Timeline for Are there infinitely many integer-valued polynomials dominated by $1.9^n$ on all of $\mathbb{N}$?

4 events

when toggle format	what		by	license	comment
Aug 12, 2013 at 18:35	comment	added	Vesselin Dimitrov		(To correct my first comment above: I meant to say that for each $C > e$ there are uncountably many such mappings $f : \mathbb{N} \to \mathbb{Z}$ --- not for $C = e$.)
Aug 12, 2013 at 18:23	comment	added	Vesselin Dimitrov		Thus, you have shown that for each $n_0 < \infty$, there are only finitely many $f \in \mathbb{Z}[x]$ with $\|f(n)\| < n!/2$ for all $n \geq n_0$. This is close to optimal upon considering $k!\binom{n}{k}$, although (especially in view of the proof) it could still be an interesting problem to see whether the coefficient $1/2$ of $n!$ is optimal, or if it could be improved to any constant smaller than $1$.
Aug 12, 2013 at 18:19	comment	added	Vesselin Dimitrov		Very neat. I had failed to fully exploit the polynomial assumption. My argument for $C < e$ was otherwise similar; it amounted to noting that $f(k)-g(k)$ was divisible by the product of the primes less than $k$. The latter argument works for arbitrary mappings $f : \mathbb{N} \to \mathbb{Z}$ with the property $f(n+p) \equiv f(n) \mod{p}$ for all $n \in \mathbb{N}$ and all primes $p$: for $C < e$ there are only finitely many such mappings (conjectured to be polynomials anyway), while for $C = e$ there are uncountably many.
Aug 12, 2013 at 17:28	history	answered	David E Speyer	CC BY-SA 3.0