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Let $f:\mathbb N \to \mathbb C$ be an arithmetic function. There are various ways to define what the Siegel–Walfisz (S–W) property is for $f(n)$. One simple way is that there exists some function $g(a,q)$ such that for every $q\leq Q(x)$ and $a$ coprime to $q$, one has $$ \sum_{n\leq x \atop n\equiv a \bmod q }f(n) =g(a,q) \left(\sum_{n\leq x \atop \gcd(n,q)=1 }f(n)\right)(1+o_{x\to\infty}(1) ) .$$ For example, in the case of primes, one can take $g(a,q)=1/\phi(q)$ and $q$ going all they way up to $$Q(x)= (\log x)^C$$ for any fixed constant $C>0$.

My question is whether there exists $f(n)$ for which $Q(x)$ cannot be taken as large as a power of $\log x$ but instead something like $\log \log x $, or even smaller?

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There are examples of real-valued functions $f$ for which this property fails even for bounded $q$. For instance, given a function $f$ let $$s_x = \sum_{n \le x,\, n \equiv 1 \bmod 3}f(n),$$ $$ t_x = \sum_{n \le x,\, n \equiv 2 \bmod 3}f(n).$$ For $a=1$ and $q=3$ your property implies $$ s_x = g(1,3)(s_x + t_x)(1+o(1)).$$ If we require $s_x \sim x$ and $t_x \sim x^2$, say, then no constant $g(1,3)$ will work here.

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  • $\begingroup$ I should have added that the assumption is that the property works for every fixed $q$. Clearly, if there is one $q$ for which equidistribution fails then there is no hope to get equidistribution modulo all small integers. $\endgroup$
    – Dr. Pi
    Commented Jan 10, 2023 at 16:20
  • $\begingroup$ @Dr.Pi My guess is that the following example will work, but haven't verified it yet: $f(n) = n^{[n]_{g(n)}-1}$ where $g$ is a slowly increasing function taking values in $\mathbb{N}$, and $[a]_{b}$ is the unique element of $\{0,1,2,\ldots,b-1\}$ congruent to $a \bmod b$. The choice $g(n)=3$ (which does not grow) recovers the example above. $\endgroup$ Commented Jan 10, 2023 at 19:49

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