8
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In Uniform Approach to Double Shuffle and Duality Relations of Various q-Analogs of Multiple Zeta Values via Rota-Baxter Algebras, Proposition 10.8, Jianqiang Zhao mentiones the sequence: $$a_n=\sum_{1\leq d,k\leq n}\sum_{\substack{x_1+\dots+x_d=d+k-1\\1\leq x_1,\dots,x_d}}x_1 x_2\cdots x_d. $$ (Actually under the first sum it reads $1\leq d\leq k\leq n$, but this is probably a typo, since in Table 8 on the same page, the sequence starts with $1,8,49,\dots$, which is what you get when you sum over ${1\leq d,k\leq n}$; see also the oeis-entry: https://oeis.org/A261668.)

Numerically, from the first few values it seems like the asymptotic of this sequence might be the following.

Edit: in a previous version, as Richard Stanley pointed out, I gave the asymptotic expansion for $a_{n+1}$. See his comments for a simple formula and generating function for $a_n$. Here the corrected guess for the asymptotics for $a_n$.

$$a_n= \frac{\sqrt{3} \left(\frac{27}{4}\right)^{n}}{6 \, \sqrt{\pi} \sqrt{n}} \left(\frac{3}{10} -\frac{223}{6000 \, n} + \frac{26689}{21600000 \, n^{2}} + \frac{31824349}{23328000000 \, n^{3}} + \frac{28917508109}{167961600000000 \, n^{4}} - \frac{84601010659543}{302330880000000000 \, n^{5}} - \frac{473846767572852311}{3265173504000000000000 \, n^{6}} + \frac{479343612166541515679}{5877312307200000000000000 \, n^{7}} + \frac{1396880611451268708008387}{16926659444736000000000000000 \, n^{8}} - \frac{4768678908408186461643400733}{274211883004723200000000000000000 \, n^{9}} \pm O\left(\frac{1}{n^{10}}\right)\right)$$

Has it been proved somewhere, that this is the correct asymptotic?

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  • $\begingroup$ I edited formula following the link OEIS, please check $\endgroup$ – Fedor Petrov Sep 13 '15 at 19:20
  • $\begingroup$ Maybe, it is better to remove $k$ in RHS and replace condition to $x_1+\dots+x_d\leq n+d-1$? $\endgroup$ – Fedor Petrov Sep 13 '15 at 19:23
  • $\begingroup$ @FedorPetrov Maybe, but I want to stay somewhat close to the notation in Zhao's paper. $\endgroup$ – Moritz Firsching Sep 13 '15 at 19:25
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    $\begingroup$ It is routine to show that $a_n=\sum_{d=1}^n{2d+n-1\choose n-1}$. Moreover, $a_n+1$ is the coefficient of $x^{2n}$ in $1/(1-x)^{n+1}(1+x)$. It shouldn't be difficult to get the asymptotics from this. $\endgroup$ – Richard Stanley Sep 13 '15 at 20:02
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    $\begingroup$ Notice that it also equals oeis.org/A225006 minus 1, which contains an empirical asymptotics based on recurrent relation (which now can be proved) and provides yet another interpretation for these quantities. $\endgroup$ – Max Alekseyev Sep 14 '15 at 21:17

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