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arXiv 1509.01752math.NT
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以$a$模$p$为阶的平均结果

Average Results on the Order of $a$ modulo $p$

Sungjin Kim

摘要:

令$a>1$为整数。用$l_a(p)$表示$a$模素数$p$的乘法阶。证明了若$\frac{x}{\log x\log\log x}=o(y)$,则$$\frac 1 y \sum_{a\leq y}\sum_{p\leq x}\frac{1}{l_a(p)}=\log x + C\log\log x+O\left(\frac x {y \log\log x}\right) $$是Felix ~ \cite{Fe}的一个定理的改进.此外,我们还证明了另外两个平均结果:如果$\log^2 x = o(ψ(x))$和$x^{1-δ}\log^3 x = o(y)$,则$$\frac1y \sum_{a<y} \sum_{\substack{{p<x} \\ {l_a(p)>\frac{x}{ψ(x)}}}} 1 = π(x) + O\left(\frac{x\log x}{ψ(x)}\right) + O\left(\frac{x^{2 - δ}\log^2 x}y\right).$$。此外,如果$x^{1-δ}\log^3 x = o(y)$,则$$\frac1y\sum_{a<y} \sum_{\substack{{p<x} \\ {p\nmid a}}}l_a(p) = c\textrm{Li}(x^2) + O\left( \frac{x^2}{\log^A x} \right) + O\left(\frac{x^{3 -δ}\log^2 x}y\right)$$,其中$$c = \prod_p \left(1-\frac p{p^3-1}\right).$$

英文摘要:

Let $a>1$ be an integer. Denote by $l_a(p)$ the multiplicative order of $a$ modulo primes $p$. We prove that if $\frac{x}{\log x\log\log x}=o(y)$, then $$\frac 1 y \sum_{a\leq y}\sum_{p\leq x}\frac{1}{l_a(p)}=\log x + C\log\log x+O\left(\frac x {y \log\log x}\right) $$ which is an improvement over a theorem by Felix ~\cite{Fe}. Additionally, we also prove two other average results If $\log^2 x = o(ψ(x))$ and $x^{1-δ}\log^3 x = o(y)$, then $$\frac1y \sum_{a<y} \sum_{\substack{{p<x} \\ {l_a(p)>\frac{x}{ψ(x)}}}} 1 = π(x) + O\left(\frac{x\log x}{ψ(x)}\right) + O\left(\frac{x^{2 - δ}\log^2 x}y\right).$$ Furthermore, if $x^{1-δ}\log^3 x = o(y)$, then $$\frac1y\sum_{a<y} \sum_{\substack{{p<x} \\ {p\nmid a}}}l_a(p) = c\textrm{Li}(x^2) + O\left( \frac{x^2}{\log^A x} \right) + O\left(\frac{x^{3 -δ}\log^2 x}y\right)$$ where $$c = \prod_p \left(1-\frac p{p^3-1}\right).$$

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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