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Timeline for Why are these sets divisible by n?

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Mar 24, 2021 at 21:07 answer added Joe Silverman timeline score: 2
Mar 24, 2021 at 16:51 answer added Asvin timeline score: 1
Mar 24, 2021 at 14:07 comment added Vlad Matei I think this conjecture is due to Odoni. See here matwbn.icm.edu.pl/ksiazki/aa/aa93/aa9317.pdf for a general discussion of irreducibility of iterates of quadratic polynomials
Mar 24, 2021 at 5:33 history edited Asvin CC BY-SA 4.0
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Mar 24, 2021 at 5:31 history undeleted Asvin
Mar 24, 2021 at 5:31 history deleted Asvin via Vote
Mar 24, 2021 at 2:29 comment added Asvin Sorry, $g_n(c) = f^{(n)}(c,0)$, I shouldn't have set that equal to 0.
Mar 24, 2021 at 2:28 history edited Asvin CC BY-SA 4.0
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Mar 24, 2021 at 2:25 comment added LSpice I'm still confused: $g_n(c) = f^{(n)}(0, c) = 0$ seems to me to say that $g_n(c) = 0$. I guess it's some implicit definition: $g_n(c)$ is the value such that $f^{(n)}(0, c) = 0$? But the value of what? Anyway, maybe it's just me, and someone more familiar with these sort of iterated systems will understand the notation immediately.
Mar 24, 2021 at 2:16 comment added Asvin I made a minor edit but what's going is that $f^{(n)}(z,c)$ is a 2 variable polynomial. We set $z = 0$ and $f^{(n)}(0,c) = 0$ to define $g_n(c)$ as a 1 variable polynomial. I hope that helps - the definition is definitely confusing at first sight.
Mar 24, 2021 at 2:15 history edited Asvin CC BY-SA 4.0
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Mar 24, 2021 at 2:13 comment added LSpice I hope it's just me, but I can't understand the definition of $g_n$. $f$ starts off as a 2-variable polynomial, then becomes a 1-variable polynomial, and then we define $g_n(c)$ by a formula that doesn't seem to involve $c$ at all.
Mar 24, 2021 at 2:11 history edited LSpice CC BY-SA 4.0
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Mar 24, 2021 at 2:08 history asked Asvin CC BY-SA 4.0