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.


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:

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

Ulam Turns

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

  • 1
    $\begingroup$ See formulas at OEIS A055086. $\endgroup$
    – BillyJoe
    Jul 5 at 17:45
  • $\begingroup$ @BillyJoe Ah this sequence is exactly what I arrived at just shifted slightly (they have a "+1" in the square root instead of a "-1"). Good find! $\endgroup$ Jul 5 at 18:13

2 Answers 2


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).


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.