What about quant exclusion?

Only n quants can occupy a spectrum with n roots at a given energy level. So is the electron excluded from the quark cubic solution? Not unless it occupies a different energy level. But doesn't that make the electron decoupled? Maybe not if the electron occupied a root three quant one integer power  below the quarks. What is the Efimov number? 22.7.  The electron has to be 22.7 times farther than the quarks are to themselves. What did we say about rest mass ratios between electron and quarks? About 21 or 22, I think. 2.7 was times 3 pi , the 3 comes from area over volume.  I always get Pi messed up because the proton does not compute that. 1836/81 = 22.666 is more likely what happened, rather than compute Pi.

But quant exclusion means the bands do not overflow the roots.  We should be able to treat the system as one digit system.
Seems like a simple issue.

Using 3/2 as the base, the electron is 18 exponents from the center of the quarks, and about 6 exponents from the first quark, and the total of all the exponents is about 108.  (3.2^18.54 is 1836). Then readjust the base, if necessary, so you get a complete sequence without overflow. In terms of bandwidth, that means:
b = (1+1/n)^n, for some maxbaud n is the finite natural base, and total rounding error accumulated for the rational approximations. The system is essentially doing Taylor series in cycles, and has to keep accumulated rounding below any overflow foe some element in the power series.

So, finite base,b, says no quant can accumulate mote than b amount of rounding in one completion it its cycle. In my system, quant q has dimension d, then
q^(1/d < b, I think.