Timeline for Prefix sums of Pascal triangle = powers of two
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2017 at 15:46 | comment | added | Aaron Meyerowitz | @PerAlexandersson Define the distance between two binary strings of length $n$ to be the number of positions in which they disagree. Then $S(n,m)$ is the volume/size of a radius $m$ sphere. Essentially this is just “choose $m$ or less out of $n$” but then $2^n/S(n,m)$ is an upper bound on the size of a set of points with minimum distance $2m+1$ so it is intriguing to have that be an integer. | |
| Oct 21, 2017 at 7:23 | comment | added | Per Alexandersson | Is there a natural combinatorial problem that gives $S_{m,n}$ as answer? I was thinking along the line of counting regions again.. | |
| Oct 21, 2017 at 2:40 | answer | added | Aaron Meyerowitz | timeline score: 6 | |
| Oct 20, 2017 at 23:56 | comment | added | Gerhard Paseman | Of course 2m either has to be 2n or less than n, and I suspect 2m has to be less than n -sqrt(n). Do you have any sense of any other solutions, say when m=2 or m=3? Gerhard "Hasn't Tried Computing This Yet" Paseman, 2017.10.20. | |
| Oct 20, 2017 at 19:32 | history | asked | Mikhail Tikhomirov | CC BY-SA 3.0 |