That’s very nice! And a little rearrangement will bring it from three to two “big multiplications”, making it faster than my routine: def fib_ring2(n): a

That’s very nice! And a little rearrangement will bring it from three to two “big multiplications”, making it faster than my routine: def fib_ring2(n): assert n >= 0 a, b = 2, 0 # invariant: phi^n = (a + b*sqrt(5)) / 2 for bit in bits(n): ab = a*b a, b = (a+b)*((a+5*b)//2) - 3*ab, ab if bit: a, b = (a + 5*b)//2, (a+b)//2 return b