Looking at digit systems

Shanon says:
2^N  - 1 = SNR
2^N is the next largest digit and is unused, 1 is the multiplicative identity.

If SNR continues to grow such that it is greater than 1.5 then the term on the left becomes:

2^N  - 2 = SNR and you are multiplying by two, making inefficient use of digits. So the efficient digit system shrinks s by one digit.

Thus, the SNR must be greater than 1/2 and less than 1 1/2 and there is always a finite number of digits in the efficient counter. The efficient digit system is one which uses all digits, and can enumerate sequentially. If the SNR is large, then the efficient system breaks into two systems, the one counts in units of the other.

That is the basic idea of a Huffman encoder. The most frequent thing gets one bit, separated by the zero.  Then the next most frequent thing can be two bits, separated by a zero, and so on, getting 101011010111010110....
The system works whenever you have a finite counter, a bandwidth limited system.

One can use this in reverse, say you want to know the digits system needed to count up three different things to the largest number.  But you need to know how SNR changes with each new digit system. In sphere packing, we want the most spheres packed per surface area, and you have three types of spheres, small to large. If we can prove small spheres go near the center, and we can compute volume to surface area each time we make a new digit system then we can sequentially determine the equivalent Avogrados number.

We pack the first spheres using SNR is r: rs^/rs^2, the is the radius of the small.  Then do the next set,but remove the volume of the first from the noise. Repeat one more time, and get three digits systems. If the number of spheres in each group are equal, then you know the sizes of each digits system, and its a matter of computing the optimum radius ratio.