Timeline for $(n-2)$-blocking sets in $AG(n,2)$

8 events

when toggle format	what		by	license	comment
Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 6, 2016 at 20:57	vote	accept	Ashot
Feb 1, 2016 at 4:23	history	edited	Anurag	CC BY-SA 3.0	added 161 characters in body
Jan 30, 2016 at 22:18	history	edited	Anurag	CC BY-SA 3.0	added 369 characters in body
Jan 26, 2016 at 22:41	history	edited	Anurag	CC BY-SA 3.0	made a correction in the Gaussian coefficient
Jan 26, 2016 at 18:33	comment	added	Anurag		You are welcome. But if we use the method that you have described for a general $q$, then shouldn't the bound be $(q^2 - 1)^{\log_2^n} = n^{\log_2^{q^2 - 1}}$, when $n$ is a power of $2$? This is because in your base step you will have $F(2) = q^2 - 1$. And then this upper bound will be worse than the quadratic upper bound we have. For example, for $q = 3$ it gives an upper bound of $O(n^3)$, while the bounds I have described are $O(n^2)$.
Jan 26, 2016 at 18:01	comment	added	Ashot		Thank you for the answer. The same method works for $q>2$ as well. Just the constant factor is changes: $q^2-1$ instead of $3$.
Jan 26, 2016 at 13:08	history	answered	Anurag	CC BY-SA 3.0