Making groups

Equal is when -iLog(i) = -jLog(j) and j+i is the complete sequence.  Then we have equality.

so, when i/(S+1) f**N = 1/(S+1) f**N and

i = f**M/f**N =  f**(M-N) and i nearest integer, we have a set , in which the larger sequence can be composed of the smaller. 

This gives the Fibonacci sequence, naturally. It tells us what the complete sequence is when we can break it up  into a subset which is N-M orders down.

Then compute the error from rounding i. The error decreases,  but the error is positive and locally minimum when i is a multiple of 3, and negative on either side.  This is mainly because 3  are the most difference from the natural ratio, the rounding process is minimal. These, points, I claim, form the sets into groups with a largest common factor. They have the maximum entropy  basis set.

If we have to ratios,f and p, what happens?

i =  f**M/p**N, and M > N. Change exponents: f**M = (p**logp(f))**M

i = p**(M*logp(f) - N) and i nearest integer, i the sequence that makes a set of p ratios within f ratios.

One can find maximum entropy basis sets by watching the error in i.

If p is a ratio of F, then the groups are aligned.  For example, making p = 3/2*f, I get groupings every 2 and 3 steps, as we expect. They actually go 1,3,2,3,1,1, then the pattern diverges. But I reference all the sampling relative to Nyquist, the common denominator of 2. This is a work in progress.

The least common denominator are the energy levels, so the matching goes as:

i = M/N and one should start with these, really locating i that have integer ratios.