I know that for a hexagonal lattice generated by (0,1) and ($\sqrt{3}/2$,1/2) (i.e., when the distance between lattice points is 1), the number of lattice points in a circle of given radius $r$ can be calculated by the equation below:
$$n(r) = \sum_{x=-\lfloor \frac{r}{\sqrt{3}}\rfloor}^{\lfloor \frac{r}{\sqrt{3}}\rfloor} (2\lfloor \sqrt{r^2-3x^2} \rfloor+1) + \sum_{x=\frac{1}{2}-\lfloor \frac{r}{\sqrt{3}}+\frac{1}{2}\rfloor}^{\lfloor \frac{r}{\sqrt{3}}+\frac{1}{2}\rfloor-\frac{1}{2}} 2\lfloor \sqrt{r^2-3x^2} +\frac{1}{2} \rfloor$$ I would like to transform this sum for the case when the distance between lattice points is $d$ where $d$ is an integer. Is the scaling $r$ by $d$ is sufficient? I mean is the following equation for counting the number of lattice points when the distance between lattice points is $d$ correct?
$$n(r) = \sum_{x=-\lfloor \frac{r}{d\sqrt{3}}\rfloor}^{\lfloor \frac{r}{d\sqrt{3}}\rfloor} (2\lfloor \sqrt{\frac{r^2}{d^2}-3x^2} \rfloor+1) + \sum_{x=\frac{1}{2}-\lfloor \frac{r}{d\sqrt{3}}+\frac{1}{2}\rfloor}^{\lfloor \frac{r}{d\sqrt{3}}+\frac{1}{2}\rfloor-\frac{1}{2}} 2\lfloor \sqrt{\frac{r^2}{d^2}-3x^2} +\frac{1}{2}\rfloor$$
Yes, if the first one is correct (which I will take your word for!) because you are simply saying how many points there are in a circle of radius $\frac rd$ instead of radius $r$, which is equivalent to what you are looking for.