Timeline for Bound on $g(n+1)/g(n)$ for Landau's function
Current License: CC BY-SA 3.0
8 events
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| Feb 9, 2019 at 21:19 | comment | added | Gagar | @AnthonyQuas Actually $g(n)$ is even except for $n=3, 8, 15$. This is proved e.g. in Nicolas, Jean-Louis Ordre maximal d'un élément du groupe Sn des permutations et "highly composite numbers''. (French) Bull. Soc. Math. France 97 1969 129–191 (available online, numdam.org/article/BSMF_1969__97__129_0.pdf p142). The fact that $g(n+1) \leq 2 g(n)$ is the Corollary p143. | |
| Jul 1, 2015 at 19:54 | comment | added | Gerhard Paseman | Nice and sharp. Once you find out how big a power of two is needed, say $c$ many, you can later establish something like $g(n) \geq g(n + 2^{c-1})/2$. Gerhard "Perhaps Already In The Literature" Paseman, 2015.07.01 | |
| Jul 1, 2015 at 19:49 | comment | added | Anthony Quas | What I had in mind is that $g(n)$ is the largest value of $p_1^{\alpha_1}\times\dots \times p_j^{\alpha_j}$ such that $p_1^{\alpha_1}+\ldots+p_j^{\alpha_j}\le n$. For all values of $n$ that are at least 4, this is obtained with one of the $p_j$'s being equal to 2. Now if $g(n+1)=2^{\alpha_1}\times\dots \times p_j^{\alpha_j}$ with $2^{\alpha_1}+\ldots+p_j^{\alpha_j}\le n+1$, then $2^{\alpha_1-1}+p_2^{\alpha_2}+\ldots+p_j^{\alpha_j}\le n$ and $g(n)\ge 2^{\alpha_1-1}\times p_2^{\alpha_2}\times\dots\times p_j^{\alpha_j}=g(n+1)/2$. | |
| Jul 1, 2015 at 19:41 | comment | added | J Fabian Meier | Can you sketch me the argument? Thanks. | |
| Jul 1, 2015 at 14:39 | comment | added | Anthony Quas | How about $g(n+1)\le 2g(n)$ for all $n$. | |
| Jul 1, 2015 at 13:10 | history | edited | GH from MO |
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| Jul 1, 2015 at 12:37 | history | edited | J Fabian Meier | CC BY-SA 3.0 |
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| Jul 1, 2015 at 12:27 | history | asked | J Fabian Meier | CC BY-SA 3.0 |