Collatz cycles with few descents

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1 ACTA ARITHMETICA XCII.2 (2000) Collatz cycles with few descents by T. Brox (Stuttgart) 1. Introduction. Let T : Z Z be the function defined by T (x) = x/2 if x is even, T (x) = (3x + 1)/2 if x is odd. T is known as the 3x + 1 function (see [4], [7]). We are interested in the cycles of T, referred to as 3x + 1 cycles (see [3] [5]). The well known Finite Cycle Conjecture asserts that T has only finitely many cycles. In this note we generalize a theorem due to R. Steiner (see [4], [6]). In order to state the result in a concise form, we consider instead of T a slightly different function T 1 defined by the unique decomposition (1.1) 2 k(x) T 1 (x) = 3x + 1 (x odd), where k(x) 1 is the multiplicity of the prime factor 2 in the number 3x+1. Note that T 1 is only defined for odd integers. For each odd argument the iteration T k(x) (x) is equal to T 1 (x). An odd integer x is called descending if k(x) 2. In what follows, a cycle of T 1 is called a Collatz cycle. By definition, we can represent a Collatz cycle by a tuple Γ = (x 1,..., x n ) consisting of distinct odd integers (n 1). By Γ we denote the number of elements in Γ, i.e. the period of the Collatz cycle. Each Collatz cycle consists of the odd elements in a 3x + 1 cycle, and conversely. If k denotes the sum of k(x) over all elements of a Collatz cycle, then k is the period of the corresponding 3x + 1 cycle. Let δ(γ ) be the number of descending elements in a Collatz cycle Γ. Theorem 1.1. The number of Collatz cycles satisfying δ(γ ) < 2 log Γ is finite. We will prove Theorem 1.1 in the following sections. In Section 4 we briefly discuss the extension of Theorem 1.1 to 3x + d mapping (see [2]). The theorem of R. Steiner states that the fixed point 1 is the only positive 2000 Mathematics Subject Classification: 11D61, 11Z05. [181]

2 182 T. Brox Collatz cycle with δ(γ ) = 1. It is easy to deduce the following corollary of Theorem 1.1, using Lemma 2.2 in Section 2. Corollary 1.2. For each fixed ν 1 the number of Collatz cycles with δ(γ ) ν is finite. 2. A divisibility problem. In this section we express the existence problem of a Collatz cycle by an appropriate periodic sequence (see Sections 2, 5 of [3]). Given an odd integer x 0, consider the iteration x m = T 1 (x m 1 ). Let k m = k(x m 1 ). Because of (1.1), we get (2.1) 2 k m x m = 3x m (m 1), where it is required that x 1, x 2,... are odd. Proceeding iteratively, we immediately obtain (2.2) 2 k k m x m = 3 m x 0 + ϕ m 1 (k 1,..., k m 1 ). Here ϕ 0 = 1 and (2.3) ϕ m (u 1,..., u m ) = m 3 m l 2 u u l ((u 1,..., u m ) R m ), l=0 where u u l is zero if l = 0. For further references see Section 2 of [3]. Assume now (x 0,..., x n 1 ) is a Collatz cycle. By construction, k i 1 for 1 i n. Since x n = x 0, (2.2) implies where Mx 0 = ϕ n 1 (k 1,..., k n 1 ), (2.4) M = 2 k 3 n, k = k k n. Consider (x 0,..., x n 1 ) and (k 1,..., k n ) as n-periodic sequences {x i } i Z, {k i } i Z. Since we can take each x i as start value, we obtain (2.5) Mx i = ϕ n 1 (k i+1,..., k i+n 1 ) (i Z). Hence M divides the right hand side of (2.5) for each i. Since the Collatz cycle consists of distinct elements, n is the smallest period of {k i }, by (2.5). Conversely, let {k i } i Z be any periodic sequence of positive integers. Let n be the smallest period of {k i }. Define M and k by (2.4). Let F i = ϕ n 1 (k i+1,..., k i+n 1 ) (i Z). Note that F i does not depend on k i. Plainly {F i } is n-periodic. Because of (2.3) and periodicity, we get by a simple computation (2.6) 2 k i F i = 3F i 1 + M (i Z).

3 Collatz cycles with few descents 183 Since each k i is positive, we conclude from (2.3) and (2.4) that (2.7) (2.8) gcd(2 3, F i ) = 1 gcd(2 3, M) = 1. (i Z), Because of (2.6) and (2.7), k i and F i are uniquely determined by F i 1. If F l = F m, then m l is a period of {k i }. Hence F 1,..., F n are distinct. We are now ready to prove the following lemmas (see also Section 5 of [3]). Lemma 2.1. Let M = 2 k 3 n. Then either M F i (1 i n) or M F i (1 i n). The latter case corresponds to a Collatz cycle (x 1,..., x n ), where x i = F i /M. P r o o f. Assume that M F i for one i. Then (2.6) (2.8) imply that for all i, M F i and x i = F i /M is odd. Dividing (2.6) by M, we arrive at x i = T 1 (x i 1 ), since each x i is odd. Plainly x 1,..., x n are distinct. Lemma 2.2. If {k i } generates a Collatz cycle, then k = k k n 2n. In particular, for each n 1, the number of Collatz cycles with Γ = n is finite. P r o o f. If {k i } generates a Collatz cycle, then x i = F i /M, by Lemma 2.1. If we multiply (2.6) from i = 1 to i = n, a simple reduction yields n 2 k = (3 + 1/x i 1 ) 4 n, i=1 since 1/x i 1. Hence k 2n. Remark 2.3. By Lemma 2.1, since k i = 1 (i Z) has period 1, there exists at most one Collatz cycle with δ(γ ) = 0, and this Collatz cycle has to be a fixed point. In fact, 1 = 1/(2 3) is a non-descending fixed point. 3. Algebraic reformulation. In order to prove Theorem 1.1, we reformulate the divisibility problem in a more convenient form. As before let {k i } i Z be a periodic sequence of positive integers. Let n be the smallest period of {k i }. Define ϕ n (u 1,..., u n ) = 2 u 1 ϕ n 1 (u 2,..., u n ) 2ϕ n 1 (u 1,..., u n 1 ) Because of (2.3), we obtain ((u 1,..., u n ) R n ). n 1 (3.1) ϕ n (u 1,..., u n ) = 2 3 n (l+1) 2 u u l {2 ul+1 1 1}. For each i Z define l=0 F i = ϕ n (k i+1,..., k i+n ).

4 184 T. Brox Then (2.6) implies (3.2) Fi = 2 k i+1 F i+1 2F i = F i + M (i Z). Let ν be the number of indices i such that 1 i n and k i 2. We assume ν 1. Define strictly ascending numbers τ(1),..., τ(ν) by 1 τ(j) n and k τ(j) 2. Put τ(0) = τ(ν) n. For 1 j ν let h j = k τ(j) 1, n j = τ(j) τ(j 1). Consider (h 1,..., h ν ), (n 1,..., n ν ) as ν-periodic sequences {h i } i Z, {n i } i Z. By construction, the two sequences consist of positive integers, and ν is the smallest common period. Also by construction, (3.3) n = n n ν, k = h + n, where (3.4) h = h h ν. Given (t 0,..., t ν 1 ) R ν and (u 1,..., u ν 1 ) R ν 1, put (3.5) ψ 2ν 1 (t 0, u 1, t 1, u 2,..., t ν 1 ) ν 1 = 2 t 0+u t l 1 +u l {2 t l 1}3 u l u ν 1, l=0 where the sum in the 2-exponent (3-exponent) is zero if l = 0 (l = ν 1). For each j Z define (3.6) H j = ψ 2ν 1 (h j, n j+1, h j+1, n j+2,..., h j+ν 1 ). Thus H j does not depend on n j. Note that H j > 0 for each j Z. Note also that {H j } is ν-periodic. We assert (3.7) Fτ(j) = 2 n j+1 H j+1 (j Z). By shift, it is enough to verify (3.7) for j = 0. Also by shift, we can assume that τ(0) = 0. Then τ(j) equals n n j (1 j ν). In (3.1) only those terms can survive which are placed at l + 1 = τ(j). Hence ν F τ(0) = 2 3 n τ(j) 2 h h j 1 +τ(j) 1 {2 h j 1}. j=1 In the last formula we take the 2-exponent n 1 1 outside the sum. After some rearrangement, the remaining sum is easily identified as H 1. Conversely, let {h j } j Z, {n j } j Z be any periodic sequences consisting of positive integers. Let ν be the smallest common period of both sequences. Define h, k, n by (3.3) and (3.4). Let H j be given by (3.6). Up to shift, {k i } is uniquely determined by both sequences, i.e. by {h 0, n 1, h 1, n 2,...}. Also n is the smallest period of {k i }. Because of (2.8), (3.2) and (3.7), we can reformulate Lemma 2.1.

5 Collatz cycles with few descents 185 Lemma 3.1. Let M = 2 k 3 n. Then either M H j (1 j ν) or M H j (1 j ν). The latter case corresponds to a Collatz cycle with Γ = n and δ(γ ) = ν. 4. Minimum of the H-sequence. In this section we prove Theorem 1.1 assuming the truth of Lemma 5.1, which will be stated and proved in Section 5. Consider a pair {h j } j Z, {n j } j Z of periodic sequences consisting of positive integers. Define ν, h, k, n as described in the preceding section. By Lemma 3.1, since each H j > 0, the two sequences cannot generate a Collatz cycle if (4.1) min 0 j ν 1 H j < M = 2 k 3 n. To prove Theorem 1.1, it is enough, by Lemma 2.2 and Remark 2.3, to show (4.1) holds for all sufficiently large n, assuming that ν < 2 log n. First we consider the left hand side of (4.1). Let θ = log 2 3 and θ 1 = θ 1. Note that θ 1 > 0. In order to estimate H j, we replace the bracket inside (3.5) by 2 h j+l. Because of periodicity, an easy rearrangement yields ν 1 3 n H j < 2 n j+s j,l (j Z), where S j,l = l=0 l (h j+i θ 1 n j+i ) (l 0). As a functional of both sequences define By an elementary estimation, We claim that E = min { n j + 0 j ν 1 max min j 3 n H j < ν2 E. 0 l ν 1 S j,l}. (4.2) E max(0, ) A ν n, where A ν = θ 1 /(θ ν 1), = h θ 1 n = k θn. But (4.2) is a consequence of Lemma 5.1 below, applied with x j = θ 1 n j, y j = h j, α = 1/θ 1. Hence we obtain min j 3 n H j < ν2 max(0, ) A νn. Let γ = 1 2 log θ = If we assume that ν < 2 log n, then (θ ν 1) 1 n > θ ν n = n 1 ν log θ/ log n > n γ,

6 186 T. Brox and it follows that (4.3) min j 3 n H j < log n 2 1+max(0, ) θ 1n γ. We now consider the right hand side of (4.1). By elementary analysis we get 3 n M = 2 1 min(, 1)2 max(0, ) 1. Note that we have n 1 and k = h + n 2. By definition, < 1 if k > θn 1 and k < θn + 1, otherwise 1. A result of A. Baker and N. Feldman (see Theorem 3.1 in [1]) implies the existence of an effectively computable constant C 0 > 0 such that > k log 2 n log 3 > {max(k, n)} C 0 (k 2, n 1). Since θn + 1 > n, and therefore > {θn + 1} C 0, (4.4) 3 n M > {θn + 1} C 0 2 max(0, ) 1. If n is large enough, the right side of (4.3) is smaller than the right side of (4.4), which gives (4.1). Remark 4.1. We briefly sketch the extension of Theorem 1.1 to the 3x+d mapping, where d is a positive integer prime to 2 and 3. Define T 1 according to (1.1), where the right hand side is replaced by 3x + d. Now k(x) is the multiplicity of 2 in the number 3x+d. Define a Collatz cycle, descending and δ(γ ) as in Section 1. After an obvious modification of (2.1), in (2.2) replace ϕ m 1 by dϕ m 1. Note that the definition of ϕ m is not affected (m 0). Hence multiply the right hand side of (2.5) with d. If (2.6) is multiplied with d, it follows that F i can be replaced by df i in Lemma 2.1. Similarly to the proof of Lemma 2.2, if {k i } generates a Collatz cycle, then n 2 k = (3 + d/x i 1 ) (3 + d) n. i=1 Hence put k C 1 n in Lemma 2.2, where C 1 = log 2 (3+d). In Lemma 3.1 put dh j instead of H j. Finally multiply the left hand side of (4.1) with d. The additional factor d causes no harm for the conclusion that the inequality is true if n is large enough and ν < 2 log n. 5. Upper value of E. Let there be given an integer ν 1 and real numbers r > 0, s > 0. Let {x j } j Z, {y j } j Z be ν-periodic sequences of non-negative real numbers such that x x ν = r, y y ν = s.

7 For each j Z and l 0 define Collatz cycles with few descents 187 S j,l = l (y j+i x j+i ). Lemma 5.1. For any real α > 0, min { αx j + max S j,l} max(0, s r) r/(β ν 1), 0 j ν 1 0 l ν 1 where β = 1 + 1/α. P r o o f. (a) Reduction to the case r = s. If r s, put ỹ j = ry j /s. Then s = r and l l S j,l = (y j+i ỹ j+i ) + (ỹ j+i x j+i ) = s r l y j+i + s S j,l. Hence max S j,l max(0, s r) + max S j,l. 0 l ν 1 0 l ν 1 Thus we have to prove the assertion for the case r = s. (b) Elimination of {y j }. We suppose r = s. Then {S 0,l } l 0 is ν-periodic. By a simple shift, we can assume If 1 j ν 1, then S 0,ν 1 = max l 0 S 0,l = 0. ν 1 max S j,l = (y i x i ) = 0 l ν 1 i=j j 1 j 1 (x i y i ) x i. Note that the last inequality is sharp if y j = 0 (0 j ν 2) and y ν 1 = r. Now we have to estimate the minimum of (5.1) αx 0, αx 1 + x 0, αx 2 + x 0 + x 1,..., αx ν 1 + x x ν 2. (c) Linear optimization. Consider x = (x 0,..., x ν 1 ) as a vector in R ν. Define linear functionals L 0,..., L ν 1 according to (5.1). Let f(x) be the minimum of L 0 x,..., L ν 1 x. Let B be the affine subspace x x ν 1 = r. We assert (5.2) sup f(x) = r/(β ν 1). x B In order to prove (5.2), consider the vector b = r(β 1)(β ν 1) 1 (β 0, β 1,..., β ν 1 ). Plainly b B. An easy computation shows (5.3) L 0 b = L 1 b =... = L ν 1 b.

8 188 T. Brox Next, f(b) equals the right hand side of (5.2). Let U be the linear subspace where the sum of coordinates equals zero. If x B, then x = b + u (u U). By (5.3), f(x) = f(b) + min j L j u = f(b) + f(u) (x B). We have to show f(u) 0. Assume the contrary, that is, each number L 0 u,..., L ν 1 u is positive. Since L 0 u > 0, we get u 0 < 0. Since u 0 < 0 and L 1 u > 0, we get u 1 < 0. Repeating the argument, each u j is negative, which contradicts u U. Acknowledgements. We thank the referee for useful hints and improvements. References [1] A. B a k e r, Transcendental Number Theory, Cambridge Univ. Press, [2] E. G. Belaga and M. Mignotte, Embedding the 3x + 1 conjecture in a 3x + d context, Experiment. Math. 7 (1998), [3] L. Halbeisen and N. Hungerbühler, Optimal bounds for the length of rational Collatz cycles, Acta Arith. 78 (1997), [4] J. C. Lagarias, The 3x + 1 problem and its generalizations, Amer. Math. Monthly 92 (1985), [5], The set of rational cycles for the 3x + 1 problem, Acta Arith. 56 (1990), [6] R. P. Steiner, A theorem on the Syracuse Problem, in: Proc. 7th Manitoba Conf. on Numerical Mathematics, 1977, Winnipeg, 1978, [7] G. J. Wirsching, The Dynamical System Generated by the 3n+1 Function, Lecture Notes in Math. 1681, Springer, Berlin, Helblingstr. 21 D Stuttgart, Germany thomas.brox@daimlerchrysler.com Received on and in revised form on (3554)