On Curling Numbers of Integer Sequences

Given a finite nonempty sequence S of integers, write it as XY k, where Y k is a power of greatest exponent that is a suffix of S: this k is the curling number of S. The curling number conjecture is that if one starts with any initial sequence S, and extends it by repeatedly appending the curling number of the current sequence, the sequence will eventually reach 1. The conjecture remains open. In this paper we discuss the special case when S consists just of 2’s and 3’s. Even this case remains open, but we determine how far a sequence consisting of n 2’s and 3’s can extend before reaching a 1, conjecturally for n ≤ 80. We investigate several related combinatorial problems, such as finding c(n, k), the number of binary sequences of length n and curling number k, and t(n, i), the number of sequences of length n which extend for i steps before reaching a 1. A number of interesting combinatorial problems remain unsolved. 1 The curling number conjecture Given a finite nonempty sequence S of integers, write it as S = XY , where X and Y are sequences of integers and Y k is a power of greatest exponent that is a suffix of S: this k is the curling number of S, denoted by cn(S). X may be the empty sequence ǫ ; there may be several choices for Y , although the shortest such Y which achieves k (which as we shall see in §3.1 is primitive) is unique. For example, if S = 01 2 2 1 2 2 1 2 2, we could write it as XY , where X = 01 2 2 1 2 2 1 and Y = 2, or as XY , where X = 0 and Y = 12 2. The latter representation is to be preferred, since it has k = 3, and as k = 4 is impossible, the curling number of this S is 3. The following conjecture was stated by van de Bult et al. [2]: Conjecture 1. The curling number conjecture. If one starts with any initial sequence of integers S, and extends it by repeatedly appending the curling number of the current sequence, the sequence will eventually reach 1. In other words, if S0 = S is any finite nonempty sequence of integers, and we define Sm+1 to be the concatenation Sm+1 := Sm cn(Sm) for m ≥ 0 , (1) then the conjecture is that for some t ≥ 0 we will have cn(St) = 1. The smallest such t is the tail length of S0, denoted by τ(S0) (and we set τ(S0) = ∞ if the conjecture is false). For example, suppose we start with S0 = 23 2 3. By taking X = ǫ, Y = 23, we have S0 = Y , so cn(S0) = 2, and we get S1 = 23 2 3 2. By taking X = 2, Y = 32 we get cn(S1) = 2, S2 = 23 2 3 2 2. By takingX = 23 2 3, Y = 2 we get cn(S2) = 2, S3 = 23 2 3 2 2 2. Again taking X = 23 2 3, Y = 2 we get cn(S3) = 3, S4 = 23 2 3 2 2 2 3. Now, unfortunately, it is impossible to write S4 = XY k with k > 1, so cn(S4) = 1, S5 = 23 2 3 2 2 2 3 1, and we have reached a 1, as predicted by the conjecture. For this example, τ(S0) = 4. (If we continue the sequence from this point, it joins Gijswijt’s sequence, discussed in §5.)