Timeline for Linear recurrences in coefficients of powers of quotients of polynomial rings
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 24 at 14:43 | comment | added | joro |
@PeterTaylor I noticed problems with my experimental data, but I still get experimental support. For k=m=2 and degree f_i=2 I got these recurrences, starting from x ZERO and recurrence order up to 10: f_i: [x0^2 - x1^2 - 4*x1 - 1, -x0^2 + 2*x1^2], [x0^2 - 2*x0 + x1^2 - x1 + 5, -x0*x1 + x1^2 + 9*x1], [x0^2 - 3*x0 + x1^2 - x1 - 3, -x0^2 + x0*x1 + 2*x0 + x1^2 + 2*x1], [-x1^2 + x1, -x0^2 - 4*x0*x1 + 5*x0 - x1^2 - x1], [x0^2 + x0 + 4*x1^2 + 5, -2*x0*x1 + x1^2 + 6*x1]
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| Jan 24 at 12:02 | comment | added | Peter Taylor | I'm curious about your experimental data. I'm experimenting with $k=m=2$, $f_i$ up to cubic terms, coefficients from $\{-1,0,1\}$ except fixed constant coefficients $-5$ and $-13$ (as a compromise because working over $\mathbb{Q}[a,b]$ and using constant coefficients $-a$ and $-b$ ran into memory trouble), and in hundreds of thousands of tests I haven't yet found a case which doesn't appear to have a linear recurrence of order at most $9$. | |
| Jan 23 at 12:57 | comment | added | joro | @PeterTaylor Thanks, very good point. Edited with lexicographic term order, allowing as an option another order. | |
| Jan 23 at 12:55 | history | edited | joro | CC BY-SA 4.0 |
defined the term order
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| Jan 23 at 12:29 | comment | added | Peter Taylor | Don't you need to specify a term ordering for Gröbner base calculations in order to define $a(n)$? Otherwise if e.g. $f_1 = x_1 - x_2 - 3$ it's arbitrary whether $t(1) = x_1 + x_2 = 2x_1 - 3 = 2x_2 + 3$ yields $a(1) = 0$ or $a(1) = 2$. | |
| Jan 23 at 8:40 | history | asked | joro | CC BY-SA 4.0 |