By Richard R. Hall
The idea of units of multiples, a topic that lies on the intersection of analytic and probabilistic quantity conception, has noticeable a lot improvement because the booklet of "Sequences" by way of Halberstam and Roth approximately thirty years in the past. the realm is wealthy in difficulties, lots of them nonetheless unsolved or bobbing up from present paintings. during this publication, the writer offers a coherent, self-contained account of the present idea, bringing the reader to the frontiers of study. one of many fascinations of the speculation is the range of tools appropriate to it, which come with Fourier research, team conception, excessive and ultra-low moments, likelihood and straight forward inequalities, and several other branches of quantity concept.
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7. 103) holds, for some c = c(sd) > 0. 109) only, for some a = a(sd) > 0. We refer to Z+ n (Tk, Tk + Vk] as a short block if Vk < Tk, else it is a long block, and we write Tk + Vk = Hk Tk in this case, so that Hk > 2. We require that Tk+1 >- Tk + Vk for all k. Sometimes we shall assume (after splitting some blocks if necessary), that Hk < Tk. Let d be Behrend and comprise long blocks. 10 and we have log Hk k=1 (log To 1 -log 2-c = 00. This is useful only if Hk is much smaller than Tk and we want a better result.
It will emerge that in an important special case we may choose y = 1 optimally. However this could be misleading: we show by example that if any open interval of the range (0, 1) be omitted then there exists a non-Behrend sequence which the theorem consequently fails to detect. 7 apart from the moot point concerning equality, best possible. Then we justify our second remark above about y. We require a lemma which is a useful variant of Erdos' law of the iterated logarithm discussed in detail in Divisors, Chapter 1.
99) with k = 1 to obtain E(fS;pv) >- E(f;pv)E(8;pv). 99). It remains to consider the cases of equality, and we leave it to the reader to check that the condition that f and g split M is sufficient for this. 99). 103) we must have E(f jgj;Mi) = E(f j;Mi)E(gj;Mi) for every j, (0 < j < v). We apply the induction hypothesis in each case. Firstly, either f or say f, is constant. In view of the monotonicity of f, this implies that f j is independent of j, that is f (pjd) = f (d) for every divisor d of Ml.