分别表示出anan+1an+2=an+an+1+an+2,an+1an+2an+3=an+1+an+2+an+3,两式相减可推断出an+3=an,进而可知数列{an}是以3为周期的数列,只要看2006是3的多少倍,然后通过a1=1,a2=2,求得a3,而2012是3的670倍余2,由此能求出S2012.
【解析】
依题意可知,anan+1an+2=an+an+1+an+2,
an+1an+2an+3=an+1+an+2+an+3,
两式相减得an+1an+2(an+3-an)=an+3-an,
∵an+1an+2≠1,
∴an+3-an=0,即an+3=an,
∴数列{an}是以3为周期的数列,
∵a1a2a3=a1+a2+a3,∴a3=3
∴S2012=670×(1+2+3)+1+2=4023
故答案为:4023.