It seems chomodynamics is a 2 and 3 exponent shift up for the gluon. Do they multiply? No. but the root is large. The proton has cut the digit size from 16 to about 6, and each of three guarkss maintain 2 and 3 digits, a 3-nary and a 2-nary finite log adder. Mainly quarks hold exactly the set 1/S for all the quark modes, and the very close Compton match in nulls. Each of the set incompatible with any other combination of null groupings.
That means the quarks can get phase in units of n-1, or exponent 0 and 1, and deliver it to the bag with a -1,-2 exponent shift. The old worm gear machine. Those are tiny amounts. Spin has to be shifted down. After that, the quarks hop around and blow bubbles into the electron orbitals.
So these exponents keep the same ratio as the 1836, which is 17 * 108, just to remind every one.
Gluons have large charge, large mass, but no unit sphere. The nulls is below unit sphere density. They are a Higgs band, and enough bandwidth to cover the electron and quark motion.
Finite sums.
This function:
x1*x2*x3 = p/a. Only p/q, a rational fr5action, is ever known.
Or x1*x2+x2*x3+x1*x3 = q/a.
Or the sum.
The inverses easily had if the definition is recursive. But you can see they are hierarchical in exponent. Even with an exponent shift between them. This is charge,spin, and fuzz?
Bumping the exponent by one adds a higher order Digit, a new spin at higher energy. Then bump it again, and continue you get a lot of atomic orbitals as the digits increase. These ration p/a are all separable, likely prime, and come with their own motion. But the inverse of that sum is collected in the quarks, for 6-9 digits, likely. They cause the quarks to follow a finite log path over the sphere surface, and the atomic orbitals follow.
So we can see the finite recursion simply traces out the points on the groups of rings that make up the quant numbers in the standard model. The rational p/a all have to exist because of dualism between groups and combinatorics. The finite sum recursion is both binary and trinary.
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