关于二元递推序列随机项的最小公倍数
数论
2020-12-15 v1
摘要
对每个正整数 n n n δ ∈ [ 0 , 1 ] \delta \in [0,1] δ ∈ [ 0 , 1 ] B ( n , δ ) B(n, \delta) B ( n , δ ) δ \delta δ { 1 , … , n } \{1, \dots, n\} { 1 , … , n } A ⊆ { 1 , … , n } A \subseteq \{1, \dots, n\} A ⊆ { 1 , … , n } ( u k ) k ≥ 0 (u_k)_{k \geq 0} ( u k ) k ≥ 0 u k = a 1 u k − 1 + a 2 u k − 2 u_k = a_1 u_{k - 1} + a_2 u_{k - 2} u k = a 1 u k − 1 + a 2 u k − 2 k ≥ 2 k \geq 2 k ≥ 2 u 0 = 0 u_0 = 0 u 0 = 0 u 1 ≠ 0 u_1 \neq 0 u 1 = 0 a 1 , a 2 a_1, a_2 a 1 , a 2 α \alpha α β \beta β ∣ α ∣ ≥ ∣ β ∣ |\alpha| \geq |\beta| ∣ α ∣ ≥ ∣ β ∣ X 2 − a 1 X − a 2 X^2 - a_1 X - a_2 X 2 − a 1 X − a 2 α / β \alpha / \beta α / β δ n / log n → + ∞ \delta n / \log n \to +\infty δ n / log n → + ∞ B ( n , δ ) B(n, \delta) B ( n , δ ) A A A 1 − o ( 1 ) 1 - o(1) 1 − o ( 1 ) log lcm ( u a : a ∈ A ) ∼ δ Li 2 ( 1 − δ ) 1 − δ ⋅ 3 log ∣ α / ( a 1 2 , a 2 ) ∣ π 2 ⋅ n 2 \log \operatorname{lcm} (u_a : a \in A) \sim \frac{\delta\operatorname{Li}_2(1 - \delta)}{1 - \delta} \cdot \frac{3\log\!\big|\alpha / \!\sqrt{(a_1^2, a_2)}\big|}{\pi^2} \cdot n^2 log lcm ( u a : a ∈ A ) ∼ 1 − δ δ Li 2 ( 1 − δ ) ⋅ π 2 3 log α / ( a 1 2 , a 2 ) ⋅ n 2 lcm \operatorname{lcm} lcm Li 2 \operatorname{Li}_2 Li 2 δ \delta δ δ = 1 \delta = 1 δ = 1 1 1 1 δ = 1 \delta = 1 δ = 1 lcm ( A ) \operatorname{lcm}(A) lcm ( A )
引用
@article{arxiv.1909.04342,
title = {On the l.c.m. of random terms of binary recurrence sequences},
author = {Carlo Sanna},
journal= {arXiv preprint arXiv:1909.04342},
year = {2020}
}