论文标题
越野 - 卢卡斯多项式与卢卡斯序列之间的关系
Relationship between Vieta-Lucas polynomials and Lucas sequences
论文作者
论文摘要
令$ w_n = w_n（p，q）$为数值序列，满足递归关系\ begin {equination*} w_ {n+2} = pw_ {n+1} -qw_n。 \ end {equation*}我们考虑两个特殊情况$（W_0，W_1）=（0,1）$和$（W_0，W_1）=（2，P）$，我们分别用$ u_n $和$ v_n $表示它们。 Vieta-lucas多项式$ v_n（x，1）$是$ n $的多项式。我们表明，一致性方程$ v_n（x，1）\等价c \ mod p $在且仅当$ u _ {（p-ε）/d}（c+2，c+2）$被$ p $划分时，$ p $，其中$ε\ in \ pm 1 \} $ c $ c $ c $ p $ n $ p $ n $ p $ n $ p $ n $ p $ n $ pp $ p $ p $ p p $ n $ c $ n和p n $ d = n $ n $ d = n $ d = n $ d = n $ d = n $ d = n $ d = n $ d = -
Let $w_n=w_n(P,Q)$ be numerical sequences which satisfy the recursion relation \begin{equation*} w_{n+2}=Pw_{n+1}-Qw_n. \end{equation*} We consider two special cases $(w_0,w_1)=(0,1)$ and $(w_0,w_1)=(2,P)$ and we denote them by $U_n$ and $V_n$ respectively. Vieta-Lucas polynomial $V_n(X,1)$ is the polynomial of degree $n$. We show that the congruence equation $V_n(X,1)\equiv C \mod p$ has a solution if and only if $U_{(p-ε)/d}(C+2,C+2)$ is divisible by $p$, where $ε\in\{\pm 1\}$ depends on $C$ and $p$, and $d=\gcd(n,p-ε)$.
