Timeline for Determinant of a matrix filled with elements of the Thue–Morse sequence
Current License: CC BY-SA 4.0
17 events
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| Nov 10, 2019 at 13:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 13, 2019 at 12:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 15, 2019 at 12:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| S Nov 25, 2018 at 8:01 | history | bounty ended | CommunityBot | ||
| S Nov 25, 2018 at 8:01 | history | notice removed | CommunityBot | ||
| Nov 17, 2018 at 18:19 | answer | added | Markus Sprecher | timeline score: 1 | |
| S Nov 17, 2018 at 7:00 | history | bounty started | Vladimir Reshetnikov | ||
| S Nov 17, 2018 at 7:00 | history | notice added | Vladimir Reshetnikov | Draw attention | |
| Nov 8, 2018 at 15:20 | comment | added | i9Fn | Some other variations: 1. Instead of starting with "0" in Thue_Morse start with any other number in any base. 2. Fill a $m \times n$ matrix instead and consider $\sqrt {\det(AA^T)}$ 3. Consider the product of non-zero eigenvalues instead of the determinant. | |
| Nov 8, 2018 at 15:20 | comment | added | i9Fn | Searching the OEIS gives 4 results of those A207039 - "Primes whose binary expansion is not palindromic" seems the most relevant. It would be interesting to try to prove that for composite numbers or non-palindromic binary numbers the determinant is zero. This also suggest the question: does there exists ternary (and higher) analog of the Thue–Morse sequence such that numbers for which determinant is non-zero is subsets of non-palindromic ternary primes? | |
| Nov 8, 2018 at 9:40 | comment | added | François Brunault | If $n \geq 4$ is even then the determinant is 0 because the columns $C_0, C_1,... C_{n-1} $ satisfy $C_{2k}+C_{2k+1}=(1,1,...,1) $ for every $k $. | |
| Nov 7, 2018 at 19:34 | comment | added | Gerhard Paseman | There are some orders n for which it is easy to prove the determinant is zero. When n is a large enough power of two, there are duplicated rows (first and fourth) and for n one more than a power of two, the first column is zero. You might see if a parity argument takes care of cases of large even orders, or even composite orders being zero. Of interest is when the matrices are non singular over F2. Gerhard "Use Power Of Binary Expansion" Paseman, 2018.11.07. | |
| Nov 7, 2018 at 18:19 | history | edited | Vladimir Reshetnikov | CC BY-SA 4.0 |
added 590 characters in body; edited title
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| Nov 7, 2018 at 13:34 | comment | added | François Brunault | Experimentally there seems to be a simple formula for the rank of your matrix in the case $n=2^k-1$, which corresponds to the iteration $n \to 2n+1$. Using $n \to 2n$ the rank seems to stabilize. Maybe you can try other iterations. | |
| Nov 7, 2018 at 13:31 | comment | added | François Brunault | I guess you can define Thue-Morse matrices using the pattern $\begin{pmatrix}A & B \\ B & A\end{pmatrix}$, like the Thue-Morse sequence is defined using the pattern $AB$. | |
| Nov 7, 2018 at 7:21 | comment | added | Alexey Ustinov | Maybe it is better to consider the differnce $n-{\rm rank }\, \mathcal D_n$? | |
| Nov 7, 2018 at 5:35 | history | asked | Vladimir Reshetnikov | CC BY-SA 4.0 |