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2.4. A block splitting pattern 43According to Proposition 2.4.10, P t,k is given byP t,k =B−1∑E=0∑P (e 1 )P (e 2 )e 1+e 2=E0≤e 1,e 2= 2 −β1−β2 (Σ e1≠0,e 2≠0+ Σ e1=0,e 2≠0 + Σ e1≠0,e 2=0 + Σ e1=0,e 2=0) ,whereΣ e1≠0,e 2≠0 =Σ e1=0,e 2≠0 =Σ e1≠0,e 2=0 =β∑1−1e 1=0⎛2 e1 − 2 −e1 β∑2−1⎝2 e2 − 2 −e233e 2=0β 1+β∑ 2−1+ 2 −e1 4β1 − 13e 1=β 1β∑2−1e 2=0β∑1−1e 1=0Σ e1=0,e 2=0 = 1 .2 e2 − 2 −e232 e1 − 2 −e13β 1+β∑2−1−e 1e 2=0β 1+β∑ 2−1+ 2 −e2 4β2 − 13e 2=β 2β 1+β∑ 2−1− 1+ 2 −e1 4β1 3e 1=β 1β 1+β∑2−1−e 1+2 −e2 4β2 − 13e 2=β 22 e2 − 2 −e23⎞⎠An easy but quite lengthy and error-prone calculation, which can be checked with a symboliccalculus software, leads to the desired expression.This result can also be seen as a consequence of Proposition 2.5.1 in Section 2.5.The graph of f 2 , computed with Sage [250], is given in Figures 2.1 and 2.2.Proposition 2.4.13. For all x, y ≥ 1 in N,Proof.f 2 (x, y) ≤ 1 2 .∂f 2∂x (x, y) = 2 9 4−x ln 4(4 −y − 1)x+ 2 ( (49 4−x −y ln 4 y − 10 )− 4 −y + 1 )33 ln 4 + 1,so that for y > 0, ∂f2∂x(x, y) ≥ 0 is equivalent tox ≤ ( 1 3 + 12 ln 2 )4y + y − 12 ln 2 − 10 34 y − 1.We denote the left hand side of that inequality by h(y). Unfortunately, it happens that h(y) > 1when y ≥ 1. However, f 2 (max(1, h(y)), y) ≤ 1 2 for y ≥ 1, so that f 2(x, y) ≤ 1 2for x, y ≥ 1 in R.We do not prove that here for the sake of simplicity, but only that f 2 (x, y) ≤ 1 2for x, y ≥ 1 in Nwhich is the case we are really interested in.