Timeline for Asymptotics of A261668
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 15, 2015 at 0:14 | comment | added | Max Alekseyev | @MoritzFirsching: Done. | |
| Sep 14, 2015 at 21:46 | comment | added | Moritz Firsching | @MaxAlekseyev How nice! Thanks for editing the oeis-entry! Maybe we should the formula with the sum of binomial coefficients as well. | |
| Sep 14, 2015 at 21:17 | comment | added | Max Alekseyev | 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. | |
| Sep 14, 2015 at 19:48 | history | edited | Moritz Firsching | CC BY-SA 3.0 |
corrected formula, implemented comments by Richard Stanley
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| Sep 14, 2015 at 19:39 | comment | added | Moritz Firsching | @RichardStanley You are right, I accidently stated the asymptotic expansion for $a_{n+1}$. I will edit the post. | |
| Sep 14, 2015 at 19:29 | comment | added | Richard Stanley | Note that $\frac{81}{40}=\frac{3}{10}\cdot \frac{27}{4}$. Perhaps the stated asymptotic expansion is for $a_{n+1}$. | |
| Sep 14, 2015 at 19:20 | comment | added | Richard Stanley | The largest term in $\sum_{d=1}^n {2d+n-1\choose n-1}$ is ${3n-1\choose n-1}$. It is asymptotic to $\frac 16\frac{\sqrt{3}}{\sqrt{\pi n}}\left(\frac{27}{4}\right)^n$. If we multiply the constant $1/6$ by $9/5$ we get $3/10$, not $81/40$. Numerically, $3/10$ seems to be the correct constant, confirming Brendan's comment. | |
| Sep 14, 2015 at 11:40 | comment | added | Moritz Firsching | @BrendanMcKay I'm not sure I understand correctly what you mean. Do you suggest a simple way to get the leading coefficient $C$ of the asymtotic, such that $a_n\sim C^n$? As stated I expect that $C=\frac{27}{4}$ and not $\frac{9}{4}$ (or $\frac{9}{5}$). | |
| Sep 14, 2015 at 9:46 | comment | added | Brendan McKay | The tail end of Richard's sum is close to geometric with ratio 9/4, so the asymptotic value is 9/5 times the last term. | |
| Sep 14, 2015 at 8:06 | comment | added | Moritz Firsching | @RichardStanley Thanks, that's definitely helpful! | |
| Sep 13, 2015 at 20:02 | comment | added | Richard Stanley | 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. | |
| Sep 13, 2015 at 19:25 | comment | added | Moritz Firsching | @FedorPetrov Maybe, but I want to stay somewhat close to the notation in Zhao's paper. | |
| Sep 13, 2015 at 19:23 | comment | added | Fedor Petrov | Maybe, it is better to remove $k$ in RHS and replace condition to $x_1+\dots+x_d\leq n+d-1$? | |
| Sep 13, 2015 at 19:20 | comment | added | Fedor Petrov | I edited formula following the link OEIS, please check | |
| Sep 13, 2015 at 19:20 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
Corrected RHS
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| Sep 13, 2015 at 19:16 | history | asked | Moritz Firsching | CC BY-SA 3.0 |