Timeline for What are the properties of this polynomial sequence?
Current License: CC BY-SA 3.0
19 events
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| Nov 6, 2017 at 9:01 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 5, 2017 at 21:50 | comment | added | Pietro Majer | Yes, thank you! It was a typo. There was a subtle anomaly too; now I think it is OK. | |
| Nov 5, 2017 at 21:45 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 5, 2017 at 9:58 | comment | added | Wolfgang | I think it must be $a_{2^n+j}=x^{2^n}b_j+\epsilon_{2^n} \delta_j a_{2^n-\color{red}{2}-j}$ in your recursion formula. | |
| Nov 5, 2017 at 9:10 | comment | added | Pietro Majer | Good point. And I think that if $xa_{m-1}$ has a non-zero term of degree less than $m$ at the same location of a zero term of $a_{m-2}$, then they must also have some non-zero term at the same location (somewhere at a higher degree). Therefore, if they have no non-zero term at the same location, all terms of degree $<m$ in $xa_{m-1}$ must vanish, that is $xa_{m-1}=x^m$ and as we know $m$ is a power of $2.$ | |
| Nov 5, 2017 at 0:23 | comment | added | Terry Tao | If $x a_{m-1}$ and $a_{m-2}$ have a non-zero term at the same location then the sign $\pm$ in $a_m = x a_{m-1} + a_{m-2}$ is forced. Presumably one can show using Lucas's theorem that this situation occurs for all $m$ that are not powers of two. Then it should be possible to show by induction that Pietro's recipe above completely parameterises all solutions, which presumably finishes off the problem. | |
| Nov 4, 2017 at 17:28 | comment | added | Pietro Majer | I'd say half mystery: certainly the free choice of $\epsilon_{m}$ is allowed by $a_{m-1}=x^{m-1}$, and this happens exactly for $m=2^k$. This is geometrically clear in terms of the Pascal triangle mod 2 (like en.wikipedia.org/wiki/Sierpinski_triangle#/media/…). But it remains mysterious why there is no free choice if $a_{m-1}\neq x^{m-1}$.. | |
| Nov 4, 2017 at 16:22 | comment | added | მამუკა ჯიბლაძე | Still it is complete mystery for me why the constraints completely free themselves up at each power of two precisely... | |
| Nov 4, 2017 at 16:02 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 4, 2017 at 15:39 | comment | added | Pietro Majer | No, sorry if it wasn't clear: $b_j$ is itself a polynomial of degree $j$, namely $a_j$ but with the sing of $\epsilon_{2^{n-1}}$ inverted. (I added the the $x$ in the notation for clarity) | |
| Nov 4, 2017 at 15:36 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 4, 2017 at 15:17 | comment | added | მამუკა ჯიბლაძე | Is there a typo in the definiton of $a_{2^n+j}$? Because the leading term must be $x^{2^n+j}$ while as it stands, degree seems to be $2^n$ only... | |
| Nov 4, 2017 at 15:00 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 4, 2017 at 14:54 | history | undeleted | Pietro Majer | ||
| Nov 4, 2017 at 14:54 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 4, 2017 at 0:10 | history | deleted | Pietro Majer | via Vote | |
| Nov 3, 2017 at 23:02 | history | undeleted | Pietro Majer | ||
| Nov 3, 2017 at 20:25 | history | deleted | Pietro Majer | via Vote | |
| Nov 3, 2017 at 16:46 | history | answered | Pietro Majer | CC BY-SA 3.0 |
