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62 Chapter 2. On a conjecture about addition modulo 2 k − 1As a corollary, we get a dependence relation.Corollary 2.5.21. For d ≥ 2, 1 ≤ n ≤ l ≤ d − 1, and ∑ nj=1 i j = l,∑d−l( ) d − la d,n+jij= 0 .1,...,i n,0,...,0j=0Proof. The proof goes by induction on d − 1 − l. For l = d − 1, this is a direct consequence of theprevious proposition. For l < d − 1, one uses the induction hypothesis and Corollary 2.5.18.Finally, here is the general expression for a d,n(i 1,...,i n) .Proposition 2.5.2. Suppose that i 1 ≥ · · · ≥ i m ≠ 0 > i m+1 = 0 = · · · = i n and m > 0. Let usdenote by l the sum l = i 1 + . . . + i n > 0 (i.e. the total degree of the monomial). Then( )a d,n(i = l1,...,i n) (−1)n+1 b d,nl,mi 1 , . . . , i ,nwithn−mb d,nl,m = ∑( n − mii=0) d−n∑( ) d − njj=0⎛⎝ ∑ k≥1Within b d,nl,m, the following notation is used:• I = {m + 1, . . . , m + i};• J = {n + 1, . . . , n + j};• S = ∑ j∈I∪J,1≤j≤m k j;• h = d − m − j − i;∑k j≥0,j∈I∪J,1≤j≤m(l + S − m)!l!2 k [ ] ⎞ h − k ⎠ ∏ A kj∏ A kj − 3 kj=0(h − k)! l + S − m k j ! k j !j∈J j∈Im∏j=1C kj−1|k j − 1|!.and⎧⎨ A j + Bj+1j+1if j > 0 ,C j = −⎩13 6if j = 0 ,1 if j = −1 .Proof. If X j = 2, then the degree of β j in SX d is zero. If X j = 0, then 4 −βj can be factored outof SX d and β j will appear in every exponential. Therefore, we can consider only X’s which verifythe following constraints to compute a d,n(i 1,...,i n) :⎧⎨ 0, 1 if 1 ≤ j ≤ m ,X j = 0, 1, 2 if m + 1 ≤ j ≤ n ,⎩1, 2 if n + 1 ≤ j ≤ d .From Proposition 2.5.14,S d X = 2j23 j2 ∏{j|X j=2}((1 − 4 −βj ) S d−j2X− Ξ d X),