Most date or time systems have an Epoch, or privileged "starting" point. For example, the Epoch of the Gregorian calendar is 1 A.D. The Epoch of Unix time January 1, 1970.
These Epochs introduce a temporal asymmetry of sorts. What would be nice if we had date systems with temporal symmetry. We define temporal symmetry for a date system as follows:
First we will define a date system. For simplicity, we will only worry about years. A date system is a function $d$ that assigns each year (which, for simplicity, is defined as a time period from the midnight of a December 31st exclusive to the midnight of the next December 31st inclusive) to a string in a formal language $L(d)$. For example, the Gregorian calendar would assign the year this post was posted the string "2019 A.D.".
A date system $d$ is said to have temporal symmetry if there exists a permutation $\sigma$ on $L(d)$'s alphabet such that
$$\forall y. d(y + \text{1 year}) = \sigma_*(d(y))$$
Where $\sigma_*(s)$ applies $\sigma$ to each letter of $s$. Note that this implies that $|d(y)|$ is equal to some constant $V(d)$, which we will call the verbosity of $d$. Unfortunately, this also implies that if $V(d)$ and $L(d)$ are finite, then $d$ is periodic. If $d$ has a large enough of period though (such as, say, $10,000$ years) this hopefully won't be a problem. We define $P(d)$ to be the fundamental period of $d$.
For any alphabet $\Sigma$, we can define a $d$ with temporal symmetry such that $\Sigma$ is the alphabet of $L(d)$. Moreover, $P(d) = g(|\Sigma|)$ (where $g$ is Landau's Function). Let $\sigma$ be some maximal order permutation of $\Sigma$. Say it decomposes into disjoint cycles $\{c_1, c_2, \dots, c_k\}$. Let $s$ be a string such that $|s| = k$ and $s_n \in c_n$ for all $n \in [1 .. k]$. Also let $y$ be an arbitrary year.
We define $$d(y + \text{n years}) := \sigma_*^n(s)$$ for all integers $n$. $\sigma$ itself witnesses that $d$ has temporal symmetry. The $P(d) = g(|\Sigma|)$ since that is defined as the order of $\sigma$, which is also the period of $d$. Additionally, $V(d) = k$.
Now, for my question. For a given (finite) alphabet $\Sigma$, what are the most efficient temporally symmetric date systems such that $L(d) = \Sigma$? We say that a temporally symmetric date system $d$ is most efficient if for all other temporally symmetric date systems $d'$ such that $L(d') = L(d)$, one of the following is true:
- $P(d) \gt P(d')$
- $V(d) \lt V(d')$
- $P(d) = P(d') \land V(d) = V(d')$
Clearly the only important thing about $\Sigma$ for the purposes of this question is its size. Something to figure out first is if the date system I gave above is efficient.
Given an alphabet $\Sigma$, a date system is completely determined by (1) the string value $d(0)$ of the year zero. (2) The alphabet permutation $\sigma:\Sigma\leftrightarrow \Sigma$. This is because of the symmetry of the date system, as you describe it.
If the alphabet has size $n$ (so $\Sigma = \{a_1,\ldots,a_n\}$), then the longest attainable period for your date system is $g(n)$, where $g$ is Landau's function. There is some alphabet permutation $\sigma^\star$ that has this period on $\Sigma$.
In this case, you could pick the initial date to be the list of all alphabet characters $d(0)=a_1\ldots a_n$ and the alphabet permutation to be $\sigma^\star$.
This date system has the longest possible period, but does not have the shortest possible date string $d(0)$. As you point out, it's enough to pick symbols $b_1,\ldots,b_k$ from each of the distinct orbits of $\sigma^\star$ and define $d(0)=b_1,\ldots,b_k$.
That date system, as you define above, is optimal in the sense that the period is as long as possible, and the initial date string is as short as possible among systems with period $g(n)$. I believe this is the best definition of optimality. (Your definition of $V(d)<V(d^\prime)$ can be cheated by picking $d(0)$ to be the empty string, in which case it's shorter than any other date system with that same alphabet.)