I've been given the following sequence: \begin{align*} &a_0 = 0; \\ &a_{n+1} = (a_n)^2+\frac{1}{4}. \end{align*}

I also have to prove that whatever I come up with is correct, but that will likely be the easy part.

Here are the first few values:

\begin{align} &a_0 = 0 \\ &a_1 = \frac{1}{4}\\ &a_2 = \frac{5}{16}\\ &a_3 = \frac{89}{256} \\ &a_4 = \frac{24305}{65536} \end{align}

I've managed to to determine that the denominators are of the form $2^{2^n}$. I've tested up to one million terms of this sequence and it appears that $\lim_{n\rightarrow\infty}a_n = \frac{1}{2}$. I spent a while trying to find something of the form $a_n = \frac{P(n)}{Q(n)}$. I haven't had any luck with this, so I started looking into some sums. I've found that \begin{align*} a_2 = \frac{1}{4} + \frac{1}{16} = \frac{5}{16} \end{align*} and, \begin{align*} a_3 = \frac{1}{4} + \frac{1}{16} + \frac{1}{32} + \frac{1}{256} = \frac{89}{256} \end{align*}

But now, \begin{align*} a_4 = \frac{1}{4} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \frac{1}{1024} + \frac{1}{2048} + \frac{1}{4096} + \frac{1}{65536}= \frac{24305}{65536}. \end{align*}

So it seems that there is some type of sum involving negative powers of 2 going on, but it isn't clear to me that there is even a pattern here. Any hints/help would be appreciated!

  • $\begingroup$ The sums over powers of 2 amount to expansions in base 2 of the fractions. They would all terminate if you showed the denominators are $2^{2^n}$ as you say, but the "pattern" of the expressions in terms of negative powers of 2 is just whatever happens to be the digits of the $a_n$ in base 2, and at least for the first four I don't see an obvious pattern in those digits. $\endgroup$ – coffeemath Jul 8 '13 at 10:40
  • $\begingroup$ The numerators seem to be those of the sequence at OEIS number A167424, which is a slightly different sequence, but only by a factor of 1/2 as I recall (just looked briefly at the sequence obtained...). $\endgroup$ – coffeemath Jul 8 '13 at 10:47
  • $\begingroup$ hmm, that page on OEIS is making me suspicious that there isn't a closed form for this recurrence relation... $\endgroup$ – stochasm Jul 8 '13 at 11:11
  • 1
    $\begingroup$ At least, the closed form of the limit is 1/2... $\endgroup$ – Julien Jul 8 '13 at 12:04
  • 2
    $\begingroup$ See this answer ... math.stackexchange.com/a/114949/442 $\endgroup$ – GEdgar Jul 8 '13 at 12:58

I just checked OEIS sequence A167424 for which $$f(1)=1/2 \\ f(n+1)=[f(n)^2+1]/2.$$ If this $f$ is divided by 2 you get your sequence $a_1,a_2,\cdots,$ since then it gives correctly $a_1=1/4$ and $a_{n+1}=a_n^2+1/4.$ The recursion on the $a_n=f(n)/2$ follows on dividing through the recursion $f(n+1)=[f(n)^2+1]/2$ by 2 to obtain $$\frac{f(n+1)}{2}=\left( \frac{f(n)}{2}\right) ^2 +\frac{1}{4}.$$

So anything like closed forms etc. should be extractable from the OEIS page, if it's there.

Side note: The Mandlebrot set intersects the real axis in the interval $[-2,1/4]$. Since it consists of those points $c$ such that the orbit of $0$ under the iteration of $f(x)=x^2+c$ is bounded, we see that your sequence is precisely the (bounded) orbit of the rightmost real point of the Mandlebrot set. (In this sense it seems not surprising there isn't a closed form, as typically Mandlebrot iterations bounce around the set unpredictably. There are periodic points, but it seems $1/4$ is not one of them. (Just thought this connection might be of interest.)


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