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34 Chapter 2. On a conjecture about addition modulo 2 k − 1Proof. For e = 1 and d and n fixed, we computen+1∑( ) l + d + 1Σ(d + 1, n + 1) = 2 −l d + 1l=0n+1∑(( ) l + d= 2 −l +dl=0n∑( ) l + d= 2 −l dThe result follows by induction.l=0+ 1 2Corollary 2.3.17. For any d ≥ 0 and e ≥ 0,if and only if n ≤ d.( )) l + dd + 1+ 2 −n−1 ( n + d + 1d( n+2 ∑( (l − 1) + d + 12 −(l−1) d + 1l=0)− 1 2 2−n−1 ( n + d + 2d + 1= Σ(d, n) + 2 −n−1 ( n + d + 1d))) )+ 1 2 Σ(d + 1, n + 1) − 1 2 2−n−1 ( n + d + 2d + 1= Σ(d, n) + 1 Σ(d + 1, n + 1)2( ) (n + d + 1+ 2 −n−1 1 − n + d + 2 )d2d + 2= Σ(d, n) + 1 2 Σ(d + 1, n + 1) + 1 ∆(d, n) .2Σ(d + e, n + e) ≥ 2 e Σ(d, n)Proof. Indeed ∆(d + i, n + i) ≥ 0 if and only if n ≤ d.As a byproduct, we get the following well-known formula [122, formula 5.20].Corollary 2.3.18. For any d positive,Proof. When n = d, the proposition becomesΣ(d, d) = 2 d .Σ(d + e, d + e) = 2 e Σ(0, 0) = 2 e .When n → ∞, the sum converges and we get the following classical result.Proposition 2.3.19. Let d be a positive integer. ThenProof. It follows from the classical identityΣ(d, ”n = ∞”) = 2 d+1 .1∞(1 − z) n+1 = ∑k=0( n + kn)z k .)