Only $x_{6n}, y_{6n}$ doesn't have the same simplest denominator

Question

I'm studying this series \begin{align*} a_1=\mathrm{i}, \quad a_{n+1}=\mathrm{i} + \frac{\mathrm{i}}{a_n}, \end{align*} where $\mathrm{i}$ is the imaginary unit, in order for me to see clearly the structure of this sequence, I separate the real part and the imaginary part as a sequence, as follows. \begin{align*} x_{n+1} = \frac{y_n}{x_n^2+y_n^2}&, \quad y_{n+1}=\frac{x_n}{x_n^2+y_n^2}+1, \\ &a_n = x_n + y_n\mathrm{i}, \end{align*} with $x_1=0, y_1=1$, when I try to find the first few patterns, I found Only $x_{6n}, y_{6n}$ doesn't have the same simplest denominator

The following are $x_n$, $y_n$ for the first 20 items ($x_n$ is on the left and $y_n$ is on the right)

and has undergone preliminary programming verification on a larger scale. But I don't really know how to prove this, Can anyone give me an idea or prove this wrong?

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Addition Information

If we consider a more complex situation, given two nonnegative integers X and Y, make the following sequence: \begin{align*} x_{n+1} = \frac{y_n}{x_n^2+y_n^2}+X&, \quad y_{n+1}=\frac{x_n}{x_n^2+y_n^2}+Y, \\ \end{align*} with $x_1=0, y_1=1$, The corresponding recursive expression of the complex sequence is \begin{align*} a_{n+1}=X+Y\mathrm{i}+\frac{\mathrm{i}}{a_n} \end{align*} For ease of description, we denote $p_k$ as the number of terms in $x_n, y_n$ where the $k^{th}$ simplest denominator is different. For instance, for the initial case of $X=0, Y=1$, we have $p_1=6, p_2=12, \cdots$. We also denote $\Delta_{p_n} = p_{n+1} - p_n$. In the initial case, our conjecture is to prove that $p_n$ forms an arithmetic sequence with a common difference of $6$, or $\Delta{p_n}=6$. To explore the general case, a natural idea is that $p_n$ could potentially be a combination of terms from multiple arithmetic sequences, for example, $p_{2k+1}=4k+3, p_{2k}=2k+2$. Of course, in this scenario, $p_n$ would not form an arithmetic sequence, but ${\Delta p_n}$ would have a periodic sequence (this is intuitively easy to prove. For two arithmetic sequences with periods $a, b$ where $a \nmid b$ and $b \nmid a$, it seems that the period of ${\Delta p_n}$ is $\frac{a+b}{gcd(a,b)}-1$).

Through programming, I tested the potential periods for $p_n \leq 5000, 0 \leq X \leq 14, 0 \leq Y \leq 16, X \neq Y$, The parameter table corresponding to periodic $\Delta p_n$ is as follows:

X	Y	period	$\Delta p_n$ items within a period
0	1	1	[6]
0	2	1	[24]
0	4	1	[15]
0	7	6	[6, 3, 3, 6, 6, 6]
0	8	2	[125, 50]
0	13	0	[6, 3, 3, 6, 6, 6]
1	9	2	[588, 504]
1	13	2	[183, 488]
2	0	7	[5, 10, 5, 5, 5, 5, 10]
2	3	23	[2, 10, 12, 8, 4, 11, 1, 12, 12, 5, 7, 8, 4, 12, 12, 2, 10, 5, 7, 12, 11, 1, 12]
3	6	1	[12]
5	2	1	[12]
5	11	2	[24, 36]
5	14	16	[5, 7, 12, 12, 12, 12, 12, 12, 12, 12, 12, 5, 7, 12, 12, 12]
6	5	180	[6, 4, 2, 6, ..., 6, 6, 3, 3, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6]
7	0	15	[6, 12, 6, 6, 6, ..., 6, 6, 6, 6, 6]
7	6	111	[6, 6, 6, ..., 6, 6, 6, 3, 3, 6]
7	15	2	[75, 100]
8	9	55	[6, 6, 12, 3, ..., 6, 6, 12]
8	11	57	[12, 12, 7, 5, ..., 12, 12, 12, 12]
8	16	2	[244, 427]
9	7	7	[5, 5, 10, 5, 10, 5, 5]
10	13	61	[12, 12, 12, ..., 12, 3, 9, 12]
11	5	2	[84, 1008]
12	14	2	[48, 120]
13	6	1	[12]
13	9	2	[24, 144]
13	16	16	[12, 12, 12, ..., 6, 6, 12]
14	6	1	[15]

For entries not included in the table, it may be the case that no period exists, or alternatively, the period might be too large to have been identified.

Additionally, regarding Eric's discovery, verification found it to be true only when $X=0, Y=1$. As for the findings of Ivan Neretin, similar properties were observed across different $X, Y$ values, particularly evident when $X=0, Y=3$ (in this case, the period of this property is $6$). However, due to difficulties in programming implementation, no tables were produced to illustrate this.

Sadly, despite all the discoveries mentioned above, a reliable method to prove them has yet to be found.

Answer 1

First of all, it is never necessary (in my experience) to separate a recurrence relation into its real and imaginary parts. Admittedly, however, you might not have made the observation about the denominators otherwise. In this answer, I'll show how I constructed a closed-form solution to this problem, while ignoring the fact that the solution is complex. To avoid confusion, I'm going to express the problem as follows

$$ f_n=a+\frac{b}{f_{n-1}},\quad f_0 \text{ given} $$

Now, assume a solution of the form $f_n=q_n/p_n$, so that

$$ \frac{q_n}{p_n}=a+\frac{bp_{n-1}}{q_{n-1}} $$

Finally, let $p_n=q_{n-1}$ so that

$$ q_n=aq_{n-1}+bq_{n-2} $$

which is a standard Fibonacci-type second-order recurrence. I have previously given the solution to this recurrence with arbitrary $q_0, q_1, a, b$ in this previous post of mine. The solution can be written as

$$ q_n=\left(q_1-\frac{aq_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+a\frac{q_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta} $$

where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$. The solution applies to negative values of $n$ as well.

Now, analysis of the initial conditions shows that we cannot isolate $q_0$, but rather can only determine the ratio $q_1/q_0$. Therefore, without any loss of generality, we can take $q_0=1$ and $q_1=a+bf_0$.

So, we finally have the completed solution as follows

$$ f_n=\frac{q_n}{p_n}=\frac{q_n}{q_{n-1}} $$

I have verified this solution numerically by comparison with the original recurrence and your $x+iy$.

Answer 2

In my previous answer I developed a closed form solution for the recurrence. My attempts on rationalization of the results, and comparison with others (for example, comment exchange with @joriki), it seemed to me that there is some ambiguity in the rationalization methods insofar as they can given different results. I have presently developed an exact, absolute, rationalization. We previously showed that the recurrence for $a$ can be expressed as the ratio of two (related) recurrences with known solutions.

$$ a_n=\frac{q_n}{q_{n-1}} $$

As it turns out, the solution to $q_n$, in the present case, at least, is in the form of Gaussian integers, which leads us to the non-ambiguous rationalization

$$ a_n=\frac{N}{D}=\frac{q_n\cdot q_{n-1}^*}{q_{n-1}\cdot q_{n-1}^*} $$

where we expect that the Numerator is a Gaussian integer and the Denominator is a real integer. Moreover, the real and imaginary parts have the same denominator, which increases monotonically. The table below shows the results for the first 34 values of $a_n$. Beyond that we get small decimal parts, probably do to machine precision.

Now this this does not address the question of why $a_{6n}$ seem to allow further reduction that other values do not. I tried different methods of visualizing the results. The figure below shows a graphic realization of the results. The (linear) size of each triangle is the absolute value of the difference of subsequent terms (e.g., $|a_n-a_{n-1}|$). Notice that the spiral turns one period ($2\pi$) per six tiles and that each group of six tiles are similarly oriented and self-similar. Perhaps this will lend some insight into the observed behavior.

Table of $N/D$ $$ \begin{align} &n\quad N\ /\ D\\ &1\quad 0+1i\ /\ 1\\ &2\quad 1+1i\ /\ 1\\ &3\quad 1+3i\ /\ 2\\ &4\quad 3+6i\ /\ 5\\ &5\quad 6+12i\ /\ 9\\ &6\quad 12+26i\ /\ 20\\ &7\quad 26+53i\ /\ 41\\ &8\quad 53+111i\ /\ 85\\ &9\quad 111+231i\ /\ 178\\ &10\quad 231+480i\ /\ 369\\ &11\quad 480+1000i\ /\ 769\\ &12\quad 1000+2080i\ /\ 1600\\ &13\quad 2080+4329i\ /\ 3329\\ &14\quad 4329+9009i\ /\ 6929\\ &15\quad 9009+18747i\ /\ 14418\\ &16\quad 18747+39014i\ /\ 30005\\ &17\quad 39014+81188i\ /\ 62441\\ &18\quad 81188+168954i\ /\ 129940\\ &19\quad 168954+351597i\ /\ 270409\\ &20\quad 351597+731679i\ /\ 562725\\ &21\quad 731679+1522639i\ /\ 1171042\\ &22\quad 1522639+3168640i\ /\ 2436961\\ &23\quad 3168640+6594000i\ /\ 5071361\\ &24\quad 6594000+13722240i\ /\ 10553600\\ &25\quad 13722240+28556241i\ /\ 21962241\\ &26\quad 28556241+59426081i\ /\ 45703841\\ &27\quad 59426081+123666803i\ /\ 95110562\\ &28\quad 123666803+257352966i\ /\ 197926885\\ &29\quad 257352966+535556412i\ /\ 411889609\\ &30\quad 535556412+1114503066i\ /\ 857150100\\ &31\quad 1114503066+2319302053i\ /\ 1783745641\\ &32\quad 2319302053+4826511631i\ /\ 3712008565\\ &33\quad 4826511631+10044062391i\ /\ 7724760338\\ &34\quad 10044062391+20901884640i\ /\ 16075373009\\ \end{align} $$

Graphic realization

Answer 3

For what it could be worth \begin{align*} a_1=\mathrm{i} \quad \quad a_{n+1}=\mathrm{i} + \frac{\mathrm{i}}{a_n} \end{align*}

leads to $$a_n=\frac {\mathrm{i}} 2+\frac{\mathrm{i}}{2} \sqrt{1-4\, \mathrm{i}}\,\, \coth \left(n \tanh ^{-1}\left(\sqrt{1-4\, \mathrm{i}}\right)\right)$$ confirms your observations.

But what happens when $n=6k$ ?

Answer 4

In this answer I shall endeavor to explain how it comes to pass that $a_{6n}$ can be expressed as rational numbers with different denominators, while others cannot.

I find that there are two conditions that are necessary for the rational numbers of $(X+iY)/D$ to have different denominators. The first is that $X,Y,D$ are all even numbers. The second is that the prime factors of $X,Y,D$ have disparate values of repeating factors such as 2 and 3. This will be shown in examples below. A short table of $X,Y,D$ values was shown in one of my previous answers on this page.

Here is what I did: First of all, $X$ and $D$ are known sequences (thank you user @Somos for pointing this out). $X$ is given exactly by OEIS A092886. You may also notice that $Y(n)=X(n+1)$. $D$ is given exactly by OEIS A105309. But you can find that A105309 (n) = A092886(n+1) - A092886(n-1). Or, $D(n)=X(n+1)-X(n-1)=Y(n)-X(n-1)$. So everything comes down to $X$.

At this point I regenerated the table of $X,Y,D$ using the OEIS integer recurrence for $X$ to avoid the vagaries attendant with large floating-point numbers. I was only able to carry out the table to $n=61$ because of limitations of Matlab’s 64-bit integers.

I then parsed the table to separate out those values of $n$ for which $X,Y,D$ are even numbers. The results showed that up to $n=54$, only those with $n$ equal to multiples of 6 meet this criterion. However, all of $55\le n \le 61$ met this criterion. This is where the second criterion comes into play.

Here are two examples that demonstrate cases where the rational numbers have different or same denominators by virtue of the prime factorization.

Case 1: $n=30$

$$ \begin{align} &X\to 2,2,3,3,14876567\\ &Y\to 2,3,3,61916837\\ &D\to 2,2,3,3,3,3,5,5,29,41,89\\ &\frac{X}{D}=\frac{X/2^23^2}{D/2^23^2},\quad \frac{Y}{D}=\frac{Y/2^13^2}{D/2^13^2} \end{align} $$

Hence different denominators.

Case 2: $n=55$

$$ \begin{align} &X\to 2,2,2,79,199,811,3221,147613\\ &Y\to 2,2,2,2,7,23,1474357,26572589\\ &D\to 2,2,59,733,126751,3539897\\ &\frac{X}{D}=\frac{X/2^2}{D/2^2},\quad \frac{Y}{D}=\frac{Y/2^2}{D/2^2} \end{align} $$

Hence the same denominators.

Similar behavior accrues for all other values of $n$. I believe this explains the $a_{6n}$ observations in the OP.