![[1] presented a deter](http://s1.studyres.com/store/data/014924405_1-eae03e3c4df5f7c19074d5b588e63597-300x300.png)
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... and again (1) from Lemma 2.2 yields (2.9) and (2.10). In the same style, for n = F2k − 1 we have ε1 (n) = ε2 (bnµc) = 1 and ε1 (n − 1) = ε2 (n) = ε2 (n − 1) = ε1 (bnµc) = 0 and (2.9), (2.10) by (1) from Lemma 2.2. Moreover, for n = F2k we have ε2 (n) = ε2 (n − 1) = ε1 (bnµc) = 0 and (2.9) again by ( ...
... and again (1) from Lemma 2.2 yields (2.9) and (2.10). In the same style, for n = F2k − 1 we have ε1 (n) = ε2 (bnµc) = 1 and ε1 (n − 1) = ε2 (n) = ε2 (n − 1) = ε1 (bnµc) = 0 and (2.9), (2.10) by (1) from Lemma 2.2. Moreover, for n = F2k we have ε2 (n) = ε2 (n − 1) = ε1 (bnµc) = 0 and (2.9) again by ( ...
