Timeline for Generating primes via composition of polynomials
Current License: CC BY-SA 3.0
6 events
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| Dec 26, 2013 at 12:44 | comment | added | Joe Silverman | @joro We seem to be talking about different problems (and mine is the one that the OP posed). You're saying that for a fixed (but arbitrarily large) $n$, Schinzel (might) imply that there exist infinitely many $x$ such that all of $f(x),\ldots,f^n(x)$ are prime. That is correct. I'm saying that there is unlikely to be a single $x$ value such that $f^n(x)$ is prime for infinitely many $n$. Both questions are certainly interesting, but I don't see that they are that closely related. (I believe that the general feeling is that there are only finitely many Fermat primes.) | |
| Dec 26, 2013 at 7:37 | comment | added | joro | I agree that your arguments are plausible. They apply to all doubly exponential sequences and in some sense the "weird" constant A is a counterexample. As for Schinzel H. Take $f=(x-1)^2+1$. Fermat numbers are f^n(3). Take the set {f(x),f(f(x)) ... f^n(x)} for $n$ arbitrary large. Schinzel H is applicable if all polynomials are irreducible (otherwise we get infinitely many Fermat composites). So Schinzel H implies iterates prime infinitely often up to $n$ for arbitrary large $n$. | |
| Dec 25, 2013 at 13:18 | comment | added | Joe Silverman | @joro I don't see the applicability of Schinzel. Probabilisitically, if $F(x)\in\mathbb{Z}[x]$ and if there is no a priori reason why $F(n)$ should always be composite, one would expect infinitely many prime values of $|F(n)|$, since $\sum 1/\log|F(n)|$ diverges. On the other hand, $\sum 1/\log|f^n(\alpha)|$ converges. As for your second point, the constant $A$ is "weird", in the sense it has to be carefully constructed. The values of $f^n(\alpha)$ feel much more random (at least to me). | |
| Dec 25, 2013 at 9:07 | comment | added | joro | Interesting. Probabilistic grounds might be misleading in my opinion. Assuming Schinzel's hypothesis H is applicable for this (it might be), it implies arbitrary large $n$ with all iterates being prime. Another argument: there is a constant A such that floor(A^3^n) is prime for all $n \ge 1$. | |
| Dec 24, 2013 at 19:53 | comment | added | Sidney Raffer | Joe, Thank you for explaining the lay of the land. | |
| Dec 24, 2013 at 16:38 | history | answered | Joe Silverman | CC BY-SA 3.0 |