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Mar 3, 2019 at 0:23 vote accept user136217
Mar 2, 2019 at 12:58 vote accept user136217
Mar 3, 2019 at 0:23
Feb 24, 2019 at 22:04 answer added kodlu timeline score: 6
Feb 24, 2019 at 11:08 comment added user136217 Changed $p(x)$ to $f(x)$. Thank you.
Feb 24, 2019 at 11:06 history edited user136217 CC BY-SA 4.0
edited body
Feb 24, 2019 at 3:32 comment added Gerry Myerson Please edit into the body of your question this explanation that your sequence comes from iterating the polynomial. Also, please don't use $p$ for both a polynomial and a prime number in the same paragraph.
Feb 24, 2019 at 1:34 history edited user136217 CC BY-SA 4.0
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Feb 24, 2019 at 1:28 comment added user136217 Yes, $p(x)$ is a function $f: \mathbb{Z}/n \to \mathbb{Z}/n$, but we get a cycle when $i \neq j$ satisfies $p(x_i) = p(x_j)$. From that point on, the sequence repeats its cycle forever.
Feb 24, 2019 at 0:12 comment added user44191 On another note: when you say a "cycle", do you mean considering $p(x)$ as a function $f: \mathbb{Z}/n \rightarrow \mathbb{Z}/n$, and looking at the number $i$ such that there is a $j,$with $f^{(j)} = f^{(j + i)}$, the composition power of the function? It may be useful to clarify the question.
Feb 24, 2019 at 0:10 comment added user44191 One possibility: when $n$ is divisible by $2$, all of the outputs are of the form $y^2 + 1$. For all odd primes $p$, this halves the number of outputs; for $N$ composite, it halves the number of outputs for each odd prime factor.
Feb 23, 2019 at 22:57 history edited YCor
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Feb 23, 2019 at 22:35 review First posts
Feb 24, 2019 at 0:14
Feb 23, 2019 at 22:31 history asked user136217 CC BY-SA 4.0