Timeline for What are the properties of this polynomial sequence?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2017 at 12:39 | comment | added | Pietro Majer | So $[x^r]a^n\neq0$ iff $n-r$ is even and $\big({{n+r\over 2} \atop {n-r\over 2}}\big)$ is odd, which is given by Lucas theorem (sorry, bad connection). This explains why $a_{n}=x^n$ for $n=2^k-1$: the binomial coefficients then fall into a one of those large even $\nabla$-shaped regions within the Pascal triangle $\Delta$. | |
| Nov 3, 2017 at 12:13 | comment | added | მამუკა ჯიბლაძე | Yet another experimental fact: total numbers of distinct polynomials that one can obtain are 1,2,1,4,2,4,1,8,4,8,2,8,4,8,1,16,8,16,4,16,8,... which seems to be A082908 (largest power of 2 dividing some entry of the $n$th row in the Pascal triangle) | |
| Nov 3, 2017 at 12:02 | comment | added | Pietro Majer | And the coefficients of $U_n(x/2)$ are binomial coefficients. | |
| Nov 3, 2017 at 11:37 | comment | added | Pietro Majer | As to what coefficients are not zero in $a_n(x)$, note that $a_n(x)=U_n(x/2) \mod 2$, (where $U_n$ are Chebyshev polynomials of second kind), since they solve the same linear recurrence mod 2. | |
| Nov 3, 2017 at 8:40 | comment | added | მამუკა ჯიბლაძე | Another experimental fact - highest power of $x$ dividing $a_n$ is $2^k-1$, where $2^k$ is the highest power of 2 dividing $n+1$. | |
| Nov 3, 2017 at 8:30 | comment | added | მამუკა ჯიბლაძე | @OleksandrKulkov just count the number of summands in $a_n$ in the table: $a_1$ has one, $a_2$ has 2, $a_3$ has 1, $a_4$ has 3, ... | |
| Nov 3, 2017 at 8:29 | comment | added | Oleksandr Kulkov | I don't see the pattern you mentioned, could you elaborate please? | |
| Nov 3, 2017 at 8:23 | comment | added | მამუკა ჯიბლაძე | @OleksandrKulkov The numbers of nonzero coefficients go like 1,2,1,3,2,3,1,4,3,5,2,5,3,4,1,5,4,7,3,8,5,...; this seems to be A002487 | |
| Nov 3, 2017 at 8:22 | comment | added | Sergey Dovgal | @მამუკაჯიბლაძე There seems to be a generating function for the sequence (though the pattern for plus/minus is not clear yet), so maybe it's possible then to proceed by induction (once the pattern is guessed) | |
| Nov 3, 2017 at 8:15 | comment | added | მამუკა ჯიბლაძე | @SergeyDovgal Well again experimentally, each $a_n$ except for $n=2^j$ has $\varepsilon_n\pm\varepsilon_k$ with $k< n$ in at least one of the coefficients, and this would imply everything one needs. But it might be not easy to prove this because this depends on simplifications from previous $n$s, and because of these exceptions at powers of 2 which I don't see how to explain. | |
| Nov 3, 2017 at 8:11 | comment | added | Sergey Dovgal | This post looks almost like a proof. By the way, the sequences of relations between $ \varepsilon_i $ are reminiscent of some "paperfolding-style" sequences, but I don't have enough expertise to recognise or make more precise statement. | |
| Nov 3, 2017 at 8:01 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
deleted 21 characters in body
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| Nov 3, 2017 at 7:59 | comment | added | Oleksandr Kulkov | Hey, that looks like A065176 ! | |
| Nov 3, 2017 at 7:56 | comment | added | მამუკა ჯიბლაძე | @OleksandrKulkov Yes it seems so... | |
| Nov 3, 2017 at 7:55 | comment | added | Oleksandr Kulkov | It's also interesting that $a_{2^k-1}=x^{2^k-1}$ | |
| S Nov 3, 2017 at 7:48 | history | answered | მამუკა ჯიბლაძე | CC BY-SA 3.0 | |
| S Nov 3, 2017 at 7:48 | history | made wiki | Post Made Community Wiki by მამუკა ჯიბლაძე |