Let $$\varphi (n) = \#(\mathbb {Z}/n\mathbb {Z})^{\times }$$φ(n)=#(Z/nZ)× (Euler’s totient function). Let $$\epsilon > 0$$ϵ>0, and let $$\alpha \in (0,1)$$α∈(0,1). We prove that for all $$x > x_0(\epsilon ,\alpha )$$x>x0(ϵ,α)… Click to show full abstract
Let $$\varphi (n) = \#(\mathbb {Z}/n\mathbb {Z})^{\times }$$φ(n)=#(Z/nZ)× (Euler’s totient function). Let $$\epsilon > 0$$ϵ>0, and let $$\alpha \in (0,1)$$α∈(0,1). We prove that for all $$x > x_0(\epsilon ,\alpha )$$x>x0(ϵ,α) and every subset $$\mathscr {S}$$S of [1, x] with $$\#\mathscr {S}\le x^{1-\alpha }$$#S≤x1-α, the number of $$n\le x$$n≤x with $$\varphi (n)\in \mathscr {S}$$φ(n)∈S is at most $$x/L(x)^{\alpha -\epsilon }$$x/L(x)α-ϵ, where $$\begin{aligned} L(x) = \exp (\log x\cdot \log _3{x}/\log _2 x). \end{aligned}$$L(x)=exp(logx·log3x/log2x).Under plausible conjectures on the distribution of smooth shifted primes, this upper bound is best possible, in the sense that the number $$\alpha $$α appearing in the exponent of L(x) cannot be replaced by anything larger.
               
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