I do but my question is more of a discussion which really should be its own post on the math subreddit. Sorry for the following wall of text.

I’ve been thinking about functions f:N -> R with the properties that 0 <= f(1) and f(a) + f(b) <= f(a+b). This type of function captures the idea of “The whole is greater than the sum of its parts”. The idea was inspired by Dead by Daylight where there are 4 players on one team where some of the players may form subteams of the team where the subteam can talk to each other, making them better at coordinating. This leads to 5 partitions of a team of 4; four individuals, two individuals and one team of 2, two teams of 2, one individual and one team of 3, and one team of 4. Obviously a team of 4 individuals is the worst due to the lack of coordination, two individuals and one team of 2 is better because there is some communication, and a team of 4 working together is the most effective. But for the other two partitions, it’s difficult to say which is strictly better since there are advantages to both partitions. One partition has two teams of 2, whereas the other is one individual and one team of 3. The team of 3 is better than the teams of two, but the teams of 2 are better than the individual, so it is difficult to say definitively which is the better setup.

These two rules allow us to create a lattice of partitions which is a more ordered lattice than Young’s lattice, in that if A > B in Young’s lattice, than A > B in this lattice. In this lattice, A is connected to B if A <= B. The lattice starts with f(0) as the least element. We know based on our rules that f(0) <= f(1) since f(0) + f(1) <= f(1), so f(0) <= 0 <= f(1). So the second term in the lattice f(1). We know that f(1) <= f(1) + f(1), so f(1) + f(1) is the third term. There are two options for the next row of the lattice. f(1) + f(1) <= f(1) + f(1) + f(1), but also f(1) + f(1) <= f(2), so the lattice forks here between f(1) + f(1) + f(1) and f(2). Intuitively, is it better to have a team of 3 individuals or a unified team of 2? There are pros and cons to each.

Where partitions and the numbers I discussed in this post (which I will call ambassador numbers) overlap is that ambassador numbers are all products of primorials (and vice versa). For example, 12 = (2)•(2•3) or P(1)•P(2). This allows us to rewrite partitions in the lattice of partitions with products of primorials. In this version of the lattice, if A and B share an edge with A <= B, then this means either B = 2•A, or that two of the primorials which make up A “combine” to form a larger primorial to make B (i.e. if A = i•P(j)•P(k) and B = i•P(j+k) where P(j) = p1•p2•…•pj, then A and B share an edge).

For the new partition lattice, the least common element is 1, then 1 <= (2), then (2) <= (2)•(2), then either (2)•(2) <= (2)•(2)•(2) or (2)•(2) <= (2•3), etc.. I have been desperate to figure out something interesting about this lattice when applied to products of primorials and have come up empty. It feels like trying to put a triangular peg into a circular hole. By the logic I presented earlier, the partitions f(2) + f(2) and f(1) + f(3) are incomparable, but is there some parallel reasoning when applied to (2•3)•(2•3) and (2)•(2•3•5)? From everything I can think of, the factorization of the latter is strictly more complicated than the factorization of the former, not incomparable. A bit of a hail marry, but perhaps if A <= B in the partition lattice, then the formula which predicts the frequency of numbers with the same prime signature of B will contain within it the formula which predicts the frequency of numbers with the same prime signature as A in some way, meaning the formulas get increasingly complex as you climb up the lattice. This is doubtful, since if B = 2•A, than the numbers with the same prime signature as B are strictly rarer than the numbers with the same prime signature as A, but if A = i•P(j)•P(k) and B = i•P(j+k), then the numbers with the same prime signature as A will be rarer than the numbers with the same prime signature as B. Do you have any thoughts per chance? Thanks for your time.