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56 Chapter 2. On a conjecture about addition modulo 2 k − 1The series is then given by the following classical formula for k ≥ 1 and |z| < 1:〈∑ k k∞∑i k z i j=0zj〉j+1=(1 − z) k+1 .i=1The formula for the truncated sum is slightly more involved as stated in the next lemma.Lemma 2.5.10. For k ≥ 1 and |z| ≠ 1,∑ kj=0 A 0(k, j)z j+1(1 − z) k+1 −( ∑ki=0( ki) ( ∑ kj=0 A i(k, j)z j+1 )n i )z nn∑i k z i =(1 − z) k+1 ,i=1〈 kwhere A i (k, j) is defined by the same recursion relation as and the initial conditionsj〉( ) iA i (i, j) = A i (i + 1, j) = (−1) j .j〈 kIn particular, A 0 (k, j) = and we have the simple recursion formula for i ≥ 1j〉A i (k, j) = A i−1 (k − 1, j) − A i−1 (k − 1, j − 1) .We are interested in the case where z = 1/4, n = β − 1 and 1 ≤ k ≤ j 2 − 1, which is writtenas (beware that we are summing up to β − 1 and not β, so the expression is slightly differentfrom the one above)(β−1 ∑ ∑k−1 (∑ke k 4 −e j=0=A 0(k, j)4 −j−1ki=0 i) ( ∑ ) )kj=0 A i(k, j)4 −j−1 β i 4 −β(3/4) k+1 −(3/4) k+1e=1( ∑k)j=0 A k(k, j)4 −j β k 4 −β−(3/4) k+1 .Moreover, we have the following identity.Lemma 2.5.11. For 0 ≤ i ≤ k,k∑k+1∑3 A i (k, j)4 −j = 4 A i+1 (k + 1, j)4 −j .Proof. Indeed,∑k+14j=0j=0k+1∑A i+1 (k + 1, j)4 −j = 4= 4= 4= 3j=0k∑j=0j=0(A i (k, j) − A i (k, j − 1))4 −jk+1∑A i (k, j)4 −j − 4 A i (k, j − 1)4 −jk∑A i (k, j)4 −j − 4j=0k∑A i (k, j)4 −j .j=0j=1k∑A i (k, j)4 −j−1j=0