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I seek for very sparse representations of positive integers. Let $$\mathbb N=\{0,1,2,\ldots\}\ \ \ \text{and}\ \ \ \mathbb Z^+=\{1,2,3,\ldots\}.$$ Recall that a polynomial $P(x,y)$ is integer-valued if $P(x,y)\in\mathbb Z$ for all $x,y\in\mathbb Z$.

Question 1. Is there an integer-valued quadratic polynomial $P(x,y)$ such that $\{P(x,y)+2^k:\ x,y\in\mathbb Z\ \text{and}\ k\in\mathbb N\}=\mathbb Z^+$?

I note that $55832434$ is the first positive integer not represented by $x^2+y(3y+1)/2+2^k$ $(x,y\in\mathbb Z\ \text{and}\ k\in\mathbb N)$.

Question 2. Whether each positive integer $n$ can be written as $$\frac{x(3x+1)}2+\frac{y(7y+1)}2+2^k$$ with $x,y\in\mathbb Z$ and $k\in\mathbb N$?

This is positive for each $n=1,\ldots,2\times10^7$, see http://oeis.org/A343503. I ever proved that any $n\in\mathbb N$ can be written as $x^2+y(3y+1)/2+z(7z+1)/2$ with $x,y,z\in\mathbb Z$, and conjectured that $$\left\{x^4+\frac{y(3y+1)}2+\frac{z(7z+1)}2:\ x,y,z\in\mathbb Z\right\}=\mathbb N.$$

Your comments are welcome! In particular, Question 2 needs further check if you are good at computation.

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    $\begingroup$ The probability a number $n$ is represented by a quadratic polynomial should be proportional to $1/\sqrt{\log n}$, so the number of $k$ so that $n -2^k$ is represented should be proportional to $\sqrt{\log n}$, which if the statistics are Poisson-like would mean the probability $n$ cannot be represented is $e^{ - O( \sqrt{ \log n})}$, whose sum is infinte, suggesting that many numbers can't be represented. But maybe statistics are not Poisson-like... $\endgroup$
    – Will Sawin
    Commented Apr 18, 2021 at 2:36
  • $\begingroup$ The oeis link does not appear to be working yet. $\endgroup$
    – Will Sawin
    Commented Apr 18, 2021 at 2:47
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    $\begingroup$ I would guess for probabilistic reasons that no finite set of expressions of the form quadratic polynomial + an exponential covers all positive integers. However, small-number effects might imply that the first counterexample is quite large. If you want to find evidence against the probabilistic model / evidence for your conjecture, it might be worthwhile to calculate the mean and variance of the number of $k$ such that there exist $x$, $y$ representing $k$. If the variance is much less than the mean then the probabilistic model is likely oversimplified. $\endgroup$
    – Will Sawin
    Commented Apr 18, 2021 at 13:58
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    $\begingroup$ Yes. I currently think this conjecture is wrong, but the first counterexamples just might be very large. $\endgroup$
    – Will Sawin
    Commented Apr 18, 2021 at 14:17
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    $\begingroup$ I get $8558169401=x(3x+1)/2+y(7y+1)/2+2^k$ for $x=51199, y=-36356, k=1$ or $x=34425, y=36383, k=31$, among more than 20 other solutions. $\endgroup$
    – Yaakov Baruch
    Commented Sep 20, 2023 at 16:45

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