On 5/26/08, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
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.
The proof I had relied on background "physics" which seemed to me unecessarily indirect. I had this feeling that there ought to be some method of dealing with such expressions that didn't rely on the special properties of tau. But I think your heroic computation confirms that these identities really are special, and there is (probably) no simple-minded algorithm for them. Sadly, in this particular case it eventually dawned on me to desist from trying to iterate the floor function, and instead replace [(x+1)/tau^2] by another variable --- at which point everything in my top-level expression cancels nicely, without any misdirected attempt at "simplification". C'est la vie. WFL