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82 Chapter 2. On a conjecture about addition modulo 2 k − 1and#E 0t,k+1 ⊔ T 0t,k+1 = ∑ i≥0#I 0t,k+1,i + #O 0t,k+1,i= ∑ i≥0#I 0t,k+1,i= 2 ∑ i≥0#I t,k,i + 2Then= 2#E t ⊔ T t,k + 2 .#S 0t,k+1 = 2 k+1 − 1 − #E 0t,k+1 ⊔ T 0t,k+1= 2(2 k − 1 − #E t ⊔ T t,k ) − 1= 2#S t,k − 1 .Unfortunately, that equality is not true for l(t) ≤ k < l(t) + w H (t), and it can even happenthat #S 0t,k+1 > 2#S t,k . However, experimental results suggest that, as soon as k ≥ l(t) + 2, thefollowing inequality is true.Conjecture 2.7.7. Let t ∈ N. For k ≥ l(t) + 2,#S 0t,k+1 ≤ 2#S t,k .2.7.3 Adding 1’sIn light of Theorem 2.4.15, one would also like to increase the size of the 1 subblocks, even “empty”ones, that is insert 1’s between 0’s. Write down t (or an equivalent one) as:t = 1---1} {{ }α 10---0...1---10---0 .In contrast with the previous section, we allow α 1 = 0, i.e. t can begin with a 0. However wewant the last 0 subblock to be non-empty. So here d will potentially denote the previous numberof blocks plus one if α 1 = 0. For the sake of clarity, d can be defined to be the number of blocksof 1t.If a ∈ O t,k,i , then 1a and 0a are in O 1t,k+1,i (except for 10---0 = 2 k , but we get 01---1 = 2 k −1instead). If a ∈ I t,k,i , then 1a ∈ O 1t,k+1,j with j ≤ i and 0a ∈ I 1t,k+1,i+1 . Hence, it may happenthat 2#S t,k > #S 1t,k+1 . It is not even true that2 B−1−α1 #S t,k ≤ #S 1...1t,k+B−1−α1where B − 1 − α 1 1’s have been added in front of t.As pointed above, it is not true in general that2#S t,k ≤ #S 1t,k+1 ,or equivalently that #I t,k,−1 < # {a ∈ I t,k,i , i ≥ 0 | 1a ∈ O 1t,k+1,j , j < 0}. However, once a blockis big enough, its behavior is known.