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I was interested in an arithmetic function satisfying a certain property, I am not sure at the moment if such thing even exists or not. But I was wondering maybe I could get some hint or idea or input from someone.

Fix $k \in \mathbb{N}$ and some $\epsilon > 0$. I would like a function $f$ defined for all $\mathbb{Z}$ such that $$| \ f(n_1) ... f(n_k) \ | \ll | \ n_1+ n_2 + .. + n_k \ |^{-\epsilon} $$
and $$| \ f(n_1) ... f(n_k) \ | \gg 1 ~~~\text{if}~~\ n_1+ n_2 + .. + n_k = 0$$

I would greatly appreciate any assistance on this matter! Thank you very much!

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  • $\begingroup$ I don't think you mean $n_1 + n_2 + \ldots n_k = 0$ if $n_1, \ldots, n_k \in \mathbb N$. What do you really mean? $\endgroup$ – Robert Israel Nov 11 '14 at 20:04
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Suppose that $f$ satisfies your conditions: $|f(n_1)\dotsb f(n_k)|\le A|n_1+\dotsb+n_k|^{-\epsilon}$ if $n_1+\dotsb+n_k\ne 0$, and $|f(n_1)\dotsb f(n_k)|\ge B$ if $n_1+\dotsb+n_k=0$, with positive $A$ and $B$ depending only on $k$ and $\epsilon$. The second condition implies that $\max\{|f(n)|,|f(-n)|\}\ge \sqrt B/|f(0)|^{k/2-1}$, for every positive integer $n$ (take $n_1=n,n_2=-n$, and $n_3=\dotsb=n_k=0$). Using now the first condition with the $n_i\in\{-n,n\}$ chosen to satisfy $|f(n_i)|\ge \sqrt B/|f(0)|^{k/2-1}$ (and all equal each other), we get $B^{k/2}/|f(0)|^{k(k/2-1)}\le A(kn)^{-\epsilon}$. This cannot be true for $n$ large enough. Consequently, no such function exists.

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