Python can do it in 1:
Code:
for z in range(N-1): print vars().setdefault('a',[1,1]) and a.append(a.pop(0)+a[0]) or a[0][0]
Stolen shamelessly from Bengt Richter. Put your term in place of N.
That being shown, this is the antithesis of the language and good coding practice. As one who has had to work with tools coded in Perl, all I will say is that obfuscation isn't nearly as cool as perlers think it is. I'd rather work with something written in a hundred lines of python than in 5 lines of perl.