Gaps between combinations of squares of integers

Question

Let $\theta$ be a positive irrational number and $S=\{\theta n^2+m^2: n, m\in \mathbb{N}\}$. The elements of $S$ can be written as a sequence of strictly increasing numbers $\{s_n\}$. My question is what is known about the difference $s_{n+1}-s_n$? Is there an estimate like $s_{n+1}-s_n\ge \frac{c}{n^\sigma}$ with some $\sigma\in (0,1)$, at least under some conditions on $\theta$?

Answer 1

For any $\theta$ there is a constant $C_{\theta}$ and infinitely many $n$ for which $s_{n+1} - s_{n} \leq \frac{C_{\theta}}{\sqrt{n}}$.

Choose a rational approximation $\frac{h}{k}$ to $\theta$ so that $\left|\frac{h}{k} - \theta\right| < \frac{1}{\sqrt{5} k^{2}}$, let $m_{1} = h+1$, $m_{2} = h-1$, $n_{2} = k+1$ and $n_{1} = k-1$. This gives $$ |4h - 4\theta k| = |m_{1}^{2} - m_{2}^{2} - \theta n_{2}^{2} + \theta n_{1}^{2}| < \frac{4}{\sqrt{5} k}. $$ Assume without loss of generality that $s_{r+1} = m_{2}^{2} + \theta n_{2}^{2}$ and $s_{r} = m_{1}^{2} + \theta n_{1}^{2}$. Since there are $\asymp n$ elements of $S$ with size less than $x$, we have that $r \asymp k^{2}$ and so $s_{r+1} - s_{r} \leq \frac{C_{\theta}}{\sqrt{r}}$.

This bound can be improved if $\theta$ has irrationality measure greater than $2$.

Answer 2

Here is an example purporting to show that for $\theta=\sqrt2,$ it happens infinitely often that $s_{n+1}-s_{n}>\frac{c}{n}$ for $c>\frac12.$ It might be that $c \approx \frac23.$ It is the result of looking for particularly small gaps. All the exceptional cases thus found belong to a particularly simple sequence with exponential growth.

Let $$(1+\sqrt{2})^n=a_n+b_n\sqrt{2}.$$ The example is $$s_n=b_{2k}^2+1\sqrt{2}$$$$s_{n+1}=b_{2k-2}^2+4b_{2k-1}^2\sqrt{2}$$ $$s_{n+1}-s_n=-2b_{4k-2}+a_{4k-2}\sqrt{2}=\sqrt{2}(\sqrt{2}-1)^{4k-2}.$$

Aside from this particular sequence, there are a few gaps as small as $\frac{1}{n^{0.75}}$ but most seem to be quite a bit larger.

It is easy to observe the first 20 or 40 cases behave as mentioned. That is not a proof but is convincing. The third equation should follow from various facts about these Pell numbers.

Another missing ingredient is a good estimate of $n$ from $s=s_n$. I will claim without proof that $n \approx 0.66s+0.88\sqrt{s}$ and simply report that for $100 \leq n \leq 6600$ the error stays between -6 and 11. This is just the result of a best quadratic fit over a certain range which turns out to be quite good over a longer range.

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Putting these things together,

$$s=s_n=b_{2k}^2+\sqrt{2}\approx \frac{(1+\sqrt{2})^{4k}}8$$ with $$n\approx\frac23 s$$ and

$$s_{n+1}-s_n=-2b_{2k-2}+a_{2k-2}\sqrt{2}=\sqrt{2}(\sqrt{2}-1)^{4k-2}.$$

So $$n(s_{n+1}-s_n)\approx \frac{(1+\sqrt{2})^2\sqrt{2}}{12} \approx 0.68$$

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Finding an upper bound on gaps would also be interesting. Here are the counts of some $s_{n+1}-s_{n}$ for $1000 \leq n \leq 6000.$ These are the ones which are at least 3, rounded to the nearest integer.

$[3, 185], [4, 258], [5, 118], [6, 60], [7, 32], [8, 13], [9, 9], [10, 2], [11, 2], [15, 1]$

Accordingly, one would expect fluctuations in any formula relating $s$ and $n.$

An approach which might validate my estimate, or give a better one starts as follows: For convenience, stipulate $s_0=0.$ We can calculate the position of $m^2+0\sqrt{2}.$ Before it are $u^2+v^2\sqrt{2}$ where $0 \leq u <m$ and for each $u,$ $0 \leq v^2\sqrt{2}<m^2-u^2$ So $m^2=s_n$ for $$n=\sum_{u=0}^{m-1}{\Big \lfloor}1+ \sqrt{\frac{m^2-u^2}{\sqrt{2}}}{\Big \rfloor}.$$ Use this to approximate $n$ as a function $f(m)$, check if $n\approx f(\sqrt{s})$ is a good enough estimate for general $s.$ If so, justify it.