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Aug 6, 2016 at 20:27 comment added Sávio So, the sequences with at least 3 classes $0$, one class $7$ and one class $3$ satisfy what I want. I'd like to show that for every sequence (with $2n-1$ elements such that all classes $\pmod p$ are represented by at most $n-1$ elements) we can take a subsequence $b_1,\dots,b_n$ and valuate it on $c_1s^{n-1} + \dots + c_{n-1}s + c_n$ (where $\{b_1,\dots,b_n\} = \{c_1,\dots,c_n\}$) returning $n+1$ distinct elements (by permuting coefficients). (I now it's a bit confusing...)
Aug 6, 2016 at 20:14 comment added Sávio Now, I would like to show that these kind of evaluations reaches at least 6 classes $\pmod{11}$. For example, in the sequence (1) above one may take $\{b_1,\dots,b_5\} = \{0,0,0,0,7\}$ and it'll just generate 5 elements $\pmod {11}$, namely $7s^4, 7s^3, 7s^2, 7s, 7$. But we instead can take $\{b_1,\dots,b_5\} = \{0,0,0,7,3\}$ and it generates $3s^4 + 7s^3 \equiv 6, 3s^3 + 7s^2 \equiv 7, 3s^2 + 7s \equiv 10, 3s + 7 \equiv 8, 3 + 7s^4 \equiv 2 \pmod{11}$ (5 distinct elements). But it also generates $7s + 3 \equiv 9 \pmod{11}$, so we generate at least 6 distinct elements $\pmod{11}$. (see next..)
Aug 6, 2016 at 19:58 comment added Sávio Sure! For example, take $n=5$, $p=11$ and $s = 4$. Suppose that we have a sequence $a_1, \dots, a_9 \in \mathbb Z_{11}$ such that each class $\pmod {11}$ is represented by at most 4 elements. Some examples of this sequences are: 1) $0,0,0,0,7,7,7,7,3$ 2) $0,1,2,3,4,5,6,7,8$ 3) $1,3,5,7,8,8,9,9,9$ A valid operation consists on taking any 5 elements $b_1,\dots,b_5$ in one of these kind of sequences and valuate it on $c_1s^4 + c_2s^3 + c_3s^2 + c_4s + c_5$, where $\{b_1,\dots,b_5\} = \{c_1,\dots,c_5\} \pmod {11}$ (we are free to choose the order of the coefficients). (next comment)
Jul 31, 2016 at 2:23 comment added WhatsUp can you please clarify what is a "valid operation"? some examples will be welcome.
Jul 26, 2016 at 17:09 review First posts
Jul 26, 2016 at 17:47
Jul 26, 2016 at 17:06 history asked Sávio CC BY-SA 3.0