Can Fibonacci be a more efficient quantizer than Shannon-Nyquist?

In an infinite world no. But when the log function is finitely calculated, than yes. Whenever the process is using a finite multiplier, then you want to builds the spectral distribution in the base of the same multiplier. The characteristic equation in the case of the proton is x^2 -x -1 = 0. So allocate spectrum in units of those solution.