I usually see the Fibonaccis start with 0 (i.e. 0, 1, 1, 2, 3, etc), but that's not crucial. As for the function: if you pass it 15, then since 15 isn't < 1, the else is executed and the result is fib(14)+fib(13). But then this produces fib(13)+fib(12)+fib(12)+fib(11), etc. It's easier to follow a simpler case: give the function 3; then you get fib(2)+fib(1) and then fib(1)+fib(0)+fib(0)+fib(-1). Since fib(1) results in fib(0)+fib(-1), the final result is 5. Your function give for 0,1,2,3,4,... the values 1,2,3,5,8,...
chas
(I shortened fibonacci to fib to save space. It does seem rather clever that the parser is able to handle functions that call themselves. I think compilers often implement the recursion in terms of loops; I don't know what Perl does.)