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2.4. A block splitting pattern 39Summing up the above formulae, we get the following theorem.Theorem 2.4.8. Let k ≥ 2 and t ∈ ( Z/(2 k − 1)Z ) ∗made of one block. ThenP t,k ={−α−β 1−2−2α23if 1 ≤ α ≤ k−12,1+2 −2β+13if k−12≤ α ≤ k − 1 .For α = 1, it reads S 1,k = 2 k−2 + 1 and for α = k − 1, it reads S −1,k = 2 k−1 .Proof. Propositions 2.4.5 and 2.4.6 say thatβ−1∑P t,k = P (e) .e=0Moreover, for such values of e, β − e is always positive so that |β − e| = β − e and M = 0. Finally,m = e as long as e ≤ α which is always true if and only if β − 1 ≤ α or equivalently k−12≤ α.If 0 < α < k−12 , then P t,k is computed asα−1∑P t,k = 2 −β +e=1= 2 −β + 2−β3= 2 −β + 2−β3e−β 1 − 4−e23α−1∑e=1∑β−1+e=αe−β 1 − 4−α23(2 e − 2 −e ) + 2−β (1 − 4 −α )3∑β−12 ee=α((2 α − 1) + 2 · (2 −α − 1) ) + 2−β (1 − 4 −α )2 α (2 β−α − 1)3= 2 −β + 2α−β − 2 −β + 2 −α−β+1 − 2 −β+13= 2 −β + 1 − 3 · 2−β − 3 · 2 −α−β − 2 −2α3−α−β 1 − 2−2α= 23If k−12≤ α < k, then the calculation is somewhat easier:.β−1∑P t,k = 2 −β e−β 1 − 4−e+ 23e=1= 2 −β + 2−β3= 2 −β + 2−β3β−1∑(2 e − 2 −e )e=1+ 1 − 2−2α − 2 α−β + 2 −α−β3((2 β − 1) + 2 · (2 −β − 1) )= 2 −β + 1 − 2−β + 2 −2β+1 − 2 −β+13= 1 + 2−2β+13.