Timeline for Count of binary matrices that avoids a certain sub-matrix
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| S Nov 9, 2013 at 9:09 | history | suggested | Abhimanyu Pallavi Sudhir | CC BY-SA 3.0 |
OEIS Links.sccx, xcl v; scpsdcpacascds
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| Nov 9, 2013 at 8:59 | review | Suggested edits | |||
| S Nov 9, 2013 at 9:09 | |||||
| Jul 3, 2010 at 5:40 | comment | added | Gerhard Paseman | Note that there can be at most (m choose 2) columns with two or more ones, and the remaining columns must have fewer than two ones. Further if one column has j ones, then that limits the number of other columns with two or more ones to (m choose 2) minus (j choose 2). So the answer will be a sum of powers of (m+1). If n > (m choose 2), then the answer will be a multiple of a power of (m+1). Gerhard "Ask Me About System Design" Paseman, 2010.07.02 | |
| Jul 3, 2010 at 4:01 | history | answered | Douglas S. Stones | CC BY-SA 2.5 |