Fred Lunnon wrote:
for integer x,
x - [[x/tau^2 + 1/tau^2]*tau^2 + 1/tau]
= ( - ([(x+3)*tau^2] - 2*[(x+2)*tau^2] + [(x+1)*tau^2]) )mod 3 - 1
The sequence concerned, starting from x = 0, is
0, 1, -1, 0, 1, 0, 1, -1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 1, -1, 0, 1, 0, 1, -1, 0, 1, 0, 1, ...
The second expression is much easier to deal with than the first, since the iterated integer part has been eliminated. But why are the two equal?
PS As previously, tau = (1 + sqrt(5))/2 and [...] denotes floor function. WFL
I'm not sure whether Fred's saying "I believe this but don't have a proof" or "I have a proof but it doesn't give any insight". In case it's the former, here's a proof that doesn't give very much insight. I suspect there's some coincidence at work here. So, the mod 3 thing is happening (in some sense) because 1/tau^2 = 3-tau^2, which lets us convert between tau^2 and 1/tau^2 at the cost of working mod 3. Let's deal with that bit first. For brevity write T for tau and y = x+1. So your identity is y - [T^2[y/T^2]+1/T] = - [T^2(y+2)] + 2[T^2(y+1)] - [T^2y] (mod 3). Since T^2 = 3-1/T^2, the RHS is congruent mod 3 to - [(3-1/T^2)(y+2)] + 2[(3-1/T^2)(y+1)] - [(3-1/T^2)y] or to - [-(y+2)/T^2] + 2[-(y+1)/T^2] - [-y/T^2] or to - [-(y+2)/T^2] - [-(y+1)/T^2] - [-y/T^2]. So much for the RHS, for now. We've made everything on both sides involve terms like y/T^2, and made the RHS more uniform. Now, what about the LHS? Since 1/T = T^2-2 it equals y - [T^2[y/T^2]+T^2] + 2 = y - [T^2(y/T^2-{y/T^2})+T^2] + 2 = y - [y-T^2{y/T^2}+T^2] + 2 = - [T^2-T^2{y/T^2}] + 2 = - [T^2{-y/T^2}] + 2 and now it looks like we no longer care that y is an integer and should write z = -y/T^2, so that we're trying to prove - [T^2{z}] + 2 = - [z-2/T^2] - [z-1/T^2] - [z] (mod 3) or, reducing the number of minus signs, [T^2{z}] - 2 = [z-2/T^2] + [z-1/T^2] + [z] (mod 3). This is now routine. The RHS is clearly unaltered when z changes by an integer, so let's write h = {z}, so that 0 <= h < 1, and the identity is now [T^2h] - 2 = [h-2/T^2] + [h-1/T^2] + [h] (mod 3). So now: 0 <= h < 1/T^2: LHS = -2; RHS = -1+-1+0 = -2 1/T^2 <= h < 2/T^2: LHS = -1; RHS = -1+0+0 = -1 2/T^2 <= h < 1: LHS = 0; RHS = 0+0+0 = 0 as required. (In some woolly sense the RHS is three terms of an infinite series but we don't need any more terms because h is always rather small. Or something like that.) -- g