Patent val/Properties

This page shows some properties of the generalized patent val (GPV).

To tell if a val is a GPV

Suppose we have a p-limit val v, to tell if it is a GPV:

For every prime q in the p-limit, solve

[math]\displaystyle{ \displaystyle \operatorname {round} (n \log_2 q) = v_{\pi (q)} }[/math]

for n. The solution is

[math]\displaystyle{ \displaystyle \frac {v_{\pi (q)} - 1/2}{\log_2 (q)} \lt n \lt \frac {v_{\pi (q)} + 1/2}{\log_2 (q)} }[/math]

Let N₁, N₂, …, N_{π (p)} denote the solution sets. Find

[math]\displaystyle{ \displaystyle \mathrm {N} = \bigcap_{i = 1}^{\pi (p)} \mathrm {N}_i }[/math]

Then v is a GPV of every edo in N if N is not empty; otherwise it is not a GPV.

Cardinality

Given a finite prime limit, the set of all GPVs are countably infinite.

Adjacent GPVs property

Given a finite prime limit, the set of all GPVs can be ordered in this way such that all but one entry in the GPV v_k and its next GPV v_{k + 1} are the same, and for the different entry, the latter increments the former by 1.

This property states that, for example, if it is known that ⟨12 19 28] is a GPV, then the next GPV is one of ⟨13 19 28], ⟨12 20 28], or ⟨12 19 29].

This also holds for any rationally independent subgroups, such as 2.3.7 and 2.9.7. It does not hold, however, for rationally dependent subgroups, such as 2.3.9.7, where at certain points of edo number N, both the mapping for 3 and 9 increment.

Application

Given a finite prime limit, the above properties offer a way to iterate through all GPVs.

Enter a GPV. Set i = 1.
Copy the input and increment its i-th entry by 1.
Test if it is a GPV.
If it is, return it and back to step 1; otherwise, increment i by 1 and back to step 2.

Notice that the all-zero val is a GPV, you can always enter it for the first one. It is guaranteed that you can iterate through all GPVs in this method. In practice it is suggested starting with the last entry and going backwards for better performance, because the largest prime is most likely to increment.