@Martin.s It's very confusing what you're looking for if not a closed form for the integral.

Unless I'm missing something (likely!), I feel like symmetric monoidal categories don't really apply here because there's no guarantee of associativity, so it seems like $\mathsf{Earth}\otimes\mathsf{Wind}\otimes\mathsf{Water}$, for instance, isn't well-defined because you have no guarantee that there's an isomorphism between $(\mathsf{Earth}\otimes\mathsf{Wind})\otimes\mathsf{Water}$ and $\mathsf{Earth}\otimes(\mathsf{Wind}\otimes\mathsf{Water})$.

The problem is very boring if you don't allow pairs to share an index! Then there's only one permutation that can be generated, the product of all the individual two-cycles in your list.

When you say that the indices are not necessarily distinct, do you mean that the sequence $\{[i_1, j_1], [i_2, j_2], \ldots\}$ of available swaps (changing your notion slightly) is potentially a multiset rather than a set?

Such a set 'should' have density proportional to $1/\left(\log(x)\log\log(x)\right)$, to give some sense of what sorts of things to look for.

This should succumb pretty easily to generating functions; you can get the $2s(2s+1)L(s-1)$ term by doing the usual differentiation (see e.g. section 2.4 of math.cmu.edu/~ploh/docs/math/2011-228/… ). This'll give you a second-order ODE for $\mathcal{L}(x)=\sum_nL(n)x^n$ that you may have more luck finding.

Your 'naive' approach is a little too naive; a better way to think of it would be to look at $\int_2^n \frac1{n-x}\frac{\mathrm{d}x}{\log x}$, the appropriately weighted integral of the term in the sum. Note that your approach would also suggest that $\sum_{p\leq n}\frac1p\approx 1$, whereas the actual value is $\Theta(\log\log n)$.

@FredHucht You might want to consider converting that into an answer (possibly summarizing the most important bits from the OEIS page); it seems to pretty much cover what OP is looking for.

@virtuolie Isn't the relationship you described pretty straightforward? The natural map from permutations of size $n$ with $k$ fixed points to derangements of size $(n-k)$ (just remove all the fixed elements and renumber) maintains the inversion count, it's clearly an $(n$ choose $k)$ to $1$ mapping, and one would certainly expect the average number of inversions in a derangement to be an increasing function of its size.

One starting point might be to take a group with center of order divisible by 3 — the first example that comes to mind is $Q_8\times C_3$ — and try the two different automorphisms on the $C_3$ portion of the center.

This will (hopefully!) get buried in the much more directly pertinent comments, but I just wanted to add a tiny note that I really like the gender edits here.

It would be nice to have a direct definition of the 'globe' permutation groups. As it is, the link in your question points to another MO question of yours, which points to a different Math.SE question of yours, which still doesn't actually give a formal definition of the group; to me, at least, it comes across a bit as self-advertisement.

@SakshamSethi One caution: you should not assume that your work at this point will lead to a paper. You are still laying foundations for yourself; writing a paper is an excellent goal to work towards but from where you're at IMHO you should be more focused on the work towards it than the paper itself.

Did you do any searching? Do you have a lower bound on $K$? (Note that $K$ has to have at least 8 composite numbers preceding it, so $K=122$ is an immediate lower bound, and there are only a handful of sufficiently large gaps for the next several hundred numbers, so it wouldn't be surprising if the minimal $K$ is fairly large.)

@CommandMaster It can, but that's actually baked into the $O(B^2)$ estimate — you're doing operations on numbers of shrinking length. Note that this also means that the necessary $B$ can be $O(r)$ or so, too — in other words, you need an approximation to $f$ of closeness roughly $\exp(-r)$ to be able to find $r$ with any confidence.

It's not too hard to show that you can only have one value of $n$ that makes $r^2$ an (exact) integer, incidentally; if $r=n+f$ with $r^2=q$, say, then $\left((m+n)+f\right)^2$ $= \left(m+(n+f)\right)^2$ $= m^2+(n+f)^2+2m(n+f)$ $=m^2+q+2m(n+f)$ and clearly the last term is irrational, since $f$ is and $m, n, q$ are all integers. But I'm not even sure if good lower bounds are known for $\min_{m,n\leq N}\left(\left|\sqrt{n}-\sqrt{m}\right|\bmod 1\right)$ as a function of $N$.

Without a bound on $r$, the answer is no; square roots are dense mod 1, so however close an approximation to $f$ you have, there are infinitely many square roots of whole numbers whose fractional part is that close to $f$.