Count the number of turns in an Ulam spiral

Question

Given a number on the Ulam Spiral, I want to know how many 90° turns were taken to arrive at that number. For example, to get to the number 10, you need to turn at 2, 3, 5, and 7 for a total of four turns.

I know how to generate the sequence of numbers where the turns occur, they follow the form:

$$y = \lfloor (n+2)^2/4 \rfloor+1$$

Here $y$ is a turning point number on the spiral and $n$ is the number of turns. For the example above, if we input $n=4$ (for 4 turns) we get:

$$y = \lfloor (4+2)^2/4 \rfloor+1 = \lfloor 36/4 \rfloor+1 = 10$$

However, I not sure how to solve for the number of turns ($n$) given some $y$ because the $floor()$ function has no inverse. Any insight would be appreciated.

EDIT

After looking at John Omielan's answer I realized if you solve the even equation for $k$, ignore the negative solution, and round up the answer it seems to work:

$$n = \left\lceil 2\sqrt{y-1}-2\right\rceil$$

I can't proof this works for every number, but it seems promising.

Answer 1

If $n$ is even, i.e., $n = 2k$, then

$$\left\lfloor \frac{(n+2)^2}{4} \right\rfloor + 1 = (k + 1)^2 + 1 = k^2 + 2k + 2 \tag{1}\label{eq1A}$$

However, if $n$ is odd, i.e., $n = 2k + 1$, then

$$\begin{equation}\begin{aligned} \left\lfloor \frac{(n+2)^2}{4} \right\rfloor + 1 & = \left\lfloor \frac{(2k + 3)^2}{4} \right\rfloor + 1 \\ & = \left\lfloor \frac{4k^2 + 12k + 9}{4} \right\rfloor + 1 \\ & = k^2 + 3k + 3 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Thus, if it's known that $y$ is an actual turning point, we then just need to solve the quadratic equation in $k$ for $y$ being equal to either \eqref{eq1A} or \eqref{eq2A} to get which one gives an integer value for $k$, and then the corresponding value of $n$. Even if it's not certain $y$ is a turning point, following this procedure will at least give us the value(s) of $n$ that are the nearest number of turn(s).

Answer 2

May be, you could use $$\color{red}{\lfloor x\rfloor=x-\frac 12+\frac 1\pi \tan ^{-1}(\cot (\pi x))}$$ which would give $$y=\frac{n^2+4 n+2 } 4+\frac 1\pi \tan ^{-1}\left(\cot \left(\frac{\pi n^2}{4}\right)\right)$$ $$-\frac 12 \leq \frac 1\pi \tan ^{-1}\left(\cot \left(n^2\frac{\pi }{4}\right)\right)\leq \frac 12$$ makes probably quite good bounds.