再来一个线性递推数列!

hongxian

设数列$\{a_n\}$满足$a_0=1$,$a_1=3$,$a_{n+2}=4a_{n+1}-a_n$,试确定$2a_n^2-a_n$ ,$n\in N^*$的末两位数字。

007

周期大约是60啊
1, 15, 31, 21, 65, 11, 91, 65, 41, 11, 15, 21, 81, 15, 51, 1, 65, 31, \
71, 65, 61, 91, 15, 41, 61, 15, 71, 81, 65, 51, 51, 65, 81, 71, 15, \
61, 41, 15, 91, 61, 65, 71, 31, 65, 1, 51, 15, 81, 21, 15, 11, 41, \
65, 91, 11, 65, 21, 31, 15, 1, 1, 15, 31, 21, 65, 11, 91, 65, 41, 11, \
15, 21, 81, 15, 51, 1, 65, 31, 71, 65, 61, 91, 15, 41, 61, 15, 71, \
81, 65, 51, 51, 65, 81, 71, 15, 61, 41, 15, 91, 61, 65, 71, 31, 65, \
1, 51, 15, 81, 21, 15, 11, 41, 65, 91, 11, 65, 21, 31, 15, 1

kuing

汗，这么大的周期……
不过周期内也有对称性的说，回文的……

hongxian

回复 1# hongxian

这样看来是非机器不能列了!

其妙

汗，这么大的周期……
不过周期内也有对称性的说，回文的……
kuing 发表于 2013-7-18 13:59

再汗！
这题居然还来回文的！
这么巧？

caijinzhi

转化成模10周期数列 不难看出 an 模10后 1 1 5 1 1 5 1 1 5…… 2an^2-an 模  1 1 5 1 1 5
故 an 末尾 1 n=1,2 mod3 ; 5 n=3 mod3

tommywong

1,3,3,1(mod4)
1,15,6,21,15,11,16,15,16,11,15,21,6,15,1(mod25)

tommywong

$\begin{pmatrix} 4 & -1 \\ 1 & 0 \end{pmatrix}^{15} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} (mod25)$

isee

矩阵也能同余啊