Harvard mathematician answers 150-year-old chess problem

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.

Given the number of significant digits, the answer is closer to stating that pi is approximately equal to 3.14

I agree, you don't have the mentality of a mathematician.

Thanks to this I know that there is exactly 1 way of positioning 0 queens on 0x0 chessboard without any two of them attacking each other.

Close enough.

- a mathematician

There is the always elided “constant coefficient”, which the press release does not even mention.

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.

I noticed the same :-)

Knowing mathematics though, it's quite likely that this result is valid only for some regimes of N, for example assymptotically

From the paper

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.

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

L(m, n) ~ A * B^{m+n} * L^{mn(1 + O(mφm))}

[1] https://tromp.github.io/go/gostate.pdf

Also wondering where the authors of the article got this formula from. Because it is not in the paper. At least not that obvious.

The paper discusses an upper bound and a lower bound, but not a singular formula.