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84 Chapter 2. On a conjecture about addition modulo 2 k − 1Proof. If t = 0, then k = 3, i.e. t = 000 and #S 1,4 = 4 = 2 · 1 + 3 = 2#S 0,3 + 1.If d = 1 and t ≠ 0, then t = 1---1} {{ }0---0} {{ }, and the formula follows from Theorem 2.4.8.α 1≥1 α 1+3Suppose now that d > 1. As in the proof of the previous proposition, I t,k,i is empty for i < 0if β d > 1 because at least B − 2 + β d ≥ B carries are lost. If β d = 1, then I t,k,−1 is not emptyand consists exactly of the following a’s:B−3t = × {1---1 × 0...10 × ,a = 0---00...1? ,with only 1’s in front of the other 0’s of t, so that #I t,k,−1 = 2 k−(B−3)−B = 2 k−2B+3 .We now enumerate the a ∈ I t,k such that 1a ∈ S 1t,k+1 . As before, a must have at least a 0 infront of a 1 of t and we get 2 k−(B−3)−B = 2 k−2B+3 different a’s with only 1’s in front of the 0’sof t except for the first one.There are also inert a’s with that 0 and another 0 just before a block of 1’s as depicted below:B−3t = × {1---1 × 0...1---10...0 × ,a = 0---00...?--?10...1 .For such an a, 1a will lose at least B − 1 carries, so it must have a 1 before the second 0 and1’s in front of any other 0 of t to be in S 1t,k+1 . There are 2 k−(B−3)−B−1 = 2 k−2B+2 such a’s foreach choice of the second 0 and d − 1 such choices, so (d − 1)2 k−2B+2 different a’s.If β 1 > 1, it is also possible to put that second 0 in front of the second 0 of t:B−3t = × {1---100 × ...0 × ,a = 0---000...1 .For such an a, at least B − 1 carries are lost for a and 1a as well, so that 1a ∈ S 1t,k+1 if and onlyif there are only 1’s in front of each other 0 of t. There are exactly 2 k−(B−3)−B = 2 k−2B+3 sucha’s.If an inert a has a 1 in front of the first 0 of t and β 1 = 1, thenB−3t = × {1---101---1 × ... × 0...0 × ,a = 0---010---0...0...1 ,so that at least B − 1 + α 2 ≥ B carries are lost for 1a and it can not be in S 1t,k+1 .If however β 1 > 1 and there is a 0 in front of the second 0 of t, thenB−3t = × {1---100 × ...0 × ,a = 0---010...1 ,so that at least B − 1 carries are lost for a and 1a as well. Hence, 1a ∈ S 1t,k+1 if and only if thereare only 1’s in front of each other 0 of t. There are exactly 2 k−(B−3)−B = 2 k−2B+3 such a’s.Adding the above quantities gives the formula of the proposition.