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2.7. An inductive approach 81• O t,k,i = { a ∈ Z/(2 k − 1)Z | r(a, t) = w H (t) − i, a + t ≥ 2 k − 1 in Z } , the overflowing modularintegers.Remember that r(0, t) = k and that 0 is considered to act as the 1...1 binary string, so that wehave 0 ∈ O t,k,wH (t)−k.We define I t,k = ⊔ i∈Z I t,k,i and O t,k = ⊔ i∈Z O t,k,i.Lemma 2.7.3. Let t ∈ N and k ≥ l(t). Then#C t,k,i = #I t,k,i + #O t,k,i .We will now study the behavior of these sets as k grows.2.7.2 Adding 0’sWe want to let k grow as t is fixed, i.e. add 0’s in front of the binary string associated with t assoon as k ≥ l(t).If a ∈ I t,k,i , then 0a and 1a are in I 0t,k+1,i . Unfortunately, the situation is more complicatedfor O t,k,i :• if a ≠ 0 and a ≠ −t is in O t,k,i , then 1a is in O 0t,k+1,i−1 , and 0a is in I 0t,k+1,j with j ≥ i;• if a = 0, then a ∈ O t,k,wH (t)−k, 0a = 0 ∈ O 0t,k+1,wH (t)−k−1, and 1a = 2 k ∈ I 0t,k+1,wH (t);• If a = −t, then a ∈ O t,k,wH (t)−k, 0a = 0t k ∈ I 0t,k+1,wH (t), and 1a = −t ∈ O 0t,k+1,wH (t)−k−1;Finally, 2 k − 1 = 0 1...1} {{ }=k∈ I 0t,k+1,j with j ≤ w H (t) − k.The following lemma summarizes the above discussion.Lemma 2.7.4. Let t ∈ N and k ≥ l(t). Then{1Ot,k,i if i < wO 0t,k+1,i−1 =H (t) − k1 (O t,k,i \ {0}) ⊔ {0} if i = w H (t) − k⊔{0Ot,k,i if i < wI 0t,k+1,j ⊃H (t) − k0 (O t,k,i \ {0}) ⊔ { 2 k} if i = w H (t) − kj≥iI 0t,k+1,i ⊃ 0I t,k,i ⊔ 1I t,k,i .,,Lemma 2.7.5. Let t ∈ N and k ≥ w H (t) + l(t). If i ≥ 0, then O t,k,i = ∅.Proof. Indeed t = 0...0} {{ }..., so a = ← 1...1} {{ }←... and r(a, t) > w H (t).≥w H (t)≥w H (t)Proposition 2.7.6. Let t ∈ N and k ≥ w H (t) + l(t). Then#S 0t,k+1 = 2#S t,k − 1 .Proof. Since k ≥ w H (t) + l(t), 2 k − 1 ∈ I 0t,k+1,i with i < 0 (t ≠ 0) and O t,k,i = ∅ for i ≥ 0 so that{0It,k,i ⊔ 1I t,k,i { for 0 ≤ i < w H (t)I 0t,k+1,i =0I t,k,i ⊔ 1I t,k,i ⊔ 2 k , t k} for i = w H (t),