中文

与类平衡序列相关的Pillai-Tijdeman型丢番图方程

数论 2021-07-19 v2

摘要

{xn}n0\{x_{n}\}_{n \geq 0} 为由 xn+1=Axnxn1x_{n+1} = A x_{n} - x_{n-1}A>2A>2)定义的类平衡序列,其中 x0=0x_0 = 0x1=1x_1 = 1。本文中,我们展示如何求出丢番图方程 C1xn1+C2xn2+C3xn3=C4xn4+C5xn5+C6xn6C_{1}x_{n_{1}} + C_{2}x_{n_{2}} + C_{3}x_{n_{3}} = C_{4}x_{n_4} + C_{5}x_{n_5} + C_{6}x_{n_{6}} 的所有解,其中固定整数 A3A \geq 3n1>n2>n30,n4>n5>n60,n_1 > n_2 > n_3\geq 0, n_4 >n_5 > n_6 \geq 0,C1xn1C4xn4C_{1}x_{n_{1}} \neq C_{4} x_{n_4},而 C1,C2,C3,C4,C5,C6C_{1}, C_{2}, C_{3}, C_{4}, C_{5}, C_{6} 是满足 C1C2C30C_{1} C_{2} C_{3} \neq 0 的给定整数。

关键词

引用

@article{arxiv.2105.15127,
  title  = {Diophantine Equation with Balancing-like Sequences Associated to the Pillai-Tijdeman-type Problem},
  author = {Bijan Kumar Patel and Prashant Tiwari},
  journal= {arXiv preprint arXiv:2105.15127},
  year   = {2021}
}

备注

12 pages. arXiv admin note: text overlap with arXiv:2105.01569