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42 Chapter 2. On a conjecture about addition modulo 2 k − 1Proof. We replace α by α i and β by β i in the expression of Definition 2.4.7 and get for 1 ≤ e ithat−|ei−βi| 1 − 4max(0,ei−βi)−eiP (e i ) = 23= 2 −βi+ei−2 max(0,ei−βi)+2 max(0,ei−βi)−2ei 4− max(0,ei−βi)+ei − 13= 2 −βi−ei 4min(ei,βi) − 1.3Considering the case e i = 0 gives the complete formula.Using that formula, it is possible to compute the exact value of P t,k for a given d and acorresponding set of β i ’s. It is also worth noting that the ordering of the β i ’s does not matterbecause each subblock behaves the same when a is in S t,k — more precisely, it overflows — whencethe following definition.Definition 2.4.11. Let k ≥ 2 and t ∈ ( Z/(2 k − 1)Z ) ∗verifyingmin(α i ) ≥iThen we define f d (β 1 , . . . , β d ) asd∑β i − 1 = B − 1 = k − w H (t) − 1 .i=1f d (β 1 , . . . , β d ) = P t,k .The value of P t,k does not depend on the particular choice of t verifying the constraint; inparticular it depends neither on the values of k and the α i ’s, nor on the exact ordering of theβ i ’s, so that the function f d is well defined.It is in fact enough to have min i (α i ) ≥ B − 2 to ensure that a carry goes through each blockwhen a ∈ S t,k , but when equality holds the blocks are not independent anymore.2.4.5 Analytic study: d = 2In this subsection we study the function f d using analytic means when d = 2.Proposition 2.4.12. The function f 2 is given byf 2 (x, y) = 11 ( 227 + 4−x 9 x − 227+ 4 −y ( 29 y − 227))+ 4 −x−y ( 2027 − 2 9 (x + y) ).Proof. Let t ∈ ( Z/(2 k − 1)Z ) ∗be any number made of two blocks corresponding to β1 and β 2and such that min(α 1 , α 2 ) ≥ β 1 + β 2 − 1. ThenP t,k = f 2 (β 1 , β 2 ) .