分析:在數(shù)列遞推式an=an-1+an-2-an-3中,以n+1替換n,得到an+1=an+an-1-an-2,作和后可得數(shù)列{an}的奇數(shù)項(xiàng)和偶數(shù)項(xiàng)均構(gòu)成等差數(shù)列,由已知求出奇數(shù)項(xiàng)的公差,代入等差數(shù)列的通項(xiàng)公式求得a2013的值.
解答:解:由an=an-1+an-2-an-3,得
an+1=an+an-1-an-2,
兩式作和得:an+1=2an-1-an-3.
即an+1+an-3=2an-1(n=4,5,…).
∴數(shù)列{an}的奇數(shù)項(xiàng)和偶數(shù)項(xiàng)均構(gòu)成等差數(shù)列,
∵a1=1,a3=9,∴奇數(shù)項(xiàng)公差為8.
則a2013=a1+8(1007-1)=1+8×1006=8049.
故答案為:8049.
點(diǎn)評(píng):本題主要考查由遞推公式推導(dǎo)數(shù)列的通項(xiàng)公式,其中滲透了周期數(shù)列這一知識(shí)點(diǎn),屬中檔題.