# What is the region of convergence of $x_n=\left(\frac{x_{n-1}}{n}\right)^2-a$, where $a$ is a constant?

The following recurrence relation came up in some research I was working on:

$$x_n=\left(\frac{x_{n-1}}{n}\right)^2-a$$

Or equivalently the map:

$$z\mapsto\frac{z^2}{n^2}-a$$

Where $$n$$ is the iteration number. Specifically, I'm interested in the size of the convergence region across the real line. Some stuff I know about this map:

• For $$a = 1$$, it's easy, the "size on the real line" is $$[-3,3]$$.

I do have an infinite radical expansion for the size of the convergence region on the real line (see Solving the infinite radical $$\sqrt{6+\sqrt{6+2\sqrt{6+3\sqrt{6+...}}}}$$):

$$\sqrt{a+2\sqrt{a+3\sqrt{a+...}}}$$

That's why it's easy for $$a=1$$ -- it's just the Ramanujan radical, and equals 3. It's also easy for $$a=0$$ -- it's $$\exp\left(-\mathrm{PolyLog}^{(1,0)}(0,1/2)\right)$$ as per Wolfram Alpha.

Has anyone seen this map before? Here's the region of convergence on the complex plane, plotted numerically (for $$a=6$$):

• For your title, instead asking "have you seen this recurrence before" to which most will say "no" and skip the question, you rather ask "what is the region of convergence for ...." and get more help. – DanielV Dec 24 '18 at 19:34
• if the sequence converges, then it must converge to $-a.$ I am not sure if any other trivial observation can be made. – dezdichado Dec 24 '18 at 22:58

• It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine). – Abhimanyu Pallavi Sudhir Dec 25 '18 at 4:38