EXPANSIONS IN NON-INTEGER BASES: LOWER ORDER REVISITED

EXPANSIONS IN NON-INTEGER BASES: LOWER ORDER REVISITED Simon Baker School of Mathematics, University of Manchester, Manchester, United Kingdom simonbaker4@gmail.com Nikita Sidorov School of Mathematics, University of Manchester, Manchester, United Kingdom sidorov@manchester.ac.uk Received:, Revised:, Accepted:, Published: Abstract Let q (, ) and x [0, q ]. We say that a sequence (ε i) i= {0, }N is an expansion of x in base q (or a q-expansion) if x = ε i q i. i= For any k N, let B k denote the set of q such that there exists x with exactly k expansions in base q. In [] it was shown that min B = q.7064, the appropriate root of x 4 = x + x +. In this paper we show that for any k 3, min B k = q f.75488, the appropriate root of x 3 = x x +.. Introduction Let q (, ) and I q = [0, q ]. Given x R, we say that a sequence (ε i) i= {0, } N is a q-expansion for x if x = i= ε i q i. () Expansions in non-integer bases were pioneered in the papers of Rényi [0] and Parry [9]. It is a simple exercise to show that x has a q-expansion if and only if x I q, when () holds we will adopt the notation x = (ε, ε,...) q. Given x I q we denote

INTEGERS: 4 (04) the set of q-expansions for x by Σ q (x), i.e., Σ q (x) = { (ε i ) i= {0, } N : i= ε } i q i = x. In [5] it is shown that for q (, + 5 ) the set Σ q (x) is uncountable for all x (0, q ); the endpoints of I q trivially have a unique q-expansion for all q (, ). In [4] it is shown that for q = + 5 every x (0, q ) has uncountably many q- expansions unless x = (+ 5)n mod, for some n Z, in which case Σ q (x) is infinite countable. Moreover, in [3] it is shown that for all q ( + 5, ) there exists x (0, q ) with a unique q-expansion. In this paper we will be interested in the set of q (, ) for which there exists x I q with precisely k q-expansions. More specifically, we will be interested in the set { ( ) } B k := q (, ) there exists x 0, satisfying #Σ q (x) = k. q It was shown in [4] that B k for any k. Similarly we can define B ℵ0 and B ℵ 0. The reader should bear in mind the possibility that the number of expansions could lie strictly between countable infinite and the continuum. By the above remarks it is clear that B = ( + 5, ). In [] the following theorem was shown to hold. Theorem.. The smallest element of B is q.7064, the appropriate root of x 4 = x + x +. The next smallest element of B is q f.75488, the appropriate root of x 3 = x x +. For each k N there exists γ k > 0 such that ( γ k, ) B j for all j k. The following theorem is the central result of the present paper. It answers a question posed by V. Komornik [7] (see also [, Section 5]). Theorem.. For k 3 the smallest element of B k is q f. The range of q > + 5 which are sufficiently close to the golden ratio is referred to in [] as the lower order, which explains the title of the present paper. During our proof of Theorem. we will also show that q f B ℵ0, which combined with our earlier remarks, Theorem., Theorem. and a result in [] which states that for q [ + 5, ) almost every x I q has a continuum of q-expansions, we can conclude the following.

INTEGERS: 4 (04) 3 Theorem.3. In base q f all situations occur: there exist x I q having exactly k q-expansions for each k =,,..., k = ℵ 0 or k = ℵ 0. Moreover, q f is the smallest q (, ) satisfying this property. Before proving Theorem. it is necessary to recall some theory. In what follows we fix T q,0 (x) = qx and T q, (x) = qx, we will typically denote an element of n=0 {T q,0, T q, } n by a; here {T q,0, T q, } 0 denotes the set consisting of the identity map. Moreover, if a = (a,..., a n ) we shall use a(x) to denote (a n... a )(x) and a to denote the length of a. We let { } Ω q (x) = (a i ) i= {T q,0, T q, } N : (a n... a )(x) I q for all n N. The significance of Ω q (x) is made clear by the following lemma. Lemma.4. #Σ q (x) = #Ω q (x) where our bijection identifies (ε i ) i= with (T q,ε i ) i=. The proof of Lemma.4 is contained within []. It is an immediate consequence of Lemma.4 that we can interpret Theorem. in terms of Ω q (x) rather than Σ q (x). An element x I q satisfies T q,0 (x) I q and T q, (x) I q if and only if x [ q, q(q ) ]. Moreover, if #Σ q(x) > or equivalently #Ω q (x) >, then there exists a unique minimal sequence of transformations a such that a(x) [ q, q(q ) ]. In what follows we let S q := [ q, q(q ) ]. The set S q is usually referred to as the switch region. We will also make regular use of the fact that if x I q and a is a sequence of transformations such that a(x) I q, then #Ω q (x) #Ω q (a(x)) or equivalently #Σ q (x) #Σ q (a(x)), () this is immediate from the definition of Ω q (x) and Lemma.4. In the course of our proof of Theorem. we will frequently switch between Σ q (x) and the dynamical interpretation of Σ q (x) provided by Lemma.4, often considering Ω q (x) will help our exposition. The following lemma is a consequence of [6, Theorem ]. Lemma.5. Let q ( + 5, q f ], if x I q has a unique q-expansion (ε i ) i=, then { (ε i ) i= 0 k (0), k (0), 0, }, where k 0. Similarly, if (ε i ) i= {0k (0), k (0), 0, } then for q ( + 5, ) x = ((ε i ) i= ) q has a unique q-expansion given by (ε i ) i=. In Lemma.5 we have adopted the notation (ε... ε n ) k to denote the concatenation of (ε... ε n ) {0, } n by itself k times and (ε... ε n ) to denote the infinite

INTEGERS: 4 (04) 4 sequence obtained by concatenating ε... ε n by itself infinitely many times, and we will use this notation throughout. The following lemma follows from the branching argument first introduced in [3]. Lemma.6. Let k, x I q and suppose #Σ q (x) = k or equivalently #Ω q (x) = k. If a is the unique minimal sequence of transformations such that a(x) S q, then #Ω q (T q, (a(x))) + #Ω q (T q,0 (a(x))) = k. Moreover, #Ω q (T q, (a(x))) < k and #Ω q (T q,0 (a(x))) < k. The following result is an immediate consequence of Lemma.4 and Lemma.6. Corollary.7. B k B for all k 3. An outline of our proof of Theorem. is as follows: first of all we will show that q f B k for all k, then by Theorem. and Corollary.7 to prove Theorem. it suffices to show that q / B k for all k 3. But by an application of Lemma.6 to show that q / B k for all k 3 it suffices to show that q / B 3 and q / B 4.. Proof that q f B k for all k To show that q f B k for all k, we construct an x I qf explicitly. satisfying #Σ qf (x) = k Proposition.. For each k the number x k = ((0000) k 0(0) ) qf #Σ qf (x k ) = k. Moreover, x ℵ0 = (0 ) q satisfies card Σ qf (x) = ℵ 0. satisfies Proof. We proceed by induction. For k = we have x = ((0) ) qf, therefore #Σ qf (x ) = by Lemma.5. Let us assume x k = ((0000) k 0(0) ) qf satisfies #Σ qf (x k ) = k, to prove our result it suffices to show that x k+ = ((0000) k 0(0) ) qf satisfies #Σ qf (x k+ ) = k +. We begin by remarking that by Lemma.5 ((0000) k 0(0) )) qf has a unique q f -expansion, therefore there is a unique q f -expansion of x k+ beginning with. Furthermore, it is a simple exercise to show that q f satisfies the equation x 4 = x 3 + x +, which implies that (0(0)(0000) k 0(0) ) is also a q f -expansion for x k+. To prove the claim, we will show that if (ε i ) i= is a q-expansion for x k+ and ε = 0, then ε =, ε 3 = and ε 4 = 0, which combined with our inductive hypothesis implies that the set of q-expansions for x k+ satisfying ε = 0 consists of k distinct elements, combining these q-expansions with the unique q-expansion of x k+ satisfying ε = we may conclude #Σ qf (x k+ ) = k +.

INTEGERS: 4 (04) 5 Let us suppose ε = 0, if ε = 0; then we would require x k+ = ((0000) k 0(0) ) qf (00() ) qf, however x k+ > q f and i=3 q < i q for all q > + 5, therefore ε =. If ε 3 = 0 then we would require which is equivalent to however x k+ = ((0000) k 0(0) ) qf (00() ) qf, (3) x k+ = q f + q 4k+3 f = q f qf i=0 qf i qf + q 4 f, q i i=0 f + q 4 f, q i i=0 f whence (3) cannot occur and ε 3 =. Now let us suppose ε 4 =, then we must have which is equivalent to x k+ = ((0000) k 0(0) ) qf (00 ) qf, (4) x k+ = q f + q 4k+3 f i=0 qf i qf + q 3 f + qf 4. (5) The left hand side of (5) is maximised when k =, therefore to show that ε 4 = 0 it suffices to show that + q f qf 7 qf i qf + qf 3 + qf 4 (6) does not hold. By a simple manipulation (6) is equivalent to i=0 q 6 f q 5 f q 4 f + q f + q f + 0, (7) but by an explicit calculation we can show that the left hand side of (7) is strictly negative, therefore (4) does not hold and ε 4 = 0. Now we consider x ℵ0, replicating our analysis for x k we can show that if (ε i ) i= is a q-expansion for x ℵ0 and ε = 0 then ε =. Unlike our previous case it is possible for ε 3 = 0, however in this case ε i = for all i 4. If ε 3 =, then as in our previous case we must have ε 4 = 0. We observe that x ℵ0 = (0 ) qf = (00() ) qf = (000 ) qf.

INTEGERS: 4 (04) 6 Clearly, there exists a unique q-expansion for x ℵ0 satisfying ε = and a unique q-expansion for x ℵ0 satisfying ε = 0, ε = and ε 3 = 0. Therefore all other q- expansions of x ℵ0 have (00) as a prefix, repeating the above argument arbitrarily many times we can determine that all the q f -expansions of x ℵ0 are of the form: x ℵ0 = (0 ) qf which is clearly infinite countable.. = (00() ) qf = (000 ) qf = (0000() ) qf = (00000 ) qf = (000000() ) qf = (0000000 ) qf, Thus, to prove Theorem., it suffices to show that q / B 3 B 4. This may look like a fairly innocuous exercise, but in reality it requires a substantial effort. 3. Proof that q / B 3 By Lemma.6 to show that q B k for all k 3 it suffices to show q / B 3 and q / B 4. To prove this we begin by characterising those x S q that satisfy #Σ q (x) =. To simplify our notation, we denote for the rest of the paper β := q and T i := T q,i for i = 0,. Proposition 3.. The only x S β that satisfy #Σ β (x) = are x = (0(0) ) β = (0000(0) ) β and x = (0(0) ) β = (00(0) ) β. Proof. It was shown in the proof of [, Proposition.4] that if + 5 < q < q f and y, y + have unique q-expansions, then necessarily q = β and either y = (0000(0) ) β and y + = ((0) ) β or y = (00(0) ) β and y + = ((0) ) β respectively. Since for either case there exists a unique x S β such that βx = y, Lemma.6 yields the claim. In what follows we shall let (ε i ) i= = 0(0), (ε i ) i= = 0000(0), (ε 3 i ) i= = 0(0) and (ε 4 i ) i= = 00(0). Remark 3.. Let ( ε i ) i= = ( ε i) i=, we refer to ( ε i) i= as the reflection of (ε i ) i=. Clearly ( ε i ) i= = (ε4 i ) i= and ( ε i ) i= = (ε3 i ) i=, this is to be expected as

INTEGERS: 4 (04) 7 every x I q satisfies #Σ q (x) = #Σ q ( q x) and mapping (ε i) i= to ( ε i) i= is a bijection between Σ q (x) and Σ q ( q x). If (ε i ) i= and (ε i ) i= were not the reflections of (ε 4 i ) i= and (ε3 i ) i= respectively then there would exist other x S β satisfying #Σ β (x) =, contradicting Proposition 3.. In this section we show that no x I β can satisfy #Σ β (x) = 3. To show that β / B 3 and β / B 4 we will make use of the following proposition. Proposition 3.3. Suppose x I β satisfies #Σ β (x) = or equivalently #Ω β (x) =, then there exists a unique sequence of transformations a such that a(x) S β. Moreover, a(x) = ((ε i ) i= ) β or a(x) = ((ε 3 i ) i= ) β. Proof. As #Ω β (x) = then there must exist a satisfying a(x) S β, otherwise #Ω β (x) =. We begin by showing uniqueness, suppose a satisfies a (x) S β and a a. If a < a then either a is a prefix of a in which case by () and Lemma.6 we have that #Ω β (x) #Ω β (a (x)) = #Ω β (T 0 (a (x))) + #Ω β (T (a (x))) 3, which contradicts #Ω β (x) =. Alternatively if a is not a prefix of a then there exists b n=0 {T 0, T } n such that b(x) S β and either b0 is a prefix for a and b is a prefix for a, or b0 is a prefix for a and b is a prefix for a. In either case it follows from () and Lemma.6 that #Ω β (x) #Ω β (b(x)) = #Ω β (T 0 (b(x))) + #Ω β (T (b(x))) 4, a contradiction. By analogous arguments we can show that if a = a or a > a then this implies #Ω β (x) >, therefore a must be unique. Now let a be the unique sequence of transformations such that a(x) S β. By Lemma.6, #Ω β (T 0 (a(x))) = #Ω β (T (a(x))) =. But it follows from Proposition 3. that this can only happen when a(x) = ((ε i ) i= ) β or a(x) = ((ε 3 i ) i= ) β. Remark 3.4. By Proposition 3.3, to show that x I β satisfies card Σ β (x) > (or equivalently, card Ω β (x) > ), it suffices to construct a sequence of transformations a such that a(x) S β with a(x) ((ε i ) i= ) β and a(x) ((ε 3 i ) i= ) β. We will make regular use of this strategy in our later proofs. Before proving β / B 3 it is appropriate to state numerical estimates for S β, ((ε i ) i= ) β and ((ε 3 i ) i= ) β. Our calculations yield S β = [0.584575..., 0.8599...], The explicit calculations performed in this paper were done using MATLAB. In our calculations we approximated β by.70644095045033, which is correct to the first fifteen decimal places.

INTEGERS: 4 (04) 8 (0 k (0) ) β + Iterates of (0 k (0) ) β + (To 6 decimal places) (0(0) ) β + Unique q-expansion by Proposition 3. (00(0) ) β +.77400,.044, 0.734788 (000(0) ) β + Unique q-expansion by Proposition 3. (0000(0) ) β +.0606, 0.84348 (00000(0) ) β +.035438, 0.7766 (000000(0) ) β +.0076, 0.74608, 0.70644 Table : Successive iterates of (0 k (0) ) β + falling into S β \ {(ε ) β, (ε 3 ) β } ((ε i ) i=) β = 0.64598... and ((ε 3 i ) i=) β = 0.76976.... These estimates will make clear when a(x) S β and whether a(x) = ((ε i ) i= ) β or a(x) = ((ε 3 i ) i= ) β. Theorem 3.5. We have β / B 3. Proof. Suppose x I β satisfies #Σ β (x ) = 3 or equivalently #Ω β (x ) = 3. Let a denote the unique minimal sequence of transformations such that a(x ) S β. By considering reflections we may assume without loss in generality that #Ω β (T (a(x ))) = and #Ω β (T 0 (a(x ))) =. Put x = T (a(x )); by a simple argument it can be shown that x 0, so we may assume that x = (0 k (0) ) β for some k. To show that β / B 3 we consider T 0 (a(x )) = x + = (0 k (0) ) β +, we will show that for each k there exists a finite sequence of transformations a such that a(x + ) S β, a(x + ) ((ε i ) i= ) β and a(x + ) ((ε 3 i ) i= ) β. By Proposition 3.3 and Remark 3.4 this implies #Ω β (x + ) >, which is a contradiction and therefore β / B 3. Table states the orbits of (0 k (0) ) β + under T 0 and T until eventually (0 k (0) ) β + is mapped into S β. Table also includes the orbit of under T 0 and T until is mapped into S β. The reason we have included the orbit of is because (0 k (0) ) β + as k, therefore understanding the orbit of allows us to understand the orbit of (0 k (0) ) β + for large values of k. By inspection of Table, we conclude that for k 6 either (0 k (0) ) β + has a unique q-expansion which contradicts #Ω β (T 0 (a(x ))) =, or there exists a such that a((0 k (0) ) β + ) S β with a((0 k (0) ) β + ) ((ε i ) i= ) β and a((0 k (0) ) β + ) ((ε 3 i ) i= ) β, which contradicts #Ω β (x + ) = by Proposition 3.3. To conclude our proof, it suffices to show that for each k 7 there exists a such that a((0 k (0) ) β + ) S β, a((0 k (0) ) β + ) ((ε i ) i= ) β and a((0 k (0) ) β +) ((ε 3 i ) i= ) β. For all k 7 (0 k (0) ) β + (, (000000(0) ) β + ), but by inspection of Table it is clear that T (x) (0.70644..., 0.74608...)

INTEGERS: 4 (04) 9 for all x (, (000000(0) ) β + ). Therefore we can infer that such an a exists for all k 7, which concludes our proof. 4. Proof that q / B 4 To prove β / B 4 we will use a similar method to that used in the previous section, the primary difference being there are more cases to consider. Before giving our proof we give details of these cases. Suppose x I β satisfies #Σ β (x ) = 4 or equivalently #Ω β (x ) = 4. Let a denote the unique minimal sequence of transformations such that a (x) S β. By Lemma.6, #Ω β (T 0 (a (x ))) + #Ω β (T (a (x ))) = 4. By Theorem 3.5, #Ω β (T 0 (a (x ))) 3 and #Ω β (T (a (x ))) 3, whence #Ω β (T 0 (a (x ))) = #Ω β (T (a (x ))) =. (8) Letting x = T (a (x )), we observe that (8) is equivalent to #Ω β (x) = #Ω β (x + ) =. (9) By Proposition 3.3, there exists a unique sequence of transformations a such that a(x) S β and a(x) = ((ε i ) i= ) β or a(x) = ((ε 3 i ) i= ) β. We now determine the possible unique sequences of transformations a that satisfy a(x) S β. To determine the unique a such that a(x) S β, it is useful to consider the interval [ β, β β ]. The significance of this interval is that T 0( β ) = β β β and T ( β ) = β. The monotonicity of the maps T 0 and T implies that if x (0, β ) and x / [ β, β β ], then there exists i {0, } and a minimal k such that Ti k(x) [ β, β β ]. Furthermore, S β [ β, β β ], in view of β > + 5. In particular, if x (0, β ), then there exists a minimal k such that T0 k (x) ( β, β β ); T 0 k (x) cannot equal β or β β as this would imply #Ω β (x) =. There are three cases to consider: either T0 k (x) S β, in which case T0 k (x) = ((ε i ) i= ) β or T0 k (x) = ((ε 3 i ) i= ) β by Proposition 3.3, or alternatively T0 k (x) ( β, β ) or T 0 k (x) ( β(β ), β β ). It is a simple exercise to show that if T0 k (x) = ((ε 3 i ) i= ) β or T0 k (x) ( ) then k. By Lemma.4 and Proposition 3.3, if T k 0 (x) S β, then β(β ), β β x = (0 k (ε i ) i=) β for some k or x = (0 k (ε 3 i ) i=) β for some k. For any q ( + 5, q f ) and y ( q, q ) there exists a unique minimal sequence a such that a (y) S q, moreover a (y) = (T q, T q,0 ) j (y) for some j and

INTEGERS: 4 (04) 0 (T q, T q,0 ) i (y) ( q, q ) for all i < k. For all y ( q, q ) we have that (T q, T q,0 )(y) = q y < q ; furthermore, it can be checked directly that β < ((ε 3 i ) i= ) β, whence if T0 k (x) ( β, β ), then x = (0 k (0) j (ε i ) i=) β, for some k and j. By a similar argument it can be shown that if T0 k (x) ( β(β ), β β ), then x = (0 k (0) j (ε 3 i ) i=) β, for some k and j. The above arguments are summarised in the following proposition. Proposition 4.. Let x be as in (9); then one of the following four cases holds: x = (0 k (ε i ) i=) β for some k, (0) x = (0 k (ε 3 i ) i=) β for some k, () x = (0 k (0) j (ε i ) i=) β for some k and j () or x = (0 k (0) j (ε 3 i ) i=) β for some k and j. (3) To prove that β / B 4 we will show that for each of the four cases described in Proposition 4. there exists a such that a(x + ) S β \ {((ε i ) i=) β, ((ε 3 i ) i=) β }, (4) which contradicts #Ω β (x + ) = by Proposition 3.3 and Remark 3.4. For the majority of our cases an argument analogous to that used in Section 3 will suffice, however in the case where k =, 3 in () and k =, 4 in (3) a different argument is required. We refer to these cases as the exceptional cases. For the exceptional cases we will also show (4), however the approach used in slightly more technical and as such we will treat these cases separately. Proposition 4.. For each of the cases described by Proposition 4. there exists a such that (4) holds. Proof of Proposition 4. for the non-exceptional cases. In the cases where x = (0 k (ε i ) i= ) β for some k or x = (0 k (ε 3 i ) i= ) β for some k it is clear that x 0 as k, therefore to understand the orbit of (0 k (ε i ) i= ) β + or (0 k (ε 3 i ) i= ) β + for large values of k it suffices to consider the orbit of. Similarly, in the cases described by () and (3) if we fix k then as j both (0 k (0) j (ε i ) i= ) β and (0 k (0) j (ε 3 i ) i= ) β converge to (0 l (0) ) for some l, therefore to understand the orbits of (0 k (0) j (ε i ) i= ) β + and (0 k (0) j (ε 3 i ) i= ) β + for large values of j it

INTEGERS: 4 (04) (0 k (ε i ) i= ) β + Iterates of (0 k (ε i ) i= ) β + (to 6 decimal places) (0(ε i ) i= ) β +.37766,.35584,.39363,.5696,.503, 0.967605, 0.6558 (00(ε i ) i= ) β +.048,.08780, 0.860857, 0.4760, 0.808484 (000(ε i ) i= ) β +.8888, 0.936, 0.5985 (0000(ε i ) i= ) β +.075344, 0.83953, 0.4364, 0.74608 (00000(ε i ) i= ) β +.044044, 0.785989 (000000(ε i ) i= ) β +.05747, 0.754688, 0.70644 Table : Successive iterates of (0 k (ε i ) i= ) β + suffices to consider the orbit of (0 l (0) ) β +, for some l. By considering these limits it will be clear when a sequence of transformations a exists that satisfies (4) for large values of k and j. We begin by considering the case x = (0 k (ε i ) i= ) β. Table plots successive (unique) iterates of (0 k (ε i ) i= ) β + until (0 k (ε i ) i= ) β + is mapped into S β for k 6. It is clear from inspection of Table that for k 6 there exists a such that a(x+) S β, a(x+) ((ε i ) i= ) β and a(x+) ((ε 3 i ) i= ) β. The case where k 7 follows from the fact that (0 k (ε i ) i= ) β + (, (000000(ε i ) i= ) β + ) for all k 7 and T (y) (0.70644..., 0.754688...) for all y (, (000000(ε i ) i= ) β + ). The case described by () follows by an analogous argument therefore the details are omitted, we just include the relevant orbits in Table 3. For the non-exceptional cases described by () and (3) an analogous argument works for the first few values of k by considering the limit of x+ as j, therefore we just include the relevant orbits in Table 3. It is clear by inspection of Table 3 that (0 k (0) j (ε i ) i= ) β + (, (0000000(ε i ) i= ) β + ) for all k 7 and j, however T (y) (0.70644..., 0.74903...) for all y (, (0000000(ε i ) i= ) β + ), therefore by inspection of Table 3 we can conclude the case described by () in the non-exceptional cases. Similarly, it is clear from inspection of Table 3 that (0 k (0) j (ε 3 i ) i= ) β + (, (0000000(0) ) β + ) for all k 8 and j, however T (y) (0.70644..., 0.746086...) for all y (, (0000000(0) ) β + ), therefore by inspection of Table 3 we can conclude the case described by (3) in the nonexceptional cases. Proof of Proposition 4. for the exceptional cases. The reason we cannot use the same method as used for the non-exceptional cases is because as j the limits of (0(0) j (ε i ) i= ) β+, (000(0) j (ε i ) i= ) β+, (00(0) j (ε 3 i ) i= ) β+ and (0000(0) j (ε 3 i ) i= ) β+ all have unique β-expansions, which follows from Proposition 3.. As a consequence of the uniqueness of the β-expansion of the relevant limit, the number of transformations required to map x + into S β becomes arbitrarily large as j. However, the following proposition shows that we can still construct an a satisfying

INTEGERS: 4 (04) (0 k (ε 3 i ) i= ) β + Iterates of (0 k (ε 3 i ) i= ) β + (to 6 decimal places) (00(ε 3 i ) i= ) β +.60388,.56076, 0.977635, 0.67385 (000(ε 3 i ) i= ) β +.56, 0.9703, 0.6609 (0000(ε 3 i ) i= ) β +.08898, 0.86860, 0.476047, 0.84348 (00000(ε 3 i ) i= ) β +.0506, 0.79966 (000000(ε 3 i ) i= ) β +.030407, 0.76660 (0000000(ε 3 i ) i= ) β +.07775, 0.7405, 0.70644 (00(0) j (ε i ) i= ) β + Iterates of (00(0) j (ε i ) i= ) β + (000(ε i ) i= ) β +.93,.03998, 0.777869 (0000(ε i ) i= ) β +.843,.070, 0.74950 (00(0) ) β +.77400,.044, 0.734788 (0000(0) j (ε i ) i= ) β + Iterates of (0000(0) j (ε i ) i= ) β + (00000(ε i ) i= ) β +.065653, 0.8954, 0.40778, 0.697570 (000000(ε i ) i= ) β +.0634, 0.8789 (0000(0) ) β +.0606, 0.84348 (00000(0) j (ε i ) i= ) β + Iterates of (00000(0) j (ε i ) i= ) β + (000000(ε i ) i= ) β +.038379, 0.77697 (00000(0) ) β +.035438, 0.7766 (000000(0) j (ε i ) i= ) β + Iterates of (000000(0) j (ε i ) i= ) β + (0000000(ε i ) i= ) β +.0435, 0.74903 (000000(0) ) β +.0076, 0.74608 (000(0) j (ε 3 i ) i= ) β + Iterates of (000(0) j (ε 3 i ) i= ) β + (0000(ε 3 i ) i= ) β +.68794, 0.99939, 0.709603 (000(0) ) β +.77400,.044, 0.734788 (00000(0) j (ε 3 i ) i= ) β + Iterates of (00000(0) j (ε 3 i ) i= ) β + (000000(ε 3 i ) i= ) β +.05768, 0.80937 (00000(0) ) β +.0606, 0.84348 (000000(0) j (ε 3 i ) i= ) β + Iterates of (000000(0) j (ε 3 i ) i= ) β + (0000000(ε 3 i ) i= ) β +.03379, 0.76836 (000000(0) ) β +.035438, 0.7766 (0000000(0) j (ε 3 i ) i= ) β + Iterates of (0000000(0) j (ε 3 i ) i= ) β + (00000000(ε 3 i ) i= ) β +.097, 0.744363 (0000000(0) ) β +.0076, 0.74608 Table 3: Successive iterates of (0 k (ε 3 i ) i= ) β +

INTEGERS: 4 (04) 3 (4) for all but three of the exceptional cases. Proposition 4.3. The following identities hold: ((T T 0 ) j (T ) 4 )((0(0) j (ε i ) i=) β + ) = β β 3 (β ) + β 0.598 for j 3, ((T T 0 ) j (T ) )((000(0) j (ε i ) i=) β + ) = β β 3 (β ) + β 0.598 for j, (5) (6) and ((T 0 T ) j (T ) 3 )((00(0) j (ε 3 i ) i=) β + ) = ((T 0 T ) j+ (T ))((0000(0) j (ε 3 i ) i=) β + ) = β β + β β 3 (β ) 0.8434 for j β β + β β 3 (β ) 0.8434 for j. (7) (8) Proof. Proving that each of the identities (5), (6), (7) and (8) hold follow by similar arguments, we will therefore just show that (5) holds. Note that (0(0) j (ε i ) i=) β + = βj+ + β β j+3 (β ) +, for all j. We observe the following: ( β ((T T 0 ) j (T ) 4 j+ + β ) ) β j+3 (β ) + =(T T 0 ) j ( β j+ + β β j (β ) + β4 β 3 β β =β j 4( β j+ + β ) β j (β ) + β4 β 3 β β = βj+ + β β 3 (β ) ) j 3 i=0 β i + β j β j β j β j 3 β j 4 βj 4 β = βj+ β 3 (β ) + βj β j β j β j 3 β j 4 βj 4 β + β β 3 (β ) + β. Therefore, to conclude our proof, it suffices to show that β j+ β 3 (β ) + βj β j β j β j 3 β j 4 βj 4 β = 0. (9)

INTEGERS: 4 (04) 4 Exceptional cases Iterates (to 6 decimal places) (00(ε i ) i= ) β +.38654,.7854,.77400,.044, 0.734788 (000(ε i ) i= ) β +.3076,.44495,.8888, 0.936, 0.5985 (000(ε 3 i ) i= ) β +.88747,.04588,.0606, 0.84348 Table 4: Remaining exceptional cases: k =, j {, } in () and k =, j = in (3) Manipulating the left hand side of (9) it is clear that satisfying (9) is equivalent to β j β j 4 + (β j β j β j β j 3 β j 4 )(β ) β = 0 or β j 3 (β )(β 4 β β ) β This is true in view of β 4 β β = 0. = 0. By Proposition 4.3 and Table 4 which displays the orbits of the exceptional cases that are not covered by Proposition 4.3 we can conclude Proposition 4. for all the exceptional cases, therefore β / B 4, and Theorem. holds. 5. Open questions To conclude the paper, we pose a few open questions: What is the topology of B k for k? In particular, what is the smallest limit point of B k? Is it below or above the Komornik-Loreti constant introduced in [8]? What is the smallest q such that x = has k q-expansions? (For k = this is precisely the Komornik-Loreti constant.) What is the structure of B ℵ0 ( + 5, q f )? In view of the results of the present paper, knowing this would lead to a complete understanding of card Σ q (x) for all q q f and all x I q. Let, as above, B = B k B ℵ0 B ℵ 0. k= By Theorem.3, q f is the smallest element of B. What is the second smallest element of B? What is the topology of B?

INTEGERS: 4 (04) 5 In [] the authors study the order in which periodic orbits appear in the set of points with unique q-expansion, they show that as q, the order in which periodic orbits appear in the set of uniqueness is intimately related to the classical Sharkovskiĭ ordering. Does a similar result hold in our case? That is, if k > k with respect to the usual Sharkovskiĭ ordering, does this imply B k B k? Acknowledgment. The authors are grateful to Vilmos Komornik for useful suggestions. References [] J,-P. Allouche, M. Clarke and N. Sidorov, Periodic unique beta-expansions: the Sharkovskiĭ ordering, Ergod. Th. Dynam. Systems 9 (009), 055 074. [] S. Baker, Generalised golden ratios over integer alphabets, arxiv:0.8397 [math.ds]. [3] Z. Daróczy and I. Katai, Univoque sequences, Publ. Math. Debrecen 4 (993), 397 407. [4] P. Erdős, I. Joó, On the number of expansions = q n i, Ann. Univ. Sci. Budapest 35 (99), 9 3. [5] P. Erdős, I. Joó and V. Komornik, Characterization of the unique expansions = i= q n i and related problems, Bull. Soc. Math. Fr. 8 (990), 377 390. [6] P. Glendinning and N. Sidorov, Unique representations of real numbers in non-integer bases, Math. Res. Letters 8 (00), 535 543. [7] V. Komornik, Open problems session of Dynamical Aspects of Numeration workshop, Paris, 006. [8] V. Komornik and P. Loreti, Unique developments in non-integer bases, Amer. Math. Monthly 05 (998), no. 7, 636 639. [9] W. Parry, On the β-expansions of real numbers, Acta Math. Acad. Sci. Hung. (960) 40 46. [0] A. Rényi, Representations for real numbers and their ergodic properties, Acta Math. Acad. Sci. Hung. 8 (957) 477 493. [] N. Sidorov, Almost every number has a continuum of β-expansions, Amer. Math. Monthly 0 (003), 838-84.

INTEGERS: 4 (04) 6 [] N. Sidorov, Expansions in non-integer bases: lower, middle and top orders, J. Number Th. 9 (009), 74 754. [3] N. Sidorov, Universal β-expansions, Period. Math. Hungar. 47 (003), 3. [4] N. Sidorov and A. Vershik, Ergodic properties of the Erdős measure, the entropy of the goldenshift, and related problems Monatsh. Math. 6 (998), 5 6.