Laver Tables A Direct Approach

Transcription

1 Laver Tables A Direct Approach Aurel Tell Adler June 6, 016

2 Contents 1 Introduction 3 Introduction to Laver Tables 4.1 Basic Definitions Simple Facts Not Primitive Recursive More on Laver Tables Reconstruction of A n Mean Period Steps

3 1 Introduction In this thesis we have a closer look at Laver tables. On the front page the heat map of the 8 8 Laver table is shown. The colors of the bottom row represent in ascending order the numbers 1 to 8, purple being 1 and red being 8. This table is the composition table of the unique system satisfying some algebraic property, namely the left self-distributivity, under the following condition: the first column starts with ascends to 8 and ends with 1. One might already notice that each row is strictly ascending until it arrives at red and then repeats. This general periodicity of the rows can be proven. Furthermore if we consider larger tables one can show that the period of a fixed row is non-decreasing. For all tables that have been computed so far the period of the first row does not exceed 16. Now a natural question comes to mind: Does the period of the first row have an upper bound or is it unbounded? Assuming some large cardinal hypothesis Laver showed that the period of each row tends to infinity. This can be looked up in [Deh00][Chapter XIII, Proposition 1.15 on page 576]. But whether this is provable in the setting of set theory remains an open question. The thesis is split up into two parts: In Chapter we give a formal introduction to Laver tables and state some common facts, which we shall use in the course of this thesis. In the last section we shall restate a proof of the fact that the period of the first row grows extremely slow compared to primitive recursive functions assuming that the period is unbounded. In Chapter 3 we prove some theorems that help us construct a Laver table from preceding ones. But it is still unclear whether some part has to actually be calculated or whether the whole table can be reconstructed by preceding ones. However, with this new knowledge one can show that the mean period grows linearly. In the last section we prove an interesting fact about steps that enables us to give equivalent characterizations about period jumps up to 8. 3

4 Introduction to Laver Tables In this chapter we give the relevant definitions and notions needed for the thesis. We recap some basic facts which can be found in [Drá94] and [Deh00]. Particularly interesting are Lemma..5, Lemma..6 and Lemma..7. In Section.3 we introduce the jump-function that indicates when the period of a row jumps. We shall translate the proof of the fact, that this function grows faster than any primitive recursive function assuming that the period is unbounded, from the assumption that some large cardinal exists to our finite case. That this translation is possible has been indicated in [Deh00][Chapter XIII, Proposition 4.13, page 596]..1 Basic Definitions Definition.1.1. Let S be a set and : S S S be a binary function on S. The pair (S, ) is called left self-distributive system (LSD-system for short) iff for all a, b, c S the following holds: a (b c) (a b) (a c) (.1.1) Definition.1.. For n, N N and N 0, we define the following: (n) N { n mod N if N n, N otherwise. (.1.) We shall denote the set {1,..., N} by S N. Let φ be a formula we define the following: { 1 if φ is true, δ φ 0 else. For an LSD-system (S, ) with x S, we define inductively: { x if N 1, x [N] x [N 1] x otherwise. { x [N] x if N 1, x x [N 1] otherwise. (.1.3) (.1.4) (.1.5) 4

5 Remark.1.1. Note that the function ( ) N is the standard modulo operation with the exception that instead of the value 0 it takes the value N. If from the context it is not clear what operation we are talking about in the case x [N] or x [N], we shall specify that by an index: x [N] or x [N].. Simple Facts We now state some facts, which can be found in [Deh00, Chapter X.1]. Lemma..1 (Lemma 1., p. 446). operation N on S N satisfying For every N N >0, there exists a unique a N 1 (a + 1) N (..1) and, for all a, b S N, a N (b N 1) (a N b) N (a N 1). (..) For a, b S N, the following relations hold in the resulting system: b if a N, a N b a + 1 else if b 1 or a N (b 1) N, > a N (b 1) otherwise. (..3) For a < N, there exists p N a and c 1 a + 1 < c <... < c p N such that, for every b S N, we have a N b c i with i (b) p, hence, in particular the following holds: a N b > a (..4) By Lemma..1 for a S N the notion of period of a is well-defined. This leads us to the following definition: Definition..1. For a S N, we shall denote the period of a in S N by p N (a) min{p S N : a N p N}. The following lemma tells us that in the study of finite monogenic LSD-systems we may concentrate on (S N, N ): Lemma.. (Lemma 1.5, p. 448). Every LSD-system admitting a single generator x satisfying x [N+1] x and x [a] x for 1 < a N is isomorphic to (S N, N ). Now we naturally ask ourselves the question whether (S N, N ) is an LSD-system. This is answered by: Lemma..3 (Proposition 1.6, p. 448). exists no LSD-system satisfying..1. If N N is not a power of, there 5

6 For each n N, there exists a unique LSD-system with domain S n satisfying..1, namely the system (S n, n) of Lemma..1. This leads us to the following definition: Definition... For n N, we shall call the LSD-system (S n, n) the Laver table of size n and denote it by A n. For a A n, we denote the period of a in S n by P n (a). Remark..1. In writing a A n we mean a S n, similarly we mean a n b while writing a n b. Table.1: The composition tables of the first five Laver tables Lemma..4. For a, b A n and k N >0, we have a n (b [k] n ) (a n b) [k] n. Proof. We show this by induction on k. For k 1 there is nothing to show. Now for 6

7 Figure.1: Heat maps of the first five Laver tables 7

8 k 1 we have using the induction hypothesis (I.H.): a n (b [k+1] n ) a n (b n b [k] n ).1.1 (a n b) n (a n b [k] n ) I.H. (a n b) n (a n b) [k] n (a n b) [k+1] n The following lemma tells us something about A n+1 given A n, in particular how to construct the second half of A n+1 out of A n : Lemma..5 (Proposition 1.16, p. 454). For every n N and 1 a n, there exists a number θ n+1 (a) with 0 θ n+1 (a) P n (a) and θ n+1 ( n ) 0 such that, for every b A n+1, we have: { a n (b) n if (b) Pn+1(a) θ n+1 (a), a n+1 b a n (b) n + n (..5) if (b) Pn+1(a) > θ n+1 (a), { (a + n b if a n, ) n+1 b a n (b) n + n (..6) else. In particular we have: { P n+1 (a + n P n (a) if a n, ) n+1 else. and P n+1 (a) {P n (a), P n (a)} So the period is always a power of. Remark... For a, b A n and n l N, with Lemma..5 we get inductively: (a n b) l (a) l l (b) l (..7) To illustrate Lemma..5 look at Figure.1. The bottom row always forms a complete rainbow in the sense that it starts with purple and increases until it ends with red. The rest of the bottom half is just the preceding table copied in twice with the colors shifted accordingly on the rainbow. Note that the color red gets copied exactly. We now state some facts, which can be found in [Drá94]. Lemma..6 (Lemma.1, p. 584). For 0 j n, n j < a < n and b A j the following holds: P n (( n j ) n) j, in particular we have: ( n j ) n n b n j + b (..8) P n (a) < j (..9) Lemma..7 (Lemma., p. 585). For 0 j n 1, n 1 j < a < n 1 and b A j+1 the following holds: P n (( n 1 j ) n) j+1, in particular we have: (..10) ( n 1 j ) n n b n 1 j + b + ( n 1 j )δ b> j (..11) P n (a) j (..1) 8

9 .3 Not Primitive Recursive In this section we elaborate the proof of [Deh00][Chapter XIII, Proposition 4.13, page 596], which needs some facts about elementary embeddings but can be translated into our finite case. It states that the period of row 1 grows extremely slowly compared to n if we assume that the period goes to infinity. Definition.3.1. For a, b, m, n N with a, b > 0, we define the jump-function of a and a b respectively: J a (m) n : (a) n+1 n+1 m n J a b (m) n : ((a) n+1 n+1 (b) n+1) n+1 n+1 m n Remark.3.1. Assume that J a (m) n holds. By Lemma..5 this means that the period of a jumps from m to m+1 between A n and A n+1. In formulas: In particular we have: P n ((a) n) m and P n+1 ((a) n+1) m+1 (.3.1) J a (m) max{k N : P k ((a) k) m } (.3.) On the other hand assume that max{k N : P k ((a) k) m } n N holds. Again by Lemma..5 we arrive at.3.. The following lemma gives us basic properties of the jump-function. Lemma.3.1. For a, b, k N with a, b > 0, the following holds: If lim n P n((a) n) holds, then we have: If lim n P n((a) n) k+1 holds, then we have: If J a b (m) n holds, then we have: J a : N N (.3.3) J a : {0,..., k} N (.3.4) N > n : J a b J (a) N N (b) N on [0, m] (.3.5) And J a is a strictly increasing function. Remark.3.. We note that the above lemma covers all cases. By Lemma..5 P n ((a) n) increases with n, so the limit exists. If the limit is finite, it cannot be equal to 1, otherwise we would have (a) n n 1 in the case n > 0, which cannot be true for all n > 0. 9

10 Proof. Assume that lim P n((a) n) k+1 holds. Because P 0 ((a) 0) P 0 (1) 0 n holds and by Lemma..5 the period doubles at most, max{n N : P n ((a) n) k } exists. By Remark.3.1 J a (k) exists. So.3.3 is established. Now assume that lim n P n((a) n) < k+1 holds. If J a (k) existed by Remark.3.1 J a (k) max{n N : P n ((a) n) k } would hold and therefore P Ja(k)+1((a) Ja(k)+1) k+1 holds contradicting our assumption. This establishes.3.4. For m, n N with m < n assume that J a (m) and J a (n) exist. By Remark.3.1 and the monotonicity of the period we get that J a is strictly increasing on its domain. Assume that J a b (m) n holds. Now for N > n and 0 c m with J a b (c) n c, we get: ((a) N N (b) N ) nc+1 nc+1 c..7 ((a) nc+1 nc+1 (b) nc+1) nc+1 nc+1 c which means that.3.5 is established. Definition.3.. For a N >0 with {k N : J a (k) k}, we define the critical integer of J a to be: crit(j a ) min{k N : J a (k) k} (.3.6) Now we show that the critical integer exists for any a N >0 : Lemma.3.. For a, k N with k a but k+1 a, we have crit(j a ) k. Proof. For l k, we have: P l ((a) l) P l ( l ) Lemma..5 l This means by Remark.3.1 that for l < k we have J a (l) l. For l N, we have (a) k+1+l {(a) k+l, (a) k+l + k+l } and by our assumption (a) k+1 k holds. So by induction one sees that there exists an l N with (a) k+1+l k+1+l k and (a) k++l k+1+l k. Now by..8 we get P k+1+l ((a) k+1+l) k and by..10 we get P k++l ((a) k++l) k+1 and therefore by Remark.3.1 J a (k) k + l + 1. The following lemma will be crucial in proving the main result. It indicates that the jump-function maps critical integers to other critical values.. Lemma.3.3. For a, b N >0 with a, b N, if J a (crit(j b )) exists and is smaller than N it is equal to crit(j a N b). Proof. Let us assume that crit(j b ) k and J a (k) n hold. So we have (a) n+1 n+1 k n and by Lemma.3. we have (b) k+1 k. Since J a is strictly increasing by Lemma.3.1 we get k n. So by Lemma..5 we have that (b) n+1 is an odd multiple of k. By.3.1 we know that P n+1 ((a) n+1) k+1 holds and we therefore..7 get (a N b) n+1 (a) n+1 n+1 (b) n+1 (a) n+1 n+1 k n. Now by Lemma.3. we get crit(j a N b) n. 10

11 Now we turn to the goal of this section: Theorem.3.1. If for all a, b N >0 we have J a, J a b : N N then the function (J 1 ) x (0) grows faster than any primitive recursive function and therefore the jumpfunction of 1 is not primitive recursive. Remark.3.3. Note that if f : N N, x f(x) is primitive recursive then g : N N, x f x (0) is as well. By direct computation one can see that J 1 (0) 1, J 1 (1) and J 1 () 4 hold. But we don t even know whether J 1 (4) exists. From now on we shall assume that for all a N >0 we have J a : N N. We will need some auxiliary lemmas, definitions and notations to show the result. Definition.3.3. For a, p 0,..., p l N with a > 0, we will use the following notations: J a : p 0... p l : for 0 i < l: J a (p i ) p i+1 J a p 0... p l : crit(j a ) p 0 and for 0 i < l: J a (p i ) p i+1 The following lemma tells us how sequences of critical values can be translated. Lemma.3.4. For a, b N >0, the following holds: J b p 0... p l K N N K : J a N b J a (p 0 )... J a (p l ) (.3.7) Proof. First we show that p i crit(j b [i+1] N ) holds for 0 i l and all N N i for some N i N. We show this by induction on i. For i 0, just take N 0 such that b N0. Now for 0 i < l and N N i+1 max{n i, J b (crit(j b [i+1] Ni )) + 1} we get: p i+1 J b (p i ) I.H. Lemma.3.3 J b (crit(j b [i+1] N )) crit(j b [i+] N ) This completes the induction on i. Now we show that for 0 i l and any N max{n l, J a (crit(j b [i+1] Ni )) + 1} with a N we have crit(j (a N b) [i+1] N ) J a(p i ). Observe that the following holds: crit(j (a N b) [i+1] N Lemma..4 Lemma.3.3 ) crit(j a N ) (b [i+1] N ) J a (crit(j b [i+1] N )) J a (p i ) For 0 i < l and N large enough, using.3.5, we get: J a N b(j a (p i )) J a N b(crit(j (a N b) [i+1] N This completes the proof. ) Lemma.3.3 crit(j (a N b) [i+] N ) J a(p i+1 ) The following lemma states that the spaces between integer sequences of a strictly increasing function are non-decreasing. Lemma.3.5. For a, p, q, r N with a > 0 and J a : p q r, we have r q q p. 11

12 Proof. If p q holds, we have r J a (q) J a (p) q and the lemma holds. If p < q holds we have q < r as well because J a is strictly increasing. The following hold: q p 1 + {b N : p < b < q} and r q 1 + {b N : q < b < r} For p < b < q we get q J a (p) < J a (b) < J a (q) r because J a is strictly increasing. So we get J a ({b N : p < b < q}) {b N : q < b < r} and because J a is injective we arrive at q p r q. Definition.3.4. For n, m N, we say that p (p 0,..., p n ) N n+1 is realizable if there exists an a N >0 that satisfies J a p 0... p n. We say that (p 0,..., p n ) has length n. We say that (p 0,..., p n ) is a base for (q 0,..., q m ) N m+1 if, for each 0 i < m, (p 0,..., p n, q i, q i+1 ) is realizable. Remark.3.4. Note that if (p 0,..., p n ) is a base for (q 0,..., q m ) then for every 0 i m we also have that (p 0,..., p n ) is a base for (q 0,..., q i ). Also observe that the existence of a base for (q 0,..., q m ) implies that this sequence is strictly increasing, and that, if (p 0,..., p n ) is such a base, for all 0 i n we have that (p i,..., p n ) is also a base: this follows by induction on i. For i 0, there is nothing to show. Now for 0 i < n we have by the induction hypothesis that for every 0 j < m there exists an a N >0 with: J a p i... p n q j q j+1 By Lemma.3.4 we conclude for some N N: J a N a p i+1... p n q j q j+1 Lemma.3.6. If (q 0,..., q m ) with m > 0 admits a base then q m m holds. Proof. Assume that (p) is a base for (q 0,..., q m ). For every 0 i < m we have by Lemma.3.5 that q i+1 q i q i p. Since q 0 > p holds, we deduce q i i + p by induction on i. The following lemmas show that a sequence that is based can be expanded in some sense. Lemma.3.7. If (q 0,..., q m ) is based on (p) and (p 0,..., p n, p, q 0 ) is realizable then there exists a sequence based on (p 0,..., p n ) that starts with p and ends with q m of length m. Proof. We show the claim by induction on m 0. For m 0, the sequence (p, q o ) is the desired one because (p 0,..., p n, p, q 0 ) is realizable by assumption and it has length 0. For m > 0, the induction hypothesis gives a sequence σ m 1 based on (p 0,..., p n ) that starts with p and ends with q m 1 of length m 1. Because (p, q m 1, q m ) is realizable, there exists an a N >0 satisfying: J a p q m 1 q m 1

13 We define the sequence σ m by extending σ m 1 with m 1 elements: σ m m 1 +i J a(σ m 1 i ) for 1 i m 1. Because σ m J a(σ m 1 m ) J m 1 a (q m 1 ) q m holds σ m starts with p and ends with q m and is of length m. Furthermore, since σ m 1 is based on (p 0,..., p n ), for 0 i < m 1 there exists a b N >0 satisfying: J b p 0... p n σ m 1 i σ m 1 i+1 As p n < p crit(j a ) holds we get by Lemma.3.4, for some N N: J a N b p 0... p n σ m m 1 +i σm m 1 +i+1. Note that for the case i 0 we use J a (σ m 1 0 ) J a (p) q m 1 σ m m 1 This shows that (p 0,..., p n ) is a base for σ m and therefore completes the proof. Definition.3.5. For n, m N, we define the following functions: m if n 0, G n (m) 0 else, if m 0, G n (m 1) + G n 1 ( Gn(m 1) ) otherwise. Remark.3.5. Note that for all n N we have G n (1) 1. (.3.8) H n (m) Gn(m) (.3.9) Lemma.3.8. For m > 0, if (q 0,..., q m ) is based on (p n, p n+1 ) and (p 0,..., p n+1, q 0 ) is realizable then there exists a sequence based on (p n+1 ) that starts with q 0 and ends with q m of length G n+1 (m). Proof. We show the claim by a double induction: induction on n 0 and for each value of n induction on m 1. For m 1, the sequence (q 0, q 1 ) is the desired one because by Remark.3.4 it is based on (p n+1 ) and starts and ends with the correct elements and is of length 1 G n+1 (1). For m, the induction hypothesis gives a sequence σ m 1 based on (p n+1 ) that starts with q 0 and ends with q m 1 of length G n+1 (m 1). Now we consider two cases to arrive at a sequence based on (p n ) that starts with p n+1 and ends with q m 1 and is of length G n ( Gn+1(m 1) ): If n 0 holds, by Lemma.3.7 we get a sequence σ m 1 that is based on (p 0 ) that starts with p 1 and ends with q m 1 and is of length G 0 ( G1(m 1) ) G1(m 1). Now simply continue with σ m 1 σ m 1. 13

14 If n > 0 holds, by Lemma.3.7 and Remark.3.4 we get a sequence σ m 1 that is based on (p n 1, p n ) that starts with p n+1 and ends with q m 1 and is of length Gn+1(m 1). Now we apply the induction hypothesis for n 1 to the sequence σ m 1. We get a sequence σ m 1 based on (p n ) that starts with p n+1 and ends with q m 1 and is of length G n ( Gn+1(m 1) ). Because (p n, p n+1, q m 1, q m ) is realizable, there exists an a N >0 satisfying: J a p n p n+1 q m 1 q m We define the sequence σ m by extending σ m 1 with the new elements: σ m G n+1(m 1)+i J a(σ m 1 i ) for 1 i G n ( Gn+1(m 1) ). The sequence σ m has length G n+1 (m 1) + G n ( Gn+1(m 1) ) G n+1 (m) and starts with q 0 and ends with σ m G n+1(m) J a(σ m 1 G n( G n+1 (m 1) ) ) J a(q m 1 ) q m. Furthermore, since σ m 1 is based on (p n ), for 0 i < G n ( Gn+1(m 1) ) there exists a b N >0 satisfying: J b p n σ m 1 i We get by Lemma.3.4, for some N N: σ m 1 i+1 J a N b p n+1 σ m G n+1(m 1)+i σm G n+1(m 1)+i+1 Note that for the case i 0 we use J a (σ m 1 0 ) J a (p n+1 ) q m 1 σ m G n+1(m 1). This shows that (p n+1 ) is a base for σ m and therefore completes the proof. Lemma.3.9. For m > 0, if (q 0,..., q m ) is based on (p n, p n+1 ) and (p 0,..., p n+1, q 0 ) is realizable then there exists a sequence based on (p 0 ) that starts with p 1 and ends with q m of length H 1 (... (H n+1 (m))...). Proof. We show the claim by induction on n 0. By Lemma.3.8 we get a sequence σ m based on (pn+1 ) that starts with q 0 and ends with q m of length G n+1 (m). Now we consider two cases: If n 0 holds, by applying Lemma.3.7 we get another sequence σ m based on (p 0 ) that starts with p 1 and ends with q m of length H 1 (m). This is the desired one. If n > 0 holds, by applying Lemma.3.7 and Remark.3.4 we get another sequence σ m based on (p n 1, p n ) that starts with p n+1 and ends with q m of length H n+1 (m). Now we are in position to apply the induction hypothesis and get the desired result. 14

15 Theorem.3.. For x 3, we have: (J 1 ) x (0) H1(...(Hx (1))...) (.3.10) Proof. Since by Remark.3.3 crit(j 1 ) 0 we get the following: In particular we get: J 1 0 J 1 (0)... (J 1 ) x (0) J (J 1 ) 3 (0) By induction on l 0 using Lemma.3.4 by taking N large enough we get: J 1 [l+1] N (J 1 ) l (0)... (J 1 ) l+3 (0) This means that ((J 1 ) x 1 (0), (J 1 ) x (0)) is based on ((J 1 ) x 3 (0), (J 1 ) x (0)) and furthermore (0, J 1 (0),..., (J 1 ) x 1 (0)) is realizable. By Lemma.3.9 we get that there exists a sequence based on (0) that starts with J 1 (0) and ends with (J 1 ) x (0) of length H 1 (... (H x (1))...). We conclude by Lemma.3.6. Now we finalize the proof of Theorem.3.1: Proof. By Theorem.3. it suffices to show that H1(...(Hx (1))...) grows faster than any primitive recursive function. This is established by comparing it with Ackermann s function. This can be looked up in [Deh00][Chapter XIII, after Definition 3.13 on page 590]. 15

16 3 More on Laver Tables Similar to Lemma..5 we shall show in Section 3.1 how some parts of A n+1 may be reconstructed by A n. In Section 3. using these results we shall show that the mean period grows linearly. In the section about steps we shall examine more closely for which b A n the following holds: a n b a n+1 b. We shall give an equivalent formulation for the period of a to jump from 4 to 8 in Theorem Reconstruction of A n+1 We recall that δ φ for any formula φ is 1 if φ is true and 0 otherwise. Now we come to our first finding, which tells us how we can construct the second quarter of A n+1 out of A n : Theorem For a, b, n N >0 with n 1 < a < n and b A n+1, the following holds: a n+1 b (a) n 1 n (b) n 1 + n 1 + n 1 δ (a) n 1 n(b) n 1 >n 1 (3.1.1) In particular we have P n+1 (a) P n ((a) n 1). Remark Note that in the case above (a) n 1 a n 1 holds. Proof. We show using a double induction: descending induction on a and for each value of a ascending induction on b. Note that we are in the case n. For n 1 < a < n and b 1 we have: a n+1 b a n a + 1 (a) n 1 + n (a) n 1 n 1 + n 1 (a) n 1 n (b) n 1 + n 1 + n 1 δ (a) n 1 n(b) n 1 > n 1 For a as above and 1 < b P n+1 (a) we have that b c + 1 c n+1 1 for some 1 c < P n+1 (a). We get by using the induction hypothesis (I.H.): a n+1 b a n+1 (c n+1 1).1.1 (a n+1 c) n+1 (a + 1) I.H. ((a) n 1 n (c) n 1 + n 1 + n 1 δ (a) n 1 n(c) n 1 > n 1) n+1 (a + 1) 16

17 Now we have to consider three cases. We will use that (c) n (c + 1) n 1 and (a) n (a + 1) n 1 hold. To see this we use that for d N >0 we have (d) n + 1 (d + 1) n n d therefore (a) n (a + 1) n 1 holds and by..1 we have P n+1 (a) n 1 so c < n If (a) n 1 n (c) n 1 < n 1 holds, we can apply the I.H. because (a) n 1 n (c) n 1 + n 1 > (a) n 1 + n 1 a holds by..4, we get: ((a) n 1 n (c) n 1 + n 1 + n 1 δ (a) n 1 n(c) n 1 > n 1) n+1 (a + 1) ((a) n 1 n (c) n 1 + n 1 ) n+1 (a + 1) I.H. ((a) n 1 n (c) n 1) n ((a) n 1 + 1) + n 1 + n δ ((a) n 1 n(c) n 1 ) n((a) n 1 +1)> n 1 (a) n 1 n ((c) n 1 + 1) + n 1 + n 1 δ (a) n 1 n((c) n 1 +1)> n 1 (a) n 1 n (b) n 1 + n 1 + n 1 δ (a) n 1 n(b) n 1 > n 1. If (a) n 1 n (c) n 1 n 1 holds, we get: ((a) n 1 n (c) n 1 + n 1 + n 1 δ (a) n 1 n(c) n 1 > n 1) n+1 (a + 1) n n+1 (a + 1)..8 n + a + 1 n 1 + (a) n n 1 + n 1..8 ((a) n 1 n (c) n 1) n ((a) n 1 + 1) + n 1 + n (a) n 1 n ((c) n 1 + 1) + n 1 + n 1..3 (a) n 1 n (b) n 1 + n 1 + n 1 δ (a) n 1 n(b) n 1 > n 1 3. Now for the last case assume that (a) n 1 n (c) n 1 > n 1 holds. We will need that (a) n 1 n (c) n 1 < n holds. To prove this, assume the contrary. We get that a n+1 c n+1 holds which would be a contradiction to c < P n+1 (a). For the second equation also note that for 1 d < n we have d n e d n (e) n 1 since P n (d) n 1 by..9. We get: ((a) n 1 n (c) n 1 + n 1 + n 1 δ (a) n 1 n(c) n 1 > n 1) n+1 (a + 1) ((a) n 1 n (c) n 1 + n ) n+1 (a + 1)..6 ((a) n 1 n (c) n 1) n ((a) n 1 + 1) + n.1.1 (a) n 1 n ((c) n 1 + 1) + n 1 + n 1..3 (a) n 1 n (b) n 1 + n 1 + n 1 δ (a) n 1 n(b) n 1 > n 1 Remark To illustrate what Theorem states, look at Figure.1 and Table.1. The second quarter is nothing but the first half of the preceding table shifted 17

18 accordingly, noting that the red part gets exactly copied. In the language of Section 3.3 it means that for n 1 < a < n holds. s n+1 (a) s n ((a) n 1) (3.1.) Now we state our second finding which tells us something about A n+1 given A n, in particular how to construct the second eighth of A n+1 out of A n and in some cases even the second sixteenth: Theorem For a, b, n N >0 with b A n+1, the following holds: If n is even, the following holds: 1. For n < a < n 1, we have a n+1 b (a) n n (b) n 1 + n 3 i0 in particular P n+1 (a) P n ((a) n ) n holds.. For a n, we have: in particular P n+1 ( n ) n. 3. For 1 a < n, we have: If n is odd, the following holds: n n+1 b (b) n + n 3 1. For n + n 3 a < n 1, we have: a n+1 b (a) n n (b) n 1 + n in particular P n+1 (a) P n ((a) n ).. For n 3 a < n + n 3, we have: δ (a) n n(b) n 1 >in, (3.1.3) i0 δ (b) n >in, (3.1.4) P n+1 (a) n 1 (3.1.5) 3 i0 δ (a) n n(b) n 1 >in, (3.1.6) a n+1 b a n (b) n + n δ a n(b) n > n +n 3, (3.1.7) in particular P n+1 (a) P n (a). For 1 a < n 1, we have: P n+1 (a) n 1 (3.1.8) 18

19 Proof. We show the claim by induction on n. In the case n 1 there is nothing to show. Now let n be even. We show using a double induction: descending induction on a and for each value of a ascending induction on b. For n < a < n 1 and b 1, we have: a n+1 b a n a + 1 (a) n + n (a) n n 1 + n 3 (a) n n (b) n 1 + n δ (a) n n(b) n 1 >i n For a as above and 1 < b P n+1 (a) we have that b c + 1 c n+1 1 for some 1 c < P n+1 (a). We get by using the induction hypothesis (I.H.): a n+1 b a n+1 (c n+1 1).1.1 (a n+1 c) n+1 (a + 1) I.H. 3 ((a) n n (c) n 1 + n δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 Now we have to consider seven cases. We will use that (c) n (c + 1) n 1 holds. To see this we use that for d N >0 we have (d) n + 1 (d + 1) n n d and c < n 1, because for c n 1 we get a n+1 c n+1 (since P n (e) n 1 holds for e < n by..9) and this contradicts c < P n+1 (a). i0 1. If (a) n n (c) n 1 < n holds, we can apply the I.H. because (a) n n (c) n 1 + n > (a) n + n a holds by..4, we get: ((a) n n (c) n 1 + n 3 δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 ((a) n n (c) n 1 + n ) n+1 (a + 1) I.H. ((a) n n (c) n 1) n ((a) n 1 + 1) + n 3 ((a) n n (c) n 1) n ((a) n + 1) i0 + n 3 i0 (a) n n ((c) n 1 + 1) + n (a) n n (b) n 1 + n δ ((a) n n(c) n 1 ) n((a) n 1 +1)>i n δ ((a) n n(c) n 1 ) n((a) n +1)>i n δ (a) n n((c) n 1 +1)>i n i0 3 i0 δ (a) n n(b) n 1 >i n 19

20 . If (a) n n (c) n 1 n holds, using for the last equality that n n ((a) n + 1)..11 n 1 we get: ((a) n n (c) n 1 + n 3 δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 n 1 n+1 (a + 1)..11 n 1 + a + 1 n + (a) n n + n..11 ((a) n n (c) n 1) n ((a) n + 1) + n + n.1.1 (a) n n ((c) n 1 + 1) + n + n..3 3 (a) n n (b) n 1 + n i0 δ (a) n n(b) n 1 >i n 3. If n < (a) n n (c) n 1 < n 1 holds, using for the last equality that for n < d < n 1 we have d n e > n 1 d n e > 3 n, which follows easily by Theorem 3.1.1, we get: ((a) n n (c) n 1 + n 3 δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 ((a) n n (c) n 1 + n 1 ) n+1 (a + 1) ((a) n n (c) n 1) n (a + 1) + n 1..1 (1 + δ ((a) n n(c) n 1 ) n(a+1)>n 1) ((a) n n (c) n 1) n ((a) n + 1) + n 1..3 (1 + δ ((a) n n(c) n 1 ) n((a) n +1)>n 1) 3 (a) n n (b) n 1 + n δ (a) n n(b) n 1 >i n i0 4. If (a) n n (c) n 1 n 1 holds, using for the last equality that n 1 n ((a) n + 1) 3 n holds by..8, we get: ((a) n n (c) n 1 + n 3 δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 n n+1 (a + 1)..8 n + a + 1 n 1 + (a) n n + n 1..8 ((a) n n (c) n 1) n ((a) n + 1) + n + n (a) n n (b) n 1 + n δ (a) n n(b) n 1 >i n i0 0

21 5. If n 1 < (a) n n (c) n 1 < n 1 + n holds, we get: ((a) n n (c) n 1 + n 3 δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 ((a) n n (c) n n ) n+1 (a + 1)..6 ((a) n n (c) n 1 n ) n (a + 1) + n ((a) n n (c) n 1 n 1 ) n 1 ((a) n + 1) + n + n..6 (1 + δ ((a) n n(c) n 1 n 1 ) n 1((a) n +1)>n ) ((a) n n (c) n 1) n ((a) n + 1) + n 1 + n..3 (1 + δ ((a) n n(c) n 1 ) n((a) n +1) n 1 >n ) 3 (a) n n (b) n 1 + n δ (a) n n(b) n 1 >i n i0 6. If (a) n n (c) n 1 n 1 + n holds, we get: ((a) n n (c) n 1 + n 3 δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 ( n + n 1 ) n+1 (a + 1)..6 n 1 n (a + 1) + n..8 n 1 + (a) n n + n..8 ((a) n n (c) n 1) n ((a) n + 1) + n..3 3 (a) n n (b) n 1 + n i0 7. If n 1 + n < (a) n n (c) n 1 holds, we get: ((a) n n (c) n 1 + n δ (a) n n(b) n 1 >i n 3 δ (a) n n(c) n 1 >i n ) n+1 (a + 1) i0 ((a) n n (c) n 1 + n ) n+1 (a + 1)..6 ((a) n n (c) n 1) n (a + 1) + n..9 ((a) n n (c) n 1) n ((a) n + 1) + n..3 3 (a) n n (b) n 1 + n i0 δ (a) n n(b) n 1 >i n This completes the proof of To show P n ((a) n ) n use The proof 1

22 of works mostly analogously using The proof of is simple: P n+1 (a) Lemma..5 P n (a) n 1 Now let n 3 be odd. The proof of is almost the same as of The careful reader will have noticed why we have to restrict ourselves to the case n + n 3 a < n 1. In the first case of the proof of we rely on 3.1.8, which is now in the case of n odd not true, because P n ( n 3 ) n 1 holds. Now that we are in the case n + n 3 a < n 1, we..4 have (a) n n (c) n 1 > (a) n n 3, so we only need to replace the remark by in the proof of the first case of to get a proof of We show using again a double induction: descending induction on a and for each value of a ascending induction on b. Try to find out where we use that n is odd. For n 3 a < n 3 + n and b 1, we have: a n+1 b a n a a n 1 a n (b) n + n δ a n(b) n > n + n 3 For a as above and 1 < b P n+1 (a) we have that b c + 1 c n+1 1 for some 1 c < P n+1 (a). We get by using the induction hypothesis (I.H.): a n+1 b a n+1 (c n+1 1).1.1 (a n+1 c) n+1 (a + 1) I.H. (a n (c) n + n δ a n(c) n > n + n 3) n+1 (a + 1) Now we have to consider three cases: 1. If a n (c) n < n + n 3, we can apply the I.H. because a n (c) n > a holds by..4, we get: (a n (c) n + n δ a n(c) n > n + n 3) n+1 (a + 1) (a n (c) n) n+1 (a + 1) I.H. (a n (c) n) n (a + 1) + n δ (a n(c) n ) n(a+1)> n + n If a n (c) n n + n 3, we get: a n (b) n + n δ a n(b) n > n + n 3 (a n (c) n + n δ a n(c) n > n + n 3) n+1 (a + 1) ( n + n 3 ) n+1 (a + 1) n 3 n (a + 1) + n 3 (a) n n 3 3 i0 Now we have to consider two subcases: i0 3 δ (a) n 1 +1>in 3 + n δ n 3 n(a+1)>i n i0 δ n 3 n(a+1)>i n The case n 3 a < n does not exist. By examining and one notices that a n b n + n 3 so we cannot be in case.

23 If n a < n + n 3, we get: (a) n n 3 3 i0 3 δ (a) n 1 +1>in 3 + n (a) n n 3 + n i0 δ n 3 n(a+1)>i n 3 δ (a) n n 3 >i n i0 (a) n n n (a + 1) n + n + 3 n n..11 n 3 n 1 (a + 1) n + n + n ( n 3 + n ) n (a + 1) n (1 + δ n 3 n 1(a+1) n > n ) + n + n (a n (c) n) n (a + 1) + n..3 a n (b) n + n δ a n(b) n > n + n 3 3. If n + n 3 < a n (c) n, we get: (a n (c) n + n δ a n(c) n > n + n 3) n+1 (a + 1) (a n (c) n + n ) n+1 (a + 1)..6 (a n (c) n) n (a + 1) + n..3 a n (b) n + n δ a n(b) n > n + n 3 This completes the proof of In the second case we used so we needed that n 1 is even. Now let s proof the final statement 3.1.8: For 1 a < n 3, we get: P n+1 (a) P n (a) < n 1 For n 3 a < n 1, we use and together with..9. It is still unclear whether the above process can be continued forever or whether some top part has to actually be calculated and cannot be reconstructed in a similar way as in the above theorems. 3. Mean Period Using the above theorems we shall show that the mean period grows linearly. Definition For n N, we define the following: P n (a) a A µ(n) n n (3..1) 3

24 We shall need some lemmas. Lemma For n 3 odd, we have: P n+1 ( n 1 + n 3 ) P n+1 ( n 1 n 3 ) n 1 (3..) P n ( n 1 + n 3 ) P n ( n 1 n 3 ) n (3..3) So in particular the periods of n 1 + n 3 and n 1 n 3 jump from n to n + 1 if n is odd. Proof. The following shows the desired statements: P n+1 ( n 1 + n 3 ) P n+1 ( n 1 n 3 ) P n ( n 1 + n 3 ) P n ( n 1 n 3 ) T heorem P n ( n 3 ) n 1 T heorem 3.1. P n ( n 3 ) n 1 Lemma..5 P n 1 ( n 3 )..10 n T heorem P n 1 ( n 3 )..10 n Lemma 3... For n N and 1 l < n, we have the following: P n ( n l 1) (3..4) P n+1 ( n l 1) 4 (3..5) In particular the period of n l 1 jumps from n to n + 1. Proof. We show the statement by induction on n. For n 0, 1, there is nothing to Lemma..5 show. For the case n note that P () P 1 () and P 3 () holds. For 1 l < n, we have: P n+1 ( n+1 l 1) P n+ ( n+1 l 1) Now for the case l n, we get: Lemma..5 P n ( n l 1) I.H. T heorem P n+1 ( n l 1) I.H. 4 P n+1 ( n+1 n 1)..10 ( n+1 n 1) n+.1.1 n n+ n..11 n+1 So, by Lemma..5, we get that the period must have doubled and is therefore 4. Now we state the main result of this section. Theorem For n N, the following holds: µ(n) > n (3..6) 4

25 Proof. We show the statement by induction on n. For n 6, the above can be achieved by direct computation. So for the following it is safe to assume n 6. For n even, using Theorem 3.1.1, Theorem 3.1., Lemma..5 and Lemma..7, we get: n+1 µ(n + 1) P n+1 (a) a A n+1 P n+1 (a) + P n+1 (a) + P n+1 (a) 1 a< n n <a< n 1 n 1 <a< n + P n+1 (a) + P n+1 ( n ) + P n+1 ( n 1 ) + P n+1 ( n+1 ) n a< n+1 P n (a) + P n (a) + P n (a) + P n (a) + n+ 1 a< n 1 a< n 1 a< n 1 1 a n n+ + n µ(n) + P n (a) + P n (a) 1 a< n 1 a< n 1 Now we do two separate calculations. Using Lemma..6, Lemma..7, Lemma 3..1, Lemma 3.. and 3.1.4, we get: 1 a< n 1 P n (a) 1 a< n 1 P n 1 (a) + n l0 P n ( n 1 l ) + P n ( n + n 4 ) + P n( n n 4 ) n 1 µ(n 1) + n P n (a) 1 a< n 1 a< n P n (a) + n 3 + n 3 l0 P n 1 (a) + Pn(n n 4 ) 1 a< n P n 1 ( n l ) n µ(n ) + n Now continuing in the case where n is even, we get: n+ + n µ(n) + P n (a) + 1 a< n So we get: + P n 1( n 4 ) + P n( n 1) 1 a< n 1 P n (a) > > + P n 1( n 3 1) n+ + n µ(n) + n 1 µ(n ) + n 1 + n 1 µ(n 1) + n µ(n + 1) I.H. > n + n + n 1 > n 3 4 > n + 1 > 5

26 This completes the case where n is even. For n odd, using Theorem 3.1.1, Theorem 3.1., Lemma..5 and Lemma..7, we get: n+1 µ(n + 1) P n+1 (a) a A n+1 P n+1 (a)+ P n+1 (a) + P n+1 (a)+ 1 a< n 3 n 3 a< n + n 3 n + n 3 a< n 1 P n+1 (a) + P n+1 (a) + P n+1 ( n 1 ) + P n+1 ( n+1 ) n 1 <a< n n a< n+1 P n (a) + P n (a) + P n (a)+ 1 a< n 3 n 3 a< n + n 3 n 3 a< n P n (a) + P n (a) + n+1 + n 1 a< n 1 1 a n P n (a) + P n (a) + P n (a) + n µ(n) + n+1 + n 1 a< n n 3 a< n + n 3 1 a< n 1 Now we do some separate calculations. Using Lemma..6, Lemma..7, Lemma 3.., Theorem 3.1.1, Theorem 3.1. and the separate calculations in the case where n is even, we get: P n (a) 1 a< n 1 1 a< n 1 P n 1 (a) + n l0 P n ( n 1 l ) + P n( n 3 ) + P n( n 1) n 1 µ(n 1) + n P n (a) P n 1 (a) + Pn(n 3 ) n µ(n ) + n 3 + n 1 a< n 1 a< n P n (a) P n (a) + P n (a) n 3 a< n + n 3 n 3 a< n n a< n + n 3 P n ( n 3 ) + P n ( n ) + 1 a< n 3 P n 1 (a) > n + n µ(n 3) + n Now continuing in the case where n is odd, we get: P n (a) + P n (a) + P n (a) + n µ(n) + n+1 + n 1 a< n n 3 a< n + n 3 1 a< n 1 n 3 + n + n µ(n ) + n + n + n µ(n 3)+ n 1 µ(n 1) + n + n µ(n) + n+1 + n > 6

27 So we get: µ(n + 1) I.H. > n + n + n 1 + n 3 > n 7 8 > n + 1 This completes the case where n is odd and therefore the statement is proven. Remark Taking a closer look at the proof, one sees that the inequality can be improved such that for sufficiently large n, we have: µ(n) 1.3n This can be achieved by observing some cases after n 6 and being more precise with µ(n),..., µ(n 3). 3.3 Steps Definition For a A n with n 1 the step of a in A n is: s n (a) min{b A n : a n b > n 1 } (3.3.1) Remark The above definition is well-defined because a n P n (a) n holds and therefore the set {b A n : a n b > n 1 } is nonempty. For a A n, note that by..5 the following holds: s n+1 (a) min{b A n : a n b a n+1 b} (3.3.) By..3 we get for n 1 that s n ( n ) n The following observation is important. Lemma For a A n and 1 i < P n (a), the following hold: 1 i < s n (a) 1 s n (a n i) > (a n 1) Pn(a ni) (3.3.3) s n (a) 1 i < P n (a) s n (a n i) (a n 1) Pn(a ni) (3.3.4) Proof. Consider the following calculation: a n (i + 1).1.1 Lemma..1 (a n i) n (a n 1) (a n i) n (a n 1) Pn(a ni) So the left hand side is greater than n 1 iff the right hand side is greater than n 1. This means by the definition of the step that s n (a) i + 1 is equivalent to s n (a n i) (a n 1) Pn(a ni), where we used that i + 1 P n (a) and (a n 1) Pn(a ni) P n (a n i) hold. So holds. And notice that is just an equivalent formulation. Lemma For 1 a < n 1 and j, k N >1, we have the following: 7

28 1. j s n+1 (a) < s n (a) is equivalent to: For 1 i < j 1, we have (a n 1) Pn(a ni) (a n+1 1) Pn+1(a n+1i) < s n (a n i), s n+1 (a n+1 i). For b a n (j 1), we have (a n 1) Pn(b) < s n (b) and if P n (b) P n+1 (b) holds we have s n+1 (b) (a n 1) Pn(b) otherwise (a n 1) Pn(b) < (a n+1 1) Pn+1(b).. j s n+1 (a) s n (a) is equivalent to: For 1 i < j 1, we have (a n 1) Pn(a ni) (a n+1 1) Pn+1(a n+1i) < s n (a n i), s n+1 (a n+1 i). For b a n (j 1), we have s n (b) (a n 1) Pn(b) and if P n (b) P n+1 (b) holds s n+1 (b) (a n 1) Pn(b) otherwise (a n 1) Pn(b) < (a n+1 1) Pn+1(b). 3. k s n+1 (a) > s n (a) j is equivalent to: For 1 i < j 1, we have (a n 1) Pn(a ni) (a n+1 1) Pn+1(a n+1i) < s n (a n i), s n+1 (a n+1 i). For b a n (j 1), we have s n+1 (b) > (a n+1 1) Pn+1(b) (a n 1) Pn(b) s n (b). For j i < k 1, we have s n+1 (a n+1 i) > (a n+1 1) Pn+1(a n+1i) (a n 1) Pn(a ni). For b a n (k 1), we have if P n (b) P n+1 (b) holds s n+1 (b) (a n 1) Pn(b) otherwise (a n 1) Pn(b) < (a n+1 1) Pn+1(b). Proof. We will use Lemma all the time. First we want to show that all cases imply the first point: For j min{s n (a), s n+1 (a)} and 1 i < j 1, by we only need to show that (a n 1) Pn(a ni) (a n+1 1) Pn+1(a n+1i) holds. If P n (a n i) P n+1 (a n+1 i) holds, this is obvious, noting that a n 1..1 a n+1 1 holds. Otherwise we have P n (a n i) P n+1 (a n+1 i) by Lemma..5. By we get: (a n+1 1) Pn+1(a n+1i) < s n+1 (a n+1 i) P n (a n i) + 1 This means that (a n 1) Pn(a ni) (a n+1 1) Pn+1(a n+1i) holds. Now we want to show that the first case implies the second point. Because of we have (a n 1) Pn(b) < s n (b). By we have s n+1 (b) (a n+1 1) Pn+1(b). So s n+1 (b) (a n 1) Pn(b) is true in the case P n (b) P n+1 (b). Otherwise because s n+1 (b) P n (b) + 1 holds we must have (a n 1) Pn(b) < (a n+1 1) Pn+1(b). Now we want to show that the second case implies the second point. Because of we have s n (b) (a n 1) Pn(b). The rest is shown by the same argument as in the first case. Now we want to show that the third case implies the other points. For the second point we only need to show (a n+1 1) Pn+1(b) (a n 1) Pn(b) because of Lemma Here we use the same argument as used for the first point. The third point works the 8

29 same way. The last point is shown the same way as in the proof that the first case implies the second point. Now we turn to the other directions. First we assume the first point. By Lemma we have j min{s n (a), s n+1 (a)}. Now we consider the first case. By that (a n 1) Pn(b) < s n (b) holds implies j < s n (a). By distinguishing the cases whether the period of b jumps or not we arrive at s n+1 (b) (a n+1 1) Pn+1(b), note that we use So by we arrive at j s n+1 (a). Now we consider the second case. By that s n (b) (a n 1) Pn(b) holds implies j s n (a). Arguing as in the first case we arrive at j s n+1 (a). Now we turn to the final case. By Lemma we arrive at s n (a) j < s n+1 (a) using the second point. The third point implies that k s n+1 (a) holds. Arguing as in the first case we arrive at k s n+1 (a) using the final point. Remark We examine the previous lemma a little closer in the case a 1. Assuming that we are in the case n > 1 we have a n 1. The following hold: 1. If j s n+1 (1) < s n (1) holds, for b 1 n (j 1) we have P n (b) because b n 1. So we cannot have () Pn(b) < () Pn+1(b). This means that the period of b has to stay the same. And that s n+1 (b) holds.. If j s n+1 (1) s n (1) holds, for b 1 n (j 1) we have P n (b) P n+1 (b) and s n+1 (b), arguing in the same way as above. 3. If j s n+1 (1) > s n (1) holds, for b 1 n (j 1) we have P n (b) P n+1 (b) and s n+1 (b), arguing in the same way as above. So in any case for b 1 n (s n+1 (1) 1), we have P n (b) P n+1 (b) and s n+1 (b). Theorem For n N and a A n+1, the following holds: For a n+1, we have: For n a < n+1, we have: For 1 a < n, we have: 1. If P n (a) < P n+1 (a) holds, we have: s n+1 (a) P n+1(a). If P n (a) P n+1 (a) holds, we have: s n+1 (a) n + 1 (3.3.5) s n+1 (a) 1 (3.3.6) + 1 and a n+1 P n+1 (a) n (3.3.7) s n+1 (a) P n(a) (3.3.8) 9

30 For a A n with a n 1 and n > 0, we have: (a n 1) n a n Pn(a) n a n P n (a) (a n 1) (n a n Pn(a) ) (3.3.9) P n (a) P n (a n ) (3.3.10) Summarizing for a n 1 we get: { Pn+1(a) s n+1 (a) + 1 if P n+1 (a) > P n ((a) n) Pn+1(a) if P n+1 (a) P n ((a) n) (3.3.11) Remark The main achievement of the theorem is This heavily depends on which at first seems unnatural. The question is whether similar statements as hold which would be needed to restrict s n+1 (a) even more because of Lemma Heuristically it seems that s n+1 (a) can only take a few values being sums of powers of. Proof. We shall show the theorem by induction on n. Statements through can be shown directly, as we shall see next. For statements and in the case of n 0 there is nothing to show. To see 3.3.5, note that n+1 n+1 b b holds by..3. For n a < n+1, since a n+1, we have a n+1 1 a + 1 > n, which shows For 1 a < n and P n (a) < P n+1 (a), by..5 we have a n+1 P n+1(a) a n+1 P n (a) { n, n+1 }. By definition of the period we cannot have a n+1 P n+1(a) n+1 therefore a n+1 P n+1(a) n holds and by..3 we get a n+1 ( Pn+1(a) + 1) > n which shows Now we turn to the hard part where 1 a < n and P n (a) P n+1 (a) holds. We show this by contraposition: For s n+1 (a) > Pn(a) we have to show that P n (a) < P n+1 (a) holds. We don t need to consider the case a n 1, because by..8 and..10 we have P n (a) 1 < P n+1 (a) and therefore there is nothing to show. So we are in the case where P n (a) is divisible by. To make things shorter we write k instead of Pn(a) in the following. Using 3.3. we can assume that for all b k the following holds: a n+1 b a n b So it suffices to show that for all b k the following holds: a n+1 (k + b) a n (k + b) (3.3.1) This is because we then get a n+1 k a n k n and therefore P n (a) < P n+1 (a). To show we proceed by induction on b. For b < k we get: a n+1 (k + b + 1)..1 a n+1 ((k + b) n+1 1).1.1 (a n+1 (k + b)) n+1 (a n+1 1) I.H. (a n (k + b)) n+1 (a n 1) 30

31 Let l N be such that P n (a) P n l ((a) n l) > P n l 1 ((a) n l 1). This is well-defined by Lemma..5 and because the period has to drop to k. We show the following by induction on l: (a n (k + b)) n+1 (a n 1) ((a) n l n l (k + b)) n l+1 ((a + 1) n l)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 For l 0, the above is obvious. For l N, using in the case s n l ((a) n l) P n l ((a) n l ) Pn(a) k for the first equality, we have: ((a) n l n l (k + b)) n l+1 ((a + 1) n l)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 ((a) n l 1 n l 1 (k + b) + n l 1 ) n l+1 ((a + 1) n l)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )>n i) i1 ((a) n l 1 n l 1 (k + b)) n l ((a + 1) n l 1)+ n l 1 (1 + δ ((a) n l 1 n l 1 (k+b)) n l ((a+1) n l 1 )> n l 1)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 ((a) n l 1 n l 1 (k + b)) n l ((a + 1) n l 1)+ l+1 n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 So the induction on l is completed. Here we have to single out the case b 0. ((a) n l n l k) n l+1 ((a + 1) n l)+ l n i (1 + δ ((a) n i n ik) n i+1((a+1) n i )>n i) i1 ( n l 1 ) n l+1 ((a + 1) n l)+ l n i..11 (1 + δ ((a) n i n ik) n i+1((a+1) n i )>n i) i1 n l 1 (1 + δ (a+1) n l > n l 1) + (a + 1) n l+ l n i (1 + δ ((a) n i n ik) n i+1((a+1) n i )> n i) i1 31

32 Now we want to use in the form: (a n 1) n l 1 (a n 1) n l. For this we need to show that n n l 1 a n k (3.3.13) holds. Seen easily as follows using the induction hypothesis 3.3.8: a n k (a) n 1 n 1 k + n 1 l l (a) n l n l k + n i n l 1 + n i n n l 1 i1 By Lemma..6 we also arrive at Now we continue in the case b 0. Using the above, all δ s drop, we get: i1 n l 1 + (a + 1) n l 1+ l n i Lemma..5 (1 + δ ((a) n i n ik) n i+1((a+1) n i )>n i) i1 n l 1 n l (a + 1) n l 1 + l n i ((a) n l n l k) n l ((a + 1) n l) + i1 By distinguishing the two cases n l a and n l a we easily get: (a) n l n l (k + 1) + Using s n l+i ((a) n l+i) k for 1 i l, we get: l i1 n i l 1 (a) n l+1 n l+1 (k + 1) + n i (a) n n (k + 1) a n (k + 1) i1 l i1 n i This completes the proof in the case b 0. For 1 b < k, we proceed as follows, using that the period drops: 3

33 ((a) n l n l (k + b)) n l+1 ((a + 1) n l)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 ((a) n l n l b + n l 1 ) n l+1 ((a + 1) n l)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )>n i) i1 ((a) n l n l b) n l ((a + 1) n l 1)+ n l 1 (1 + δ ((a) n l n l b) n l ((a+1) n l 1 )> n l 1)+ l i1 n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )>n i) ((a) n l n l b) n l ((a + 1) n l)+ n l 1 (1 + δ ((a) n l n l b) n l ((a+1) n l 1 )> n l 1)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 Now we have to consider two cases: 1. If n l a, using b + 1 k P n l((a) n l ) n l 1 all δ s get annihilated, we get: n l n l (b + 1)+ n l 1 (1 + δ n l n l (b+1))> n l 1)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 l+1 l+1 n l n l (b + 1) + n i (a) n l n l (b + 1) + i1 i1 n i. If n l a, we have (a+1) n l (a) n l n l 1 and as above b+1 P n l((a) n l ) holds and therefore all δ s get annihilated, so we get: (a) n l n l (b + 1) + n l 1 (1 + δ ((a) n l n l (b+1)> n l 1)+ l n i (1 + δ ((a) n i n i(k+b)) n i+1((a+1) n i )> n i) i1 l+1 (a) n l n l (b + 1) + i1 n i 33

34 This completes the two cases. Using s n l+i ((a) n l+i) k for 1 i l, we get: l+1 (a) n l n l (b + 1) + n i (a) n l n l (k + b + 1) + i1 l n i l 1 (a) n l+1 n l+1 (k + b + 1) + n i (a) n n (k + b + 1) a n (k + b + 1) i1 This completes the induction on b and shows and therefore is achieved. Now we want to prove the final statement We consider some cases: 1. If a n+1 holds, we get: i1 (a n+1 1) n+1 a n+1 P n+1 (a) (1) n (1) n+1 (a n+1 1) ( n+1 a n+1 P n+1 (a) ). If n a < n+1 1 holds, noting that by we have n (a) n n P n((a) n ) n 1 and n (a) n n P n((a) n ) is a power of, we get: (a n+1 1) n+1 a n+1 P n+1 (a) ((a) n n 1) n (a) n n Pn((a) n )..6 ((a) n n 1 + n ) n+1 Pn((a) (a) n n ) n n I.H...6 ((a) n n 1) ( n Pn((a)) a n n ) ((a) n n 1) ((a) ( n+1 P a n+1 (a) n+1 ) n n 1 + n..6 ) ( n+1 P a n+1 (a) n+1 ) (a n+1 1) P ( n+1 a n+1 (a) n+1 ) 3. If 1 a < n holds, we consider three cases: If a n 1 holds, we get: (a n+1 1) n+1 a n+1 P n+1 (a)..10 ( n ) n If P n (a) < P n+1 (a) holds, we get: ( n ) n+1 (a n+1 1) ( n+1 a n+1 P n+1 (a) ) (a n+1 1) n+1 a n+1 P n+1 (a) (a n+1 1) n+1 n (a n+1 1) n+1 (a n+1 1) ( n+1 a n+1 P n+1 (a) ) If P n (a) P n+1 (a) holds, we get: (a n+1 1) n+1 a n+1 P n+1 (a) I.H. (a n 1) n+1 Pn(a) a n n (a n 1) ( n+1 Pn(a) a n (a n ) n+1 1) P ( n+1 a n+1 (a) n+1 ) 34

35 This completes the proof of And therefore Theorem is proved. The following states that the period cannot be stuck at. Corollary For n N and 1 a < n with P n (a), we have P n+1 (a) 4. Proof. Because a n 1..1 a n+1 1 holds, we have s n+1 (a) > 1 Pn(a). So by the period of a must jump. Remark Now the second statement of Lemma 3.. becomes a consequence of Corollary The following corollary uses in action. Corollary For n 4 and 1 b n+1, we have the following: If n is even, we have: ( n 1) n+1 b n (b) 8 in particular P n+1 ( n 1) 8. (3.3.14) If n is odd, we have: ( n 3 1) n+1 b n 3 (b) 16 in particular P n+1 ( n 3 1) 16. (3.3.15) Proof. For n 5 odd, by Lemma 3.., we have: 4 P n 1 ( n 3 1) Lemma..5 P n ( n 3 Lemma..5 1) P n+1 ( n 3 1) 16 We first show for the first four values of b: 1. For b 1, this case is obvious by..1.. For b, we have: 3. For b 3, we have: 4. For b 4, we have: ( n 3 1) n and..1 n 3 n+1 n n 3 n n 3 + n δ n 3 n n 3 > n + n n ( n 3 1) n and..1 n n+1 n n n n 3 + n δ n n n 3 > n + n n + n 3 ( n 3 1) n+1 4 n 3 n n 3 + n and..1 ( n + n 3 ) n+1 n i0 δ n 3 n n 3 >i n n 1 35

36 By..5 we have n 1 ( n 3 1) n+1 4 ( n 3 1) n 4 ( n 3 1) n 1 4. So follows by Lemma..5. Now follows by The following lemma gives an equivalent characterization for the period to be. This will be important for the next theorem. Lemma For a A n, we have the following: P n (a) either n 1 and a or 0 l < n : a n l 1 (3.3.16) Proof. For, note that P 1 () and use Lemma 3.. and Lemma..6. For, we proceed by induction on n. For n 0, we have P 0 (1) 1, so we don t have to show anything. For n 1, we have P 1 (1) 1 and P 1 (). Now for n 1, we consider the following cases: If 1 a < n holds, because we assume that P n+1 (a) holds, by Corollary we have P n (a) 1 and therefore a n 1 n+1 n 1 holds. If n a < n+1 Lemma..5 holds, we have P n+1 (a) P n ((a) n). By induction hypothesis and because (a) n n (note that if we are in the case n 1 it is true that (a) n n and then a n n holds) we get that (a) n n l 1 for some 0 l < n. Therefore a n + (a) n n+1 l 1 holds. We don t need to consider the case a n+1 because P n+1 ( n+1 ) n+1 >. This completes the induction on n and therefore the lemma is proven. The following theorem states that the period of a jumps from 4 to 8 iff the period of a + 1 passes a certain level. Theorem For a N >0 with P n0 ((a) n 0 ) and P n0+1((a) n 0 +1) 4 the following equivalence holds for any n: P n ((a) n) 4 and P n+1 ((a) n+1) 8 P n ((a + 1) n) n0 and P n+1 ((a + 1) n+1) n0+1 This means that the period of a jumps from 4 to 8 between n and n + 1 iff the period of a + 1 jumps from n0 to n0+1 between n and n + 1. Proof. We will show that for l 0 as long as P n0+1+l((a + 1) n0+1+l) n0 holds we have P n0+1+l((a) n0+1+l) 4. And in a second step that if the period of a + 1 goes up to n0+1 the period of a has to go up to 8. In both cases we will need Lemma and Theorem First we notice that by Lemma we have (a) n 0 n0 l 1 for some 0 l < n 0 or n 0 1 and (a). First we consider the case n 0 1 and (a). Since we assume that P ((a) ) 4 holds we get that (a) 4. We have P ((a + 1) ) P (1) by Lemma..7. Now we consider two cases: 36

37 1. If (a) 3 8 holds, we have P 3 ((a) 3) P 3 (8) 8 and P 3 ((a + 1) 3) P 3 (1) 4 by Corollary If (a) 3 4 holds, we have P 3 ((a) 3) P 3 (4) P (4) 4 and P 3 ((a + 1) 3) P 3 (5) P (1) by Lemma..5. Now for 0 l with (a) 3 <... < (a) 3+l (a) 3+l+1, we have by Lemma..5 that 4 P 3 ((a) 3) P 3+l ((a) 3+l) and P 3 ((a+1) 3) P 3+l ((a+1) 3+l) hold. By Corollary we have P 3+l+1 ((a+ 1) 3+l+1) 4. We get using that (a) 3+l+1 is divisible by 4 and (a) 3+l+1 < 3+l+1 : s 3+l+1 ((a) 3+l+1 3+l+1 1) s 3+l+1 ((a + 1) 3+l+1) ((a) 3+l+1 3+l+1 1) P3+l+1 ((a) 3+l+1 3+l+1 1) ((a + 1) 3+l+1) P3+l+1 ((a+1) 3+l+1 ) 1 Now by Lemma we have < s 3+l+1 ((a) 3+l+1). So by we must have P 3+l+1 ((a + 1) 3+l+1) 8. So we showed the claim in the case n 0 1 and (a). Now we turn to the case were (a) n 0 n0 l 1 for some 0 l < n 0 holds. Note that since the period of a jumps we must have (a) n 0 (a) n 0 +1 by Lemma..5. By Lemma..7 we have P n0+1((a + 1) n 0 +1) P n0+1( n0 l ) l+1. Note that l + 1 n 0 holds. It is easily seen by induction that for all n N (a) n 0 +1+n n ɛ i n0+i + n0 l 1 i1 holds, where for 1 i n we have ɛ i {0, 1}. For l + 1 k n 0, we have for all n N: k 1 ((a + 1) n 0 +1+n) k Now let us assume that P n0+1+n((a + 1) n 0 +1+n) n0 for n 1. We consider two cases: 1. If P n0+1+n((a + 1) n 0 +1+n) l+1 holds, we have by 3.3.8: il i s n0+1+n((a + 1) n 0 +1+n) l. If n0 k P n0+1+n((a + 1) n 0 +1+n) > l+1 holds, we have by and 3.3.7: s n0+1+n((a + 1) n 0 +1+n) k Now putting the above together we get for n 1 and P n0+1+n((a + 1) n 0 +1+n) n0 : s n0+1+n((a) n 0 +1+n n0+1+n 1) ((a) n 0 +1+n n0+1+n 1) Pn0 +1+n((a) n 0 +1+n n0 +1+n1) By Lemma we get s n0+1+n((a) n 0 +1+n). So inductively by we have P n0+1+n((a) n 0 +1+n n0+1+n 1) 4. 37

38 Now assume that for n 1 we have P n0+1+n((a+1) n 0 +1+n) n0+1 and the period just jumped from n0. We get by using in the second line: ((a) n 0 +1+n n0+1+n 1) Pn0 +1+n((a) n 0 +1+n n0 +1+n1) ((a + 1) n 0 +1+n) n 0 +1 s n0+1+n((a) n 0 +1+n n0+1+n 1) s n0+1+n((a + 1) n 0 +1+n) n0 + 1 By Lemma we get < s n0+1+n((a) n 0 +1+n), so by we must have: This ends the proof of the theorem. P n0+1+n((a) n 0 +1+n) 8 n 0 1 il i 38

39 Acknowledgments I thank Dr. Lorenz Halbeisen for his guidance. I appreciated the close mentoring and the valuable time he took to look after me in our weekly meetings. He made everything possible I needed to focus on my work, I got access to the necessary computation power and he encouraged me to contact Patrick Dehornoy the author of my main reference. I thank Patrick Dehornoy for his time in corresponding with me about a specific problem I had with a proof. Finally I thank all the great people that helped me on my path through life and made this thesis possible. I want to highlight my muse Lena. 39

40 Bibliography [Deh00] Patrick Dehornoy. Braids and self-distributivity, volume 19 of Progress in Mathematics. Birkhäuser Verlag, Basel, 000. [Drá94] Aleš Drápal. Homomorphisms of primitive left distributive groupoids. Comm. Algebra, (7):579 59,

41