Looking at the actual result, you can really only expect that (92^(1/8))/8 should be close to 0.143. It works out to about 0.22, which is less than double. The result actually says this ratio tends to 1 as n tends to infinity.
When used in that equation it's meant to implicitly depend on n as edflsafoiewq points out. o(1) stands for o(f(n)) where f(n) = 1. Hence any function g(x) in the family of functions represented by o(1) must be less than c * f(x) for every positive c and all x greater than some m. This is exactly the statement that any function in the family of functions o(1) must tend to zero.
Whenever you see big-O/little-O/theta notation there is always an implied dependent variable, even for o(1)/O(1)/Theta(1).
I think a nuance people are missing is that for big O, it is sufficient for the existence of any positive C and x, but for small o there must always exist an x for every c.
The (0.143n)^n leaves out some smaller order factors that would help to give more accurate results. The situation is very similar to the number of legal Go positions on an nxn board [1], which Theorem 6 in that paper states as
It does make me chuckle the way this is reported. We answered this problem! Ok what's the answer "well about (0.143n)^n". I've figured out the value of Pi! It's about 3. Firstly, that's not an answer that's's an approximate answer, and secondly, it's wrong for a practical values of n. I really don't have the mentality of a mathematician.
You say "practical values of n", but the answer to n = 27 is already known and I would be very surprised if the final result had any practical application. The point of these problems is rarely the numerical answer, but the development of new techniques: The problem of determining the asymptotic growth of this number is studied because it is a good playground for sharpening mathematical tools and methods.
In this sense, the asymptotic problem is an interesting one, and it was solved. Now that the main term is characterized, people can investigate sharper estimates.
The problem of determining an efficient algorithm for every n, on the other hand, is just hopeless, which means that it is not interesting for the purpose I described above. Studying "small" n (up to 40, say) is not useless, though, if it leads to algorithmic breakthroughs (in SAT solving, for example). It is, however, an "orthogonal" problem, in the sense that it requires a completely different set of techniques.
To illustrate my point: If the answer for n = 50 appeared out of nowhere, would people care about the number itself? I bet people would be much more interested in how it was done.
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[ 2.9 ms ] story [ 70.8 ms ] thread- a mathematician
Knowing mathematics though, it's quite likely that this result is valid only for some regimes of N, for example assymptotically
We show that there exists a constant α = 1.942±3×10−3 such that Q(n) = ((1 ± o(1))ne^−α)^n
Dunno what o(1) is.
For example x^2 is in O(x^2) but is not in o(x^2).
Whenever you see big-O/little-O/theta notation there is always an implied dependent variable, even for o(1)/O(1)/Theta(1).
L(m, n) ~ A * B^{m+n} * L^{mn(1 + O(mφm))}
[1] https://tromp.github.io/go/gostate.pdf
The paper discusses an upper bound and a lower bound, but not a singular formula.
[1] https://www.quantamagazine.org/mathematician-answers-chess-p...
The problem of determining an efficient algorithm for every n, on the other hand, is just hopeless, which means that it is not interesting for the purpose I described above. Studying "small" n (up to 40, say) is not useless, though, if it leads to algorithmic breakthroughs (in SAT solving, for example). It is, however, an "orthogonal" problem, in the sense that it requires a completely different set of techniques.
To illustrate my point: If the answer for n = 50 appeared out of nowhere, would people care about the number itself? I bet people would be much more interested in how it was done.
I agree, you don't have the mentality of a mathematician.
Related (N-Queens is NP-hard, 4 years ago): https://news.ycombinator.com/item?id=15168867