bm_fib.rb

require "benchmark"

N = (ARGV.shift || 40).to_i

def fib(n)
  return n if n < 2
  fib(n-2) + fib(n-1)
end

Benchmark.bmbm do |x|
  x.report("fib(#{N})") { fib N }
end

fib.lua

function fib(n)
  if n < 2 then return 1
  else return fib(n-2) + fib(n-1) end
end

print(fib(40))

gistfile3.txt

gauss:rubinius brian$ time lua -v fib.lua 
Lua 5.1.4  Copyright (C) 1994-2008 Lua.org, PUC-Rio
165580141

real	0m49.459s
user	0m46.261s
sys	0m0.217s

LuaJIT 2.0.0-beta5 -- Copyright (C) 2005-2010 Mike Pall. http://luajit.org/
165580141

real	0m12.862s
user	0m12.090s
sys	0m0.054s

gauss:rubinius brian$ rbx -v work/bench/bm_fib.rb 
rubinius 1.2.1dev (1.8.7 aa7535ab 2010-12-21 JI) [i686-apple-darwin9.8.0]
Rehearsal -------------------------------------------
fib(40)   9.221481   0.035743   9.257224 (  9.767241)
---------------------------------- total: 9.257224sec

              user     system      total        real
fib(40)   9.203364   0.038571   9.241935 (  9.770584)
gauss:rubinius brian$ rbx -v -Xint work/bench/bm_fib.rb 
rubinius 1.2.1dev (1.8.7 aa7535ab 2010-12-21) [i686-apple-darwin9.8.0]
Rehearsal -------------------------------------------
fib(40)  29.933660   0.140400  30.074060 ( 32.344898)
--------------------------------- total: 30.074060sec

              user     system      total        real
fib(40)  29.278165   0.127529  29.405694 ( 31.091665)

Please learn about some Lua basics before attempting to benchmark it. Use "local function fib(n)" and it runs 30% faster with LuaJIT.
Oh, and performance on this benchmark is completely irrelevant unless your application is computing Fibonacci numbers all day long.

Interesting. But like you say, this is completely irrelevant. ;)

Isn't rather unfair to benchmark lua with time while benchmarking rbx in process?

Yep

Reverse the recursion and the LuaJIT version becomes many times faster (fib(n-1) + fib(n-2)). Down-recursive trace can be unrolled then, Mike said.