Counting quants in the Afimov root

I think this is how it works.  Given spheres with three charges, take the sphere with the small charge. It counts three angles pi/6, each with offset pi/12. These are all stable points, but he counts only once per carry from his left digits.  When it has counted to two, its next count is zero and he carries. At zero he accumulates up to 1/3 of the phase error from all units spheres sharing the quant, and these errors are added to its fractional part. If we do that, then I think we meet all consistency rerquirements, and remain at maximum entropy. We are sticking with one energy level at the moment. ere is wiki

The Efimov effect is an effect in the quantum mechanics of few-body systems predicted by the Russian theoretical physicist V. N. Efimov^[1][2] in 1970. Efimov’s effect refers to a scenario in which three identical bosons interact, with the prediction of an infinite series of excited three-body energy levels when a two-body state is exactly at the dissociation threshold.

I held of on reading this because the puzzle of 22.7 was too fun to extinguish.  The 22.7 I uses is (1/2 + sqrt(3) + 1/(1/2 + sqrt(3)). I missed the 1/f in the previous post on Afimov, and almost never get things right the first time, so always beware. If we are stable to the sample rate of multiply, then the accumulated error from the total of all fractions at the Afimov root should never exceed 1.  It is accumulated from the fractional digits of all unit spheres that share the quant.  I think this work. The power spectrum in the fractional error give you the probability of state transition. The Afimov number they give, 21, probably includes the uncertainty of multiply, the coupling constant, which likely is (1/2+sqrt(3) ) - 1/(3/2 + 3*sqrt(3)). It is related to the full spectrum,
F + 1/F, which includes spin.

I think this will all work. There is a metric on baud:
M(e^baud) =  some r+i*theta. 
And the orbitals at each baud should be given by the direct sum of each Unit sphere wave equation: U1(baud) + U2(baud).... Baud will count the length of the number line, one whole complete sequence. The zero function insures we are a complete Shannon space, it gives space for baud uncertainty to do imprecise multiply.