$对m归纳[分隔符]证明a_n(\mod 3^m)最小[分隔符]正[分隔符]周期为3^m$
$m=1显[分隔符]然$
$假[分隔符]设m时[分隔符]成立,m+1时$
$而[分隔符]由[分隔符]归纳[分隔符]假设,\{a_n,a_{n+1},...,a_{n+3^m-1}\}\equiv \{1,2,...,3^m\}(\mod 3^m)$
$又[分隔符]所[分隔符]有a_n均[分隔符]为[分隔符]奇[分隔符]数$
$故\{a_n,a_{n+1},...,a_{n+3^m-1}\}\equiv\{1,3,5...2\cdot 3^m-1\}(\mod 2\cdot 3^m)$
$由[分隔符]欧拉[分隔符]定理,\{2^{a_n},2^{a_{n+1}},...,2^{a_{n+3^m-1}}\}\equiv\{2^1,2^3,2^5,...,2^{2\cdot 3^m-1}\}(\mod 3^{m+1})$
$则a_{n+3^m}=a_n+\sum_{i=n}^{n+3^m-1}2^{a_i}\equiv a_n+\sum_{i=1}^{3^m}2^{2i-1}$
$=a_n+2\cdot \frac{4^{3^m}-1}{3}(\mod 3^{m+1})$
$而[分隔符]由[分隔符]升[分隔符]幂[分隔符]定理,v_3(4^{3^m}-1)=v_3(4-1)+v_3(3^m)=1+m$
$故v_3(2\cdot \frac{4^{3^m}-1}{3})=v_3(2)+v_3(4^{3^m}-1)-v_3(3)=0+1+m-1=m$
$故[分隔符]可[分隔符]设2\cdot \frac{4^{3^m}-1}{3}=3^m\cdot s,其[分隔符]中(s,3)=1$
$则a_{n+3^m}\equiv a_n+s\cdot 3^m(\mod 3^{m+1})$
$故a_{n+3^{m+1}}=a_{n+3\cdot 3^m}\equiv a_n+3\cdot s\cdot 3^m\equiv a_n(\mod 3^{m+1})$
$则3^{m+1}是a_n(\mod 3^{m+1})的[分隔符]周期,设T是a_n(\mod 3^{m+1})的最小[分隔符]正[分隔符]周期,则T|3^{m+1}$
$又T显[分隔符]然[分隔符]是a_n(\mod 3^m)的[分隔符]周期,由[分隔符]归纳[分隔符]假设,3^m|T$
$而a_{n+3^m}\equiv a_n+s\cdot 3^m\not\eqiuv a_n(\mod 3^{m+1})$
$a_{n+2\cdot 3^m}\equiv a_n+2s\cdot 3^m\not\equiv a_n(\mod 3^{m+1})$
$故T=3^{m+1}归纳[分隔符]成立$
$由a_n(\mod 3^m)最小[分隔符]正[分隔符]周期为3^m知a_1,a_2,...,a_{3^m}(\mod 3^m)互[分隔符]不[分隔符]相同$
$(否[分隔符]则[分隔符]若a_i\equiv a_j,则[分隔符]由[分隔符]递[分隔符]推[分隔符]式[分隔符]知$
$a_{i+1}\equiv a_{j+1},a_{i+2}\equiv a_{j+2}...,从[分隔符]而|i-j|为[分隔符]周期,这[分隔符]不可能)$