Timeline for Non-trivial alternating sums of binomial coefficients
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2019 at 15:00 | answer | added | Richard Stanley | timeline score: 2 | |
| Dec 16, 2019 at 11:47 | vote | accept | Michal | ||
| Dec 16, 2019 at 7:58 | answer | added | Tom De Medts | timeline score: 1 | |
| Dec 16, 2019 at 7:20 | answer | added | Brendan McKay | timeline score: 3 | |
| Dec 16, 2019 at 5:50 | comment | added | Noam D. Elkies | The number of solutions for $n=1,2,3,\ldots,25$ is $$ 2,2,4,2,8,2,16,6,32,2,64,2,144,14,256,2,512,2,1024,6,2048,2,4096,50,8192. $$ As the OP notes, it is easy to construct $2$ solutions for $n$ even, and $2^k$ solutions for $n=2k-1$; but note that there are $16$ more for $n=13$. The four extra solutions for $n=8$ are obtained from $001001100$ (the identity ${8 \choose 2} - {8 \choose 5} + {8 \choose 6} = 0$ noted by LeechLattice) by reflection and 1's complement; the twelve for $n=14$ are the orbits of $000001001110000, 000001100110000, 000010000110000$. | |
| Dec 15, 2019 at 15:54 | answer | added | Claude Chaunier | timeline score: -1 | |
| Dec 15, 2019 at 14:52 | comment | added | Michal | @TomDeMedts Yes, you are totally right. So my question trivially reduces to an already addressed one. I guess I cannot mark a comment as a solution, but I would if you only post it as an answer. | |
| Dec 14, 2019 at 5:41 | comment | added | Ira Gessel | @Gerry you're right—I deleted my comment. | |
| Dec 13, 2019 at 23:03 | comment | added | Gerry Myerson | @Ira, the question asks for a binary vector, which I think means a vector whose entries are just zeros & ones. | |
| S Dec 13, 2019 at 16:33 | history | suggested | jeq | CC BY-SA 4.0 |
Fixed link, which had been going to the edit page.
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| Dec 13, 2019 at 14:24 | review | Suggested edits | |||
| S Dec 13, 2019 at 16:33 | |||||
| Dec 13, 2019 at 13:44 | answer | added | LeechLattice | timeline score: 2 | |
| Dec 13, 2019 at 13:39 | comment | added | Tom De Medts | I'm not sure I understand your last comment. If you have a solution with $a_i \in \{ -1, 1\}$, then you also have a solution with $a_i \in \{ 0, 1\}$ simply by replacing each $a_i$ with $(a_i + 1)/2$ (using the fact that setting all $a_i = 1$ is also a valid solution). Am I misunderstanding something? | |
| Dec 13, 2019 at 13:25 | review | First posts | |||
| Dec 13, 2019 at 14:02 | |||||
| Dec 13, 2019 at 13:24 | history | asked | Michal | CC BY-SA 4.0 |