The logic behind a sequence

Question

I am trying to get the logic behind the sequence: for $n=2,3,\ldots$ $$\left(\frac{\log (2)}{\log \left(\frac{3}{2}\right)},\frac{\log (3)}{\log \left(\frac{17}{9}\right)},\frac{\log (4)}{\log \left(\frac{71}{32}\right)},\frac{\log (5)}{\log \left(\frac{1569}{625}\right)},\frac{\log (6)}{\log \left(\frac{899}{324}\right)},\frac{\log (7)}{\log \left(\frac{355081}{117649}\right)},\frac{\log (8)}{\log \left(\frac{425331}{131072}\right)},\frac{\log (9)}{\log \left(\frac{16541017}{4782969}\right)},\frac{\log (10)}{\log \left(\frac{5719087}{1562500}\right)},\frac{\log (11)}{\log \left(\frac{99920609601}{25937424601}\right)},\frac{\log (12)}{\log \left(\frac{144619817}{35831808}\right)},\ldots\right)$$ for $n=30$ it is $$\frac{\log (30)}{\log \left(\frac{53774416559964522337191179}{16}\right)-16 \log (3)-23 \log (5)}$$ $\textbf{Background}$: This is part of a project to figure out how slowly the errors of power law distributed sums obey the law of large numbers. So it may not be necessary to find the logic of the sequence above, but these correspond to the exponent $\left\{{\alpha:\frac{MD(n)}{MD(1)}=\left(\frac{1}{n}\right)^{1-\frac{1}{\alpha }}}\right\}$,where $MD(n)$ is the mean absolute deviation of an n-summed Student T distributed variable with tail exponent/degrees of freedom equal 3 (and a mean of $0$). Numerically we get $$\{1.70951,1.72741,1.73951,1.74855,1.7557,1.76158,1.76655,1.77084,1.7746,1.77795,1.78095,1.78367,1.78615,1.78843,1.79054,1.79249,1.79432,1.79602,1.79762,1.79913,1.80056,1.80191,1.80319,1.80441,1.80557,1.80669,1.80775,1.80877,1.80974\},$$ a slow convergence to 2, which is the case with a Normal Distribution.

Answer 1

It looks like we have $$I(n)=\frac{\log (n)}{\log \left(\frac{f(n)}{z(n)}\right)}$$ where $f(n)= \sum _k^n \frac{n! n^{-k+n-1}}{(n-k)!}$ and $z(n)=n^{n-1}$ (thanks to Mike Lawler via https://oeis.org/search?q=355081&language=english&go=Search)

Answer 2

Let me call $b(n)$ the $n$-th term of your sequence.

It is easy(?) to see that $$ b(n) = \frac{\log(n)}{\log a(n)-(n-1)\log(n)} $$ where $$ a(n)=\sum_{k=1}^n\frac{n!\cdot n^{n-k-1}}{(n-k)!}=\frac{1}{n}\left[n^n+\sum_{k=1}^{n-1}{n\choose k}(n-k)^{n-k}k^k\right]=\frac{e^n\cdot \Gamma(n+1,n)}{n}-n^{n-1}\,, $$ where $$\Gamma(a,z)=\int_z^\infty t^{a-1}e^{-t}\,dt$$ is the incomplete Gamma function.

Actually $a(n)$ is the sequence A001865 (Number of connected functions on n labeled nodes) in the OEIS, so if you follow the link, you will find more details, asymptotics and other combinatorial interpretations for $a(n)$. Hope this helps.