TITLE: The least number of quadratic polynomials needed to cover $[1,N]$
QUESTION [6 upvotes]: Let $r = r(N)$ denote the least number of quadratic polynomials $f_i \in \mathbb{Z}[x]$, $1 \leq i \leq r$, each with non-negative coefficients, such that for every integer $n \in [1,N]$, there exists $1 \leq i \leq r$ such that $n = f_i(m)$ for some integer $m \in \mathbb{N}$. Clearly, $r(N) \rightarrow \infty$, since for each quadratic polynomial $f$ with non-negative coefficient the set of integral values in $[1,N]$ that can be represented by $f$ is $O(N^{1/2})$, so for any number $c$, the set of values that can be covered by $f_1, \cdots, f_c$ is also $O(N^{1/2})$. This also shows that $r(N) \gg N^{1/2}$.
In fact, $r(N) = O(N^{1/2})$ as well, since the polynomials $f_1(x) = x^2, f_2(x) = x^2 + 1, \cdots, f_r(x) = x^2 + r$ with $r = N - \lfloor N^{1/2} \rfloor^2$ will cover cover $[1,N]$.
My question is what is the smallest positive value $c$ such that $r(N) \geq cN^{1/2}$ for all $N$ sufficiently large?
Thanks for any input.

REPLY [2 votes]: That's not an answer too, just a result of a quick and naive computation: For $k\ge1$ let $n_k$ be the largest integer such that $r(n_k)=k$. Of course then $r(m)=k$ for all $n_{k-1}+1\le m\le n_k$. We have $n_1, n_2, \ldots = 1, 2, 6, 9, 13, 19, 23, 32, 32, 37, 45, 54, 61, 72, 82, 91, 101, 114, 128, 137\ldots$. And no, OEIS does not know this sequence ...