of u different from heyareimplicitlyorderedwith respect to their first appearance in u If x, y, 0 {a 1,,a m }, we let i x,i y,i 0 {1,,m} respectively denote the indices such that x = x, y = y and 0 =0 For all i {1,,m}, we let u i denote the prefix of u that ends with the first occurrence of the letter in u, and we let denote the subtree of (u) withroot u i First, observe that the subtree of (u) withroot is equal to the subtree of (xu) withrootx, tothesubtreeof (yu) withrooty and also to the subtree of (xyu) withrootxy Suppose that x and y Using the same reasoning as in the proof of Lemma 1, the situation is depicted in Figure 3 he claimed formula holds since 2+2# # # + =4+4# +4 # +#x +2#y +#0 i i x,i y,i 0 # +2#x +#y +#0 i i x,i y,i 0 # +#x +#y i i x,i y,i 0 # +3#x +2#y +3#0 i i x,i y,i 0 Suppose that x and y = he situation is depicted in Figure 4 he

9 INEGERS: 18A (2018) 9 x,i i x,i 0 y,i i y,i 0 (a) he tree (xu) (b) he tree (yu),i i 0 R 1 a 1 a 2 y x y,i i x,i y,i 0,i i y R 2 (c) he tree (u) (d) he tree (xyu) Figure 3: Schematic structure of the trees (xu), (yu), (u) and (xyu) when x, y and 0

10 INEGERS: 18A (2018) 10 x,i i x,i 0,i i 0 (a) he tree (xu) (b) he tree (yu),i i 0 R 1 a 1 a 2 x,i i x,i 0 R 2 (c) he tree (u) (d) he tree (xyu) Figure 4: Schematic structure of the trees (xu), (yu), (u) and (xyu) when x, y = and 0

11 INEGERS: 18A (2018) 11 claimed formula holds since 2+2# +2 # +#x +#0 i i x,i # +2 # +2#x +#0 i i x,i # + =4+2# +4 # +#x i i x,i 0 # +3#x +3#0 i i x,i 0 Suppose that x = and y he situation is depicted in Figure 5 he claimed formula holds since 2+# +2 # +2#y +#0 i i y,i # +2 # +#y +#0 i i y,i # + =4+3# +4 # +#y i i y,i 0 # +2#y +3#0 i i y,i 0 As a second case, suppose that = 0 It is useful to note that rep b (val b ( )) : {0,,b 1} L b plays a normaliation role and removes leading eroes Consequently, rep b (val b (u)) = rep b (val b (u)) hen we must prove that the following formula holds # {v L b v xy0u} = #{v L b v x0u} +2 # {v L b v y0u} 2 # {v L b v rep b (val b (u))}

12 INEGERS: 18A (2018) 12,i i 0 y,i i y,i 0 (a) he tree (xu) (b) he tree (yu),i i 0 R 1 a 1 a 2 y y,i i y,i 0,i i y R 2 (c) he tree (u) (d) he tree (xyu) Figure 5: Schematic structure of the trees (xu), (yu), (u) and (xyu) when x =, y and 0

13 INEGERS: 18A (2018) 13 If u =0 n,withn 0, then rep b (val b (u)) = ε and the tree (rep b (val b (u))) has only one node he trees (x0u) and (y0u) bothhaven +3nodesandthe tree (xy0u) has3(n + 2) + 1 nodes and the claimed formula holds Now suppose that u contains other letters than 0 We let a 1,,a m denote all the pairwise distinct letters of u different from 0 hey are implicitly ordered with respect to their first appearance in u If x, y {a 1,,a m }, we let i x,i y {1,,m} respectively denote the indices such that x = x and y = y For all i {1,,m}, we let u i denote the prefix of rep b (val b (u)) that ends with the first occurrence of the letter in rep b (val b (u)), and we let denote the subtree of (rep b (val b (u))) with root u i he situation is depicted in Figure 6 Observe that the subtree of (y0u) with root y0 is equal to the subtree of (x0u) withrootx0 andtothesubtreeof (xy0u) withrootxy0 he claimed formula holds since 2+# +2 # +#x +2#y i i x,i y +2 2+# +2 # +2#x +#y i i x,i y 2 1+ # +#x +#y i i x,i y =4+3# +4 # +3#x +2#y i i x,i y hose five lemmas can be translated into recurrence relations satisfied by the sequence (S b (n)) n 0 using Definition 1 Proof of Proposition 1 he first part is clear using able 1 Let x, y {1,,b 1} with x y Proceedbyinductiononl 1 Let us first prove (2) If l =1,thenr = 0 and (2) follows from able 1 Now suppose that l 2 and assume that (2) holds for all l <lletr {0,,b l 1 1}, and let u be a word in {0,,b 1} such that u 1andrep b (xb l + r) =x0u he proof is divided into two parts according to the first letter of u Ifu =0u with

14 INEGERS: 18A (2018) 14 x,i i x y,i i y 0 0 (a) he tree (x0u) (b) he tree (y0u),i 1 a 1 R 1 a 1 a 1 0 y x 0 y,i i x,i y,i i y a 2 0 R 2 (c) he tree (rep b (val b (u))) (d) he tree (xy0u) Figure 6: Schematic structure of the trees (x0u), (y0u), (rep b (val b (u))) and (xy0u)

15 INEGERS: 18A (2018) 15 u {0,,b 1}, then, using Lemma 1 and the induction hypothesis twice, S b (xb l + r) = 2S b (xb l 1 + r) S b (xb l 2 + r) = 2(S b (xb l 2 + r)+s b (r)) S b (xb l 2 + r) = S b (xb l 2 + r)+s b (r)+s b (r) = S b (xb l 1 + r)+s b (r), which proves (2) Now if u = u with {1,,b 1} and u {0,,b 1}, then (2) directly follows from Definition 1 and Lemma 3 Let us prove (3) If l =1,thenr = 0 and (2) follows from able 1 Now suppose that l 2 and assume that (3) holds for all l <lletr {0,,b l 1 1}, and let u be a word in {0,,b 1} such that u 1andrep b (xb l + xb l 1 + r) =xxu he proof is divided into two parts according to the first letter of u Ifu =0u with u {0,,b 1},then,usingLemma2and(2), S b (xb l + xb l 1 + r) = S b (xb l 1 + r)+s b (xb l 2 + r) = S b (xb l 2 + r)+s b (r)+s b (xb l 2 + r) = 2(S b (xb l 2 + r)+s b (r)) S b (r) = 2S b (xb l 1 + r) S b (r), which proves (3) Now if u = u with {1,,b 1} and u {0,,b 1}, then (3) directly follows from Definition 1 and Lemma 4 Let us finally prove (4) If l =1,thenr =0and(2)followsfromable1 Now suppose that l 2 and assume that (4) holds for all l <lletr {0,,b l 1 1}, let be a letter in {0,,b 1} and let u be a word in {0,,b 1} such that rep b (xb l + yb l 1 + r) =xyu Using Definition 1 and Lemma 5, we directly have that S b (xb l + yb l 1 + r) =S b (xb l 1 + r)+s b (yb l 1 + r) 2S b (r) since rep b (r) =rep b (val b (u)), which proves (4) 3 Regularity of the Sequence (S b (n)) n 0 he sequence (S 2 (n)) n 0 is known to be 2-regular; see [13] b-kernel of a sequence s =(s(n)) n 0 is the set We recall that the K b (s) ={(s(b i n + j)) n 0 i 0and0 j<b i } Asequences = (s(n)) n 0 Z N is b-regular if there exists a finite number of sequences (t 1 (n)) n 0,, (t l (n)) n 0 such that every sequence in the Z-module K b (s) generated by the b-kernel K b (s) is a Z-linear combination of the t r s In

16 INEGERS: 18A (2018) 16 this section, we prove that the sequence (S b (n)) n 0 is b-regular As a consequence, one can get matrices to compute S b (n) in a number of matrix multiplications proportional to log b (n) o prove the b-regularity of the sequence (S b (n)) n 0 for any base b, we first need a lemmnvolving some matrix manipulations Lemma 6 Let I and 0 respectively be the identity matrix of sie b 2 b 2 and the ero matrix of sie b 2 b 2 LetM b be the block-matrix of sie b 3 b 3 I I 2I 2I 2I 3I 3I 4I 4I 4I M b := 4I 3I 2I 3I 4I 4I More precisely, the matrix M b is the block-matrix (B i,j ) 1 i,j b,whereb i,j is the matrix of sie b 2 b 2 such that I, if i =1and j {1, 2}; 2I, if (i =1and j 3) or (j =1and i 2); B ij = 3I, if (j =2and i 2) or (j = i +1 3); 4I, otherwise his matrix is invertible and its inverse is given by 3I 2I 2I (2b 3)I 2I 0 0 I M 1 0 I b := I I More precisely, the matrix M 1 b is the block-matrix (C i,j ) 1 i,j b,wherec i,j is the matrix of sie b 2 b 2 such that 3I, if i = j =1; (2b 3)I, if i =1and j = b; 2I, if i =1and 2 j<b; C ij = 2I, if i =2and j =1; I, if j = b and i 2; I, if i = j +1 2; 0, otherwise

17 INEGERS: 18A (2018) 17 rep b (r) ε x b 1 x0 (b 1)0 xx a r 1 2 2b 3 2 4b 4 1 rep b (r) (b 1)(b 1) xy (b 1)x x(b 1) a r 4b 3 2 4b 4 2b 3 able 2: Values of a r for 0 r<b 2 with x, y {1,,b 2} and x y rep b (r) ε x b 1 x0 (b 1)0 xx c r, rep b (r) (b 1)(b 1) xy (b 1)x x(b 1) c r, able 3: Values of c r,0 for 0 r<b 2 with x, y {1,,b 2} and x y For the proof of the previous lemma, simply proceed to the multiplication of the two matrices Using this lemma, we prove that the sequence (S b (n)) n 0 is b-regular heorem 1 For all r {0,,b 2 1}, wehave b 2 S b (nb 2 + r) =a r S b (n)+ c r,s S b (nb + s) for all n 0, (5) s=0 where the coefficients a r and c r,s are unambiguously determined by the first few values S b (0), S b (1),, S b (b 3 1) and s in able 2, able 3 and able 4 In particular, the sequence (S b (n)) n 0 is b-regular Moreover, a set of generators for K b (s) is given by the b sequences (S b (n)) n 0, (S b (bn)) n 0, (S b (bn +1)) n 0,, (S b (bn + b 2)) n 0 Proof We proceed by induction on n 0 For the base case n {0, 1,,b 2 1}, we first compute the coefficients a r and c r,s using the values of S b (nb 2 + r) for n {0,,b 1} and r {0,,b 2 1} hen we show that (5) also holds with these coefficients for n {b,, b 2 1} Base case Let I denote the identity matrix of sie b 2 b 2 he system of b 3 equations (5) when n {0,,b 1} and r {0,,b 2 1} can be written as

18 INEGERS: 18A (2018) 18 rep b (r) ε x b 1 x0 (b 1)0 xx s x x x c r,s rep b (r) (b 1)(b 1) xy x(b 1) (b 1)x s x y x x c r,s able 4: Values of c r,s for 0 r<b 2 and 1 s b 2withx, y, {1,,b 2} pairwise distinct MX = V where the matrix M Z b3 b is equal to 3 S b (0)I S b (0)I S b (1)I S b (b 2)I S b (1)I S b (b)i S b (b +1)I S b (2b 2)I S b (b 1)I S b (b(b 1))I S b (b(b 1) + 1)I S b (b(b 1) + b 2)I and the vectors X, V Z b3 are respectively given by X = ( a 0 a b 2 1 c 0,0 c b 2 1,0 c 0,b 2 c b 2 1,b 2 V = ( S b (0) S b (1) S b (b 3 1) ) ), Observe that in the vector X, thecoefficientsc r,s are first sorted by s then by r Using able 1, the matrix M is equal to the matrix M b of Lemma 6 By this lemma, the previous system has a unique solution given by X = M 1 b V Consequently, using Lemma 6, we have, for all r {0,,b 2 1} and all s {1,,b 2}, b 2 a r = 3S b (r)+2 S b (jb 2 + r) (2b 3) S b ((b 1)b 2 + r), j=1 c r,0 = 2S b (r)+s b ((b 1)b 2 + r), c r,s = S b (sb 2 + r)+s b ((b 1)b 2 + r) he values of the coefficients can then be computed using able 1 and are stored in able 2, able 3 and able 4 For n {b,, b 2 1}, the values of S b (nb 2 + r) arestoredinable5,able6 and able 7 according to whether rep b (n) is of the form x0, xx or xy with x y he proof that (5) holds for each n {b,, b 2 1} only requires easy computations that are left to the reader Inductive step Consider n b 2 and suppose that the relation (5) holds for all m<n hen rep b (n) 3 Like for the base case, we need to consider several

19 INEGERS: 18A (2018) 19 rep b (r) ε x y x0 y0 xx yy xy yx y S b (nb 2 + r) able 5: Values of S b (nb 2 + r) forb n<b 2 with rep b (n) =x0 andx, y, {1,,b 1} pairwise distinct rep b (r) ε x y x0 y0 xx yy xy yx y S b (nb 2 + r) able 6: Values of S b (nb 2 + r) forb n<b 2 with rep b (n) =xx and x, y, {1,,b 1} pairwise distinct cases according to the form of the base-b expansion of n Moreprecisely,weneedto consider the following five forms, where u {0,,b 1}, x, y, {1,,b 1}, x, andt {0,,b 1}: x00u or xx0u or x0yu or xxyu or xtu Let us focus on the first form of rep b (n) sincethesamereasoningcanbeapplied for the other ones Assume that rep b (n) =x00u where x {1,,b 1} and u {0,,b 1} For all r {0,,b 2 1}, thereexistr 1,r 2 {0,,b 1} rep b (r) ε x y x0 y0 0 xx yy S b (nb 2 + r) rep b (r) xy x yx y x y t S b (nb 2 + r) able 7: Values of S b (nb 2 + r) forb n<b 2 with rep b (n) =xy and x, y,, t {1,,b 1} pairwise distinct

20 INEGERS: 18A (2018) 20 such that val b (r 1 r 2 )=r Using Lemma 1 and the induction hypothesis, we have which proves (5) S b (nb 2 + r) = S b (val b (x00ur 1 r 2 )) = 2S b (val b (x0ur 1 r 2 )) S b (val b (xur 1 r 2 )) b 2 = a r 2S b (val b (x0u)) + c r,s 2S b (val b (x0us)) s=0 b 2 a r S b (val b (xu)) c r,s S b (val b (xus)) s=0 b 2 = a r S b (val b (x00u)) + c r,s S b (val b (x00us)) s=0 b 2 = a r S b (n)+ c r,s S b (nb + s), s=0 b-regularity From the first part of the proof, we directly deduce that the Z-module K b (S b ) is generated by the (b +1)sequences (S b (n)) n 0, (S b (bn)) n 0, (S b (bn +1)) n 0,,(S b (bn + b 1)) n 0 We now show that we can reduce the number of generators o that aim, we prove that b 2 S b (nb + b 1) = (2b 1)S b (n) S b (nb + s) for all n 0 (6) We proceed by induction on n 0 As a base case, the proof that (6) holds for each n {b,, b 2 1} only requires easy computations that are left to the reader (using able 1) Now consider n b 2 and suppose that the relation (6) holds for all m<n hen rep b (n) 3 Mimicking the first induction step of this proof, we need to consider several cases according to the form of the base-b expansion of n More precisely, we need to consider the following five forms, where u {0,,b 1}, x, y, {1,,b 1}, x, andt {0,,b 1}: s=0 x00u or xx0u or x0yu or xxyu or xtu Let us focus on the first form of rep b (n) sincethesamereasoningcanbeapplied for the other ones Assume that rep b (n) =x00u where x {1,,b 1} and

21 INEGERS: 18A (2018) 21 u {0,,b 1} Using Lemma 1 and the induction hypothesis, we have S b (nb + b 1) = S b (val b (x00u(b 1))) = 2S b (val b (x0u(b 1))) S b (val b (xu(b 1))) b 2 = (2b 1) 2S b (val b (x0u)) 2S b (val b (x0us)) s=0 b 2 (2b 1)S b (val b (xu)) + S b (val b (xus)) s=0 b 2 = (2b 1)S b (val b (x00u)) S b (val b (x00us)) s=0 b 2 = (2b 1)S b (n) S b (nb + s), which proves (5) he Z-module K b (S b ) is thus generated by the b sequences (S b (n)) n 0, (S b (bn)) n 0, (S b (bn +1)) n 0,,(S b (bn + b 2)) n 0 s=0 Example 4 Let b =2 Usingable2,able3andable4,wefindthata 0 = 1, a 1 =1,a 2 =4,a 3 =5,c 0,0 =2,c 1,0 =1,c 2,0 = 1 andc 3,0 = 2 In this case, there are no c r,s with s>0 Applying heorem 1 and (6), we get S 2 (2n +1) = 3S 2 (n) S 2 (2n), S 2 (4n) = S 2 (n)+2s 2 (2n), S 2 (4n +1) = S 2 (n)+s 2 (2n), S 2 (4n +2) = 4S 2 (n) S 2 (2n), S 2 (4n +3) = 5S 2 (n) 2S 2 (2n) for all n 0 his result is a rewriting of [13, heorem 21] Observe that the third and the fifth identities are redundant: they follow from the other ones Example 5 Let b =3 Usingable2,able3andable4,thevaluesofthe coefficients a r, c r,0 and c r,1 can be found in able 8 Applying heorem 1 and (6),

22 INEGERS: 18A (2018) 22 r a r c r, c r, able 8: he values of a r,c r,0,c r,1 when b =3andr {0,,8} we get S 3 (3n +2) = 5S 3 (n) S 3 (3n) S 3 (3n +1), S 3 (9n) = S 3 (n)+2s 3 (3n), S 3 (9n +1) = 2S 3 (n)+2s 3 (3n)+S 3 (3n +1), S 3 (9n +2) = 3S 3 (n)+s 3 (3n) S 3 (3n +1), S 3 (9n +3) = 2S 3 (n)+s 3 (3n)+2S 3 (3n +1), S 3 (9n +4) = S 3 (n)+2s 3 (3n +1), S 3 (9n +5) = 3S 3 (n) S 3 (3n)+S 3 (3n +1), S 3 (9n +6) = 8S 3 (n) S 3 (3n) 2S 3 (3n +1), S 3 (9n +7) = 8S 3 (n) 2S 3 (3n) S 3 (3n +1), S 3 (9n +8) = 9S 3 (n) 2S 3 (3n) 2S 3 (3n +1) for all n 0 his result is a proof of [13, Conjecture 26] Observe that the fourth, the seventh and the tenth identities are redundant Remark 2 Combining (5) and (6) yield b 2 +1 identities to generate the Z-module K b (S b ) However, as illustrated in Example 4 and Example 5, only b 2 b +1 identities are useful: the relations established for the sequences (S b (b 2 n + br + b 1)) n 0,withr {0,,b 1}, canbededucedfromtheotheridentities Remark 3 Using heorem 1 and (6) and the set of b generators of the Z-module K b (S b ) being {(S b (n)) n 0, (S b (bn)) n 0, (S b (bn +1)) n 0,,(S b (bn + b 2)) n 0 }, we get matrices to compute S b (n) in a number of steps proportional to log b (n) For all n 0, let S b (n) S b (bn) V b (n) = S b (bn +1) Z b S b (bn + b 2)

23 INEGERS: 18A (2018) 23 Consider the matrix-valued map µ b : {0, 1,,b 1} Z b b defined as follows If s {0,,b 2}, thenweset µ b (s) := ( A(s) C 0 (s) C s 1 (s) C s (s) C s+1 (s) C b 2 (s) ) where the vectors A(s),C 0 (s),,c b 2 (s) Z b are given by A(s) = ( 0 a bs a bs+1 a bs+b 2 ) ; C i (s) = ( 0 c bs,i c bs+1,i c bs+b 2,i ) for all i {0,,b 2}\{s}; C s (s) = ( 1 c bs,s c bs+1,s c bs+b 2,s ) We also set µ b (b 1) := (2b 1) a b(b 1) c b(b 1),0 c b(b 1),1 c b(b 1),b 2 a b(b 1)+1 c b(b 1)+1,0 c b(b 1)+1,1 c b(b 1)+1,b 2 a b(b 1)+b 2 c b(b 1)+b 2,0 c b(b 1)+b 2,1 c b(b 1)+b 2,b 2 Observe that the number of generators explains the sie of the matrices above For each s {0,,b 2}, exactly b 1 identities from heorem 1 are used to define the matrix µ b (s) If s, s {0,,b 2} are such that s s,thentherelationsusedto define the matrices µ b (s) andµ b (s )arepairwisedistinct Finally,thefirstrowof the matrix µ b (b 1) is (6) and the other rows are b 1 identities from heorem 1, which are distinct from the previous relations Consequently, (b 1)(b 1) + b identities are used, which corroborates Remark 2 Using the definition of the map µ b,wecanshowthatv b (bn + s) =µ b (s)v b (n) for all s {0,,b 1} and n 0 Consequently, if rep b (n) =n k n 0,then S b (n) = ( ) µ b (n 0 ) µ b (n k ) V b (0) For example, when b =2,thematricesµ 2 (0) and µ 2 (1) are those given in [13, Corollary 22] When b =3,weget µ 3 (0) = ,µ 3 (1) = , µ 3 (2) = he class of b-synchronied sequences is intermediate between the classes of b- automatic sequences and b-regular sequences he map rep b is extended to N N as follows For all m, n N, rep b (m, n) := (0 M rep b (m) rep b (m), 0 M rep b (n) rep b (n)) where M =max{ rep b (m), rep b (n) } heideaisthattheshorterwordispadded with leading eros to get two words of the same length A sequence s =(s(n)) n 0 Z N is b-synchronied if the language {rep b (n, s(n)) n N} is accepted by some finite automaton reading pairs of digits hese sequences were first introduced in [8]

24 INEGERS: 18A (2018) 24 Proposition 3 he sequence (S b (n)) n 0 is not b-synchronied Proof he proof is exactly the same as [13, Proposition 24] o conclude this section, the following result shows that the sequence (S b (n)) n 0 has a partial palindromic structure For instance, the sequence (S 3 (n)) n 0 is depicted in Figure 7 inside the interval [2 3 4, 3 5 ] Figure 7: he sequence (S 3 (n)) n 0 inside the interval [2 3 4, 3 5 ] Proposition 4 Let u be a word in {0, 1,,b 1} Define ū by replacing in u every letter a {0, 1,,b 1} by the letter (b 1) a {0, 1,,b 1} hen # {v L b v (b 1)u} =#{v L b v (b 1)ū} In particular, there exists a palindromic substructure inside of the sequence (S b (n)) n 0,ie,foralll 1 and 0 r<b l, S b ((b 1) b l + r) =S b ((b 1) b l + b l r 1) Proof he trees ((b 1)u) and ((b 1)ū) areisomorphic Indeed,ontheone hand, each node of the form (b 1)x in the first tree corresponds to the node (b 1) x in the second one and conversely On the other hand, if there exist letters a {1,,b 2} in the word (b 1)u, the position of the first letter n the word (b 1)u is equal to the position of the first letter (b 1) n the word (b 1)ū and conversely Consequently, the node of the form ax in the first tree corresponds to the node of the form ((b 1) a) x in the second tree and conversely For the special case, note that for every word of length l, there exists r {0,,b l 1} such that rep b ((b 1) b l + r) =(b 1) and val b ( ) =b l 1 r {0,b l 1} Hence, (b 1) =rep b ((b 1) b l + b l 1 r) Using (1), we obtain the desired result

25 INEGERS: 18A (2018) 25 4 Asymptotics of the Summatory Function (A b (n)) n 0 In this section, we consider the summatory function (A b (n)) n 0 of the sequence (S b (n)) n 0 ; see Definition 1 he aim of this section is to apply the method introduced in [14] to obtain the asymptotic behavior of (A b (n)) n 0 As an easy consequence of the b-regularity of (S b (n)) n 0,wehavethefollowingresult Proposition 5 For all b 2, thesequence(a b (n)) n 0 is b-regular Proof his is a direct consequence of heorem 1 and of the fact that the summatory function of a b-regular sequence is also b-regular; see [2, heorem 1641] Remark 4 From a linear representation with matrices of sie d d associated with a b-regular sequence, one can derive a linear representation with matrices of sie 2d 2d associated with its summatory function; see [9, Lemma 1] Consequently, using Remark 3, one can obtain a linear representation with matricesofsie2b 2b for the summatory function (A b (n)) n 0 In order to prove heorem 2, the goal is to decompose (A b (n)) n 0 into linear combinations of powers of (2b 1) We need the following two lemmas Lemma 7 For all l 0 and all x {1,,b 1}, wehave A b (xb l )=(2x 1) (2b 1) l Proof We proceed by induction on l 0 If l =0andx {1,,b 1}, then using able 1, we have x 1 A b (x) =S b (0) + S b (j) =2x 1 j=1 If l =1andx {1,,b 1}, thenwehave x 1 b 1 A b (xb) =A b (b)+ S b (yb + j) y=1 Using able 1, we get A b (xb) =(2x 1)(2b 1) Now suppose that l 1 and assume that the result holds for all l l oprove the result, we again proceed by induction on x {1,,b 1} When x =1,we must show that A b (b l+1 )=(2b 1) l+1 Wehave b 1 A b (b l+1 )=A b (b l )+ b l 1 y=1 S b (yb l + j)

26 INEGERS: 18A (2018) 26 By decomposing the sum into three parts accordingly to Proposition 1, we get b 1 A b (b l+1 ) = A b (b l )+ b 1 + y=1 and, using Proposition 1, A b (b l+1 ) = A b (b l ) b 1 + y=1 b 1 + y=1 b 1 + y=1 y=1 1 b 1 y b l 1 1 b l 1 1 By observing that for all y, we obtain b l 1 1 b l b 1 y b l 1 1 b l 1 1 b 1 S b (yb l + j)+ y=1 b l 1 1 S b (yb l + b l 1 + j), S b (yb l + yb l 1 + j) (S b (yb l 1 + j)+s b (j)) (7) (2S b (yb l 1 + j) S b (j)) (8) b l 1 1 (S b (yb l 1 + j)+2s b (b l 1 + j) 2S b (j))(9) S b (yb l 1 + j) =A b ((y +1)b l 1 ) A b (yb l 1 ), (10) S b (j)) = A b (b l 1 ), (11) b 1 ( Ab ((y +1)b l 1 ) A b (yb l 1 ) ) = A b (b l ) A b (b l 1 ), (12) y=1 (7) = A b (b l )+(b 2)A b (b l 1 ), (8) = 2A b (b l ) (b +1)A b (b l 1 ), (9) = 3(b 2)(A b (b l ) A b (b l 1 )) 2(b 1)(b 2)A b (b l 1 ) = 3(b 2)A b (b l ) (b 2)(2b +1)A b (b l 1 ), and finally A b (b l+1 )=(3b 2)A b (b l ) (2b 2 3b +1)A b (b l 1 )

27 INEGERS: 18A (2018) 27 Using the induction hypothesis, we obtain A b (b l+1 )=(3b 2)(2b 1) l (2b 2 3b +1)(2b 1) l 1 =(2b 1) l+1, which ends the case where x =1 Now suppose that x {2,,b 1} and assume that the result holds for all x <x he proof follows the same lines as in the case x =1withthedifference being that we decompose the sum into b l+1 1 A b (xb l+1 ) = A b ((x 1)b l+1 )+ S b ((x 1)b l+1 + j) b l 1 = A b ((x 1)b l+1 )+ S b ((x 1)b l+1 + j) b l 1 + S b ((x 1)b l+1 +(x 1)b l + j) + 1 y b 1 y x 1 b l 1 S b ((x 1)b l+1 + yb l + j) Applying Proposition 1 and using (10), (11) and (12) lead to the equality A b (xb l+1 ) = A b ((x 1)b l+1 )+(b 1)A b (xb l ) (b 1)A b ((x 1)b l ) +2A b (b l+1 ) 2(b 1)A b (b l ) he induction hypothesis ends the computation Lemma 8 For all l 1 and all x, y {1,,b 1}, wehave { A b (xb l + yb l 1 (4xb 2x +4y 2b) (2b 1) l 1, if y x; )= (4xb 2x +4y 2b 1) (2b 1) l 1, if y>x Proof he proof of this lemms similar to the proof of Lemma 7 so we only prove the formula for A b (xb l + xb l 1 ), the others being similarly handled We proceed by induction on l 1 If l =1,theresultfollowsfromable1 Assumethatl 2 and that the formulas hold for all l <lwehave b l 1 1 A b (xb l + xb l 1 )=A b (xb l )+ S b (xb l + j)+ x 1 y=1 b l 1 1 S b (xb l + yb l 1 + j) Applying Proposition 1 and using (10), (11) and (12) leads to the equality A b (xb l +xb l 1 )=A b (xb l )+xa b ((x+1)b l 1 )+(2 x)a b (xb l 1 )+(1 2x)A b (b l 1 ) Using Lemma 7 completes the computation

28 INEGERS: 18A (2018) 28 Lemma 7 and Lemma 8 give rise to recurrence relations satisfied by the summatory function (A b (n)) n 0 as stated below his is a key result that permits us to introduce (2b 1)-decompositions (Definition 4 below) of the summatory function (A b (n)) n 0 and allows us to easily deduce heorem 2; see [14] for similar results in base 2 It permits us to express A b (n) as a linear combination of a power of (2b 1) and elements of the form A b (m) withm<n Proposition 6 For all x, y {1,,b 1} with x y, alll 1 and all r {0,,b l 1 }, A b (xb l + r) = (2b 2) (2x 1) (2b 1) l 1 + A b (xb l 1 + r) +A b (r); (13) A b (xb l + xb l 1 + r) = (4xb 2x 2b +2) (2b 1) l 1 +2A b (xb l 1 + r) A b (xb l + yb l 1 + r) = A b (r); (14) (4xb 4x 2b +3) (2b 1) l 1 +A b (xb l 1 + r) +2A b (yb l 1 + r) 2A b (r), if y<x; (4xb 4x 2b +2) (2b 1) l 1 (15) +A b (xb l 1 + r)+2a b (yb l 1 + r) 2A b (r), if y>x Proof We first prove (13) Let x {1,,b 1}, l 1andr {0,,b l 1 } If r = 0, then (13) holds using Lemma 7 Now suppose that r {1,,b l 1 } Applying successively Proposition 1 and Lemma 7, we have r 1 A b (xb l + r) = A b (xb l )+ S b (xb l + j) r 1 = A b (xb l )+ (S b (xb l 1 + j)+s b (j)) = A b (xb l )+(A b (xb l 1 + r) A b (xb l 1 )) + A b (r) = (2b 2)(2x 1)(2b 1) l 1 + A b (xb l 1 + r)+a b (r), which proves (13) he proof of (14) and (15) are similar, thus we only prove (14) Let x {1,,b 1}, l 1andr {0,,b l 1 } If r = 0, then (14) holds using Lemma 8 Now

29 INEGERS: 18A (2018) 29 suppose that r {1,,b l 1 }ApplyingProposition1,wehave r 1 A b (xb l + xb l 1 + r) = A b (xb l + xb l 1 )+ S b (xb l + xb l 1 + j) Using Lemma 7 and Lemma 8, we get r 1 = A b (xb l + xb l 1 )+ (2S b (xb l 1 + j) S b (j)) = A b (xb l + xb l 1 )+2(A b (xb l 1 + r) A b (xb l 1 )) A b (r) A b (xb l + xb l 1 + r) = (4xb +2x 2b)(2b 1) l 1 2(2x 1)(2b 1) l 1 which proves (14) +2A b (xb l 1 + r) A b (r) = (4xb 2x 2b +2)(2b 1) l 1 +2A b (xb l 1 + r) A b (r), he following corollary was conjectured in [14] Corollary 1 For all n 0, wehavea b (nb) =(2b 1)A b (n) Proof Let us proceed by induction on n 0 It is easy to check by hand that the result holds for n {0,,b 1} hus consider n b and suppose that the result holds for all n <n he reasoning is divided into three cases according to the form of the base-b expansion of n As a first case, we write n = xb l + r with x {1,,b 1}, l 1and0 r<b l 1 By Proposition 6, we have A b (nb) (2b 1)A b (n) = (2b 2) (2x 1) (2b 1) l + A b (xb l + br) +A b (br) (2b 2) (2x 1) (2b 1) l (2b 1)A b (xb l 1 + r) (2b 1)A b (r) We conclude this case by using the induction hypothesis he other cases can be handled using the same technique Using Proposition 6, we can define (2b 1)-decompositions as follows Definition 4 Let n b decomposition of the form Iteratively applying Proposition 6 provides a unique A b (n) = l b (n) i=0 d i (n)(2b 1) l b(n) i where d i (n) are integers, d 0 (n) 0andl b (n) standsfor log b n 1 We say that the word d 0 (n) d lb (n)(n)

30 INEGERS: 18A (2018) 30 is the (2b 1)-decomposition of A b (n) For the sake of clarity, we also write (d 0 (n),,d lb (n)(n)) Also notice that the notion of (2b 1)-decomposition is only valid for integers in the sequence (A b (n)) n 0 Example 6 Let b =3 Letuscomputethe5-decompositionofA 3 (150) = 1665 We have rep 3 (150) = and l 3 (150) = 3 Applying once Proposition 6 leads to A 3 (150) = A 3 ( ) = A 3 ( )+2A 3 ( ) 2A 3 (15) (16) Applying Proposition 6 on terms of the form A 3 (m) that have just appeared in the rhs of (16), we get A 3 ( ) = A 3 ( ) = A 3 (3 2 +6) A 3 (6), A 3 ( ) = A 3 ( ) = A 3 ( )+2A 3 (3 2 +6) 2A 3 (6), A 3 (15) = A 3 ( ) = A 3 (3 1 )+2A 3 (2 3 1 ) 2A 3 (0) Using again Proposition 6 on the new terms of the form A 3 (m), we find A 3 (3 2 +6) = A 3 ( ) = A 3 (3 1 )+2A 3 (2 3 1 ) 2A 3 (0), A 3 ( ) = A 3 ( ) = A 3 (2 3 1 ) A 3 (0), A 3 (6) = A 3 (2 3 1 ) = A 3 (2 3 0 )+A 3 (0) = Using Lemma 7, we have A 3 (3 1 )=5 1 and A 3 (2 3 1 )=3 5 1,andtheprocedure halts Plugging all those values together in (16), we finally have A 3 (150) = he 5-decomposition of A 3 (150) is thus (4, 32, 82, 45) he proof of the next result follows the same lines as the proof of [14, heorem 1] herefore we only sketch it heorem 2 here exists a continuous and periodic function H b of period 1 such that, for all large enough n, A b (n) =(2b 1) log b n H b (log b n)

31 INEGERS: 18A (2018) 31 Figure 8: he function H 3 over one period As an example, when b =3,thefunctionH 3 is depicted in Figure 8 over one period Sketch of the proof of heorem 2 Let us start by defining the function H b Given any integer n 1, we let φ n denote the function A b (e n (α)) φ n (α) =, α [0, 1) (2b 1) log b (en(α)) where e n (α) = b n+1 + b b n α + 1 With a proof analogous to the one of [14, Proposition 20], the sequence of functions (φ n ) n 1 uniformly converges to a function Φ b As in [14, heorem 5], this function is continuous on [0, 1] and such that Φ b (0) = Φ b (1) = 1 Furthermore, it satisfies ( A b (b k + r) =(2b 1) log b r ) (bk +r) Φ b b k k 1, 0 r<b k ; (17) see [14, Lemma 24] Using Corollary 1, we get that, for all n = b j (b k + r), j, k 0 and r {0,,b k 1}, ( A b (n) =(2b 1) log b r ) (n) Φ b b k he function H b is defined by H b (x) =Φ b (b {x} 1) for all real x ({ } stands for the fractional part) Remark 5 As stated in [5, Remark 922], observe that since the 1-periodic function Φ b is continuous, it is completely defined in the interval [0, 1] by the values taken on the dense set of points of the form r/b k Having no error term for these values, see (17), there is no error term in heorem 2 Remark 6 he sequence (S 2 (n)) n 0 turns out to be the subsequence with odd indices of the Stern Brocot sequence [13] Hence the present paper could motivate the quest for generalied Stern Brocot sequences and analogues of the Farey tree [4, 6, 7, 10, 11, 15] Namely can one reasonably define a tree structure, or some other combinatorial structure, in which the sequence (S b (n)) n 0 naturally appears?

32 INEGERS: 18A (2018) 32 References [1] J-P Allouche and J Shallit, he ring of k-regular sequences, heoret Comput Sci 98 (1992), [2] J-P Allouche and J Shallit, Automatic Sequences heory, Applications, Generaliations, Cambridge University Press, 2003 [3] J-P Allouche and J Shallit, he ring of k-regular sequences II, heoret Comput Sci 307 (2003), no 1, 3 29 [4] B Bates, M Bunder, and K ognetti, Locating terms in the Stern Brocot tree, European J Combin 31 (2010), [5] V Berthé, M Rigo (Eds), Combinatorics, automata and number theory, Encycl of Math and its Appl 135, CambridgeUniversityPress,2010 [6] N J Calkin and HS Wilf, Binary partitions of integers and Stern Brocotlike trees (1998), unpublished Updated version August 5, 2009, 19 pages, [7] I Canakci and R Schiffler, Cluster algebras and continued fractions, unpublished Updated version September 26, 2016, 28 pages, [8] A Carpi and C Maggi, On synchronied sequences and their separators, heor Inform Appl 35 (2001), [9] P Dumas, Joint spectral radius, dilation equations, and asymptotic behavior of radix-rational sequences, Linear Algebra Appl 438 (2013), no 5, [10] Garrity, A multidimensional continued fraction generaliation of Stern s diatomic sequence, J Integer Seq 16 (2013), no 7, Article 1377, 24 [11] S P Glasby, Enumerating the rationals from left to right, Amer Math Monthly 118 (2011), [12] J Leroy, M Rigo, and M Stipulanti, Generalied Pascal triangle for binomial coefficients of words, Adv in Appl Math 80 (2016), [13] J Leroy, M Rigo, and M Stipulanti, Counting the number of non-ero coefficients in rows of generalied Pascal triangles, Discrete Math 340 (2017), [14] J Leroy, M Rigo, and M Stipulanti, Behavior of digital sequences through exotic numeration systems, Electron J Combin 24 (2017), no 1, Paper 144, 36 pp [15] S Morier-Genoud, V Ovsienko, and S abachnikov, SL 2 (Z)-tilings of the torus, Coxeter Conway friees and Farey triangulations, Enseign Math 61 (2015), [16] N J A Sloane, he On-Line Encyclopedia of Integer Sequences Published electronically at