Piatetski-Shapiro sequences N_(c) contain thin subbases A of order h>=5 (for 1<c<2) or h>=(floor(2c)+1)(floor(2c)+2)+1 (for c>2), with r_{A,h}(n) ~ F(n) for regularly varying F satisfying the stated growth bounds.
Representation functions with prescribed rates of growth
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abstract
Fix an integer $h \geq 2$, and let $b_1, \ldots, b_h$ be (not necessarily distinct) positive integers with $\gcd(b_1, \ldots, b_h) = 1$. For any subset $A \subseteq \mathbb{N}$, let $r_A(n)$ denote the number of solutions $(k_1, \ldots, k_h) \in A^h$ to the equation \[ b_1 k_1 + \cdots + b_h k_h = n. \] Given a function $F$ satisfying $F(n) \leq r_{\mathbb{N}}(n)$, we ask: when does there exist a set $A \subseteq \mathbb{N}$ such that $r_A(n) \sim F(n)$? We prove that this is always possible when $F$ is regularly varying and satisfies $\lim_{n\to\infty} F(n)/\log n = \infty$. If one only requires $r_A(n) \asymp F(n)$, much weaker regularity conditions suffice: we show such a set $A$ exists for every increasing function $F$ satisfying $F(2x) \ll F(x)$ and $\log x \ll F(x) \ll x^{h-1}$. Finally, we give a probabilistic heuristic supporting the following: if $A \subseteq \mathbb{N}$ satisfies $\limsup_{n\to\infty} r_A(n)/\log n < 1$, then $r_A(n) = 0$ for infinitely many $n$.
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math.NT 1years
2026 1verdicts
UNVERDICTED 1representative citing papers
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Thin subbases of Piatetski-Shapiro sequences
Piatetski-Shapiro sequences N_(c) contain thin subbases A of order h>=5 (for 1<c<2) or h>=(floor(2c)+1)(floor(2c)+2)+1 (for c>2), with r_{A,h}(n) ~ F(n) for regularly varying F satisfying the stated growth bounds.