中文

佩尔数与佩尔-卢卡斯数作为两个同码数之差

数论 2023-10-13 v1

摘要

{Pn}n0\{P_{n}\}_{n\geq 0} 为由 P0=0P_0=0P1=1P_1 =1 且对所有 n0n\geq 0Pn+2=2Pn+1+PnP_{n+2}= 2P_{n+1} +P_n 定义的佩尔数序列,并设 {Qn}n0\{Q_{n}\}_{n\geq 0} 为其伴随序列,即由 Q0=Q1=2Q_0=Q_1 =2 且对所有 n0n\geq 0Qn+2=2Qn+1+QnQ_{n+2}= 2Q_{n+1} +Q_n 定义的佩尔-卢卡斯数。本文中,我们求出所有可写为两个同码数之差的佩尔数与佩尔-卢卡斯数。结果表明,可写为两个同码数之差的最大佩尔数与佩尔-卢卡斯数分别为 P6=70=777Q7=478=55577.P_6=70= 77-7 \quad\quad \hbox{和} \quad\quad Q_7 = 478=555-77.

关键词

引用

@article{arxiv.2310.08422,
  title  = {Pell and Pell-Lucas numbers as difference of two repdigits},
  author = {Bilizimbeye Edjeou and Bernadette Faye},
  journal= {arXiv preprint arXiv:2310.08422},
  year   = {2023}
}

备注

to appear in Afrika Matematika