By Alina Carmen Cojocaru

Brief yet candy -- by way of some distance the simplest advent to the topic, which would organize you for the firehose that's the huge Sieve and its functions: mathematics Geometry, Random Walks and Discrete teams (Cambridge Tracts in arithmetic)

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**Example text**

Let = n ≤ x n = ai bj for some i j and = t t prime power fa t ≤ y where y = y x is some parameter to be chosen later. By Exercise 6, is a finite set. 1. 1 implies that t − log 2x # ≤ t∈ t∈ t − log 2x fa t provided that the denominator is positive. 4) To this end, let us remark that if all the powers of a and b are distinct, then the set has cardinality 2 log x (see Exercise 7). 4), and so we conclude that for some i0 j0 we have ai0 = bj0 We may even suppose that i0 j0 = 1 for otherwise we can take i0 j0 -th roots of both sides of the above equality.

19. Show that the implied constant in the Turán–Kubilius inequality can be taken to be 32. 20. Show that the factor 32x implied by the previous exercise for the Turán–Kubilius inequality can be replaced by ⎞1/2 ⎛ a b⎟ ⎜ x = 2x + ⎝ p q ⎠ pa q b ≤x p=q 1 +4 a pa ≤x p Deduce that lim sup x→ x ≤2 x 1/2 q q b ≤x b 4 The Turán sieve In 1934, Paul Turán (1910–76) gave an extremely simple proof of the classical theorem of Hardy and Ramanujan about the normal number of prime factors of a given natural number.

2 The normal number of prime divisors of a polynomial 37 We can now complete our analysis of the normal order of f n . 3)) and the corollary above, = x log log x + O x fn n≤x Also, 2 = fn n≤x 2 y f n +O n≤x y fn n≤x = 2 y f n + O x log log x n≤x We find by the Chinese remainder theorem that 2 y x = fn p q≤y p=q n≤x f p pq f q +O 1 + O x log log x where the latter error term arises from terms where p = q. Since f p p p2 2 =O 1 we have f p q≤y p=q p f q = pq f p≤y p p 2 +O 1 Thus 2 y fn = x log log x 2 + O y2 + O x log log x n≤x Now we recall that y = x with 0 < < 1/2, and so the O y2 error term is dominated by O x log log x .