Meh, if you don't have to use perl, you can do much better: Re: Fibonacci numbers (again).

Update: let's put them under each other to see which one is longer.

         1         2         3
123456789012345678901234567890123
dc -e1d[pdsd+ldrlxx]dsxx
ruby -e'x=i=1;p x+=i=x-i while+1'
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