I was playing around with number sequences and came across the following interesting type of sequences of positive rational numbers: The sequence starts with any rational number $x_1$. Each subsequent term $x_n$ is defined by $x_n=\frac{a+b}{a+1}$ when the previous term in simplest form is $x_{n-1}=\frac{a}{b}$, where a and b are coprime.
Any sequence in which any term $x_i$ can be written in either of the following forms: $\frac{a}{1}, \frac{1}{b}$ will have every subsequent term be $x_{j>i}=1$. This result is trivial.
Every other sequence that I tried that didn't converge into the above result or one of the following loops: $...\frac{3}{2},\frac{5}{4},\frac{3}{2},\frac{5}{4},...$ or $...\frac{29}{18},\frac{47}{30},\frac{77}{48},\frac{125}{78},\frac{29}{18},\frac{47}{30},\frac{77}{48},\frac{125}{78},...$.
Is there any way of proving that every starting point for such a sequence will enter into loop or of predicting what loop will be entered?
Comments have mentioned that many such sequences seem to never become eventually periodic (based on computing a "large" number of terms). Here's a possible approach to proving this.
Consider the following recursion on ordered pairs of positive integers: $$(a_{k+1},b_{k+1}) = (a_k+b_k,a_k+1),\quad k=1,2,3,...$$ in which the initial pair $(a_1,b_1)$ determines the whole sequence.
Claim: If it happens that $(a_1,b_1)$ is such that $a_k, b_k$ are coprime for all $k$, then the sequence $\left({a_k\over b_k}\right)_k$ is one of your sequences $(x_k)_k$ with $x_1={a_1\over b_1}$, and hence $\lim_{k\to\infty}x_k=\lim_{k\to\infty}{a_k\over b_k}=\varphi,$ where $\varphi={1+\sqrt{5}\over 2}=1.618...$ is the Golden Mean.
Proof of Claim: The first part is clear, because if all $a_k, b_k$ are coprime then every ${a_k\over b_k}$ is an irreducible fraction, so starting with $x_1={a_1\over b_1}$, no reduction occurs in any iteration of your mapping. Furthermore, by inspection of the recursion it is readily seen that $(a_k,b_k)=(G_{k+1}-1,G_k)$, where $G_k=(a_1+1)\,F_k+b_1\,F_{k-1}$, and $F_k$ is the $k$th Fibonacci number.Then $${a_k\over b_k}= {G_{k+1}-1\over G_k} = {F_{k+1}\over F_k}{(a_1+1)+b_1\,{F_{k}\over F_{k+1}}-{1\over F_{k+1}} \over (a_1+1)+b_1\,{F_{k-1}\over F_{k}} }\to \varphi{(a_1+1)+b_1\,{1\over \varphi}-0 \over (a_1+1)+b_1\,{1\over \varphi} }=\varphi$$ using the known fact that ${F_{k+1}\over F_k}\to\varphi.$
Therefore, proving the following conjecture would establish that some of your sequences never enter a cycle:
Conjecture 1: There exist initial pairs $(a_1,b_1)$ such that $a_k, b_k$ are coprime for all $k$ (and hence $\lim_{k\to\infty}{a_k\over b_k}=\varphi$). (I suspect that there are infinitely many such initial pairs.)
For example, with $(a_1,b_1)=(5,12),$ computations show that all $(a_k,b_k)$ are coprime for $1\le k\le 10^6.$ (Thus, no reductions occur in generating the first $10^6$ terms of your sequence starting with $x_1={5\over 12}$.)
EDIT: Conjecture 1 has since been proven, as it is a consequence of this answer. (That there are infinitely many such pairs also follows from this.)
For example, $x_1={5\over 12}$ is one of the proven cases for which no reductions occur among the terms in the sequence $(x_1,x_2,x_3,...)=({5\over 12},{17\over 6},{23\over 18},...).$ But there are infinitely many other values of $x_1$ giving the same tail of this sequence $(x_2,x_3,...).$ This is due to the easily-proved fact that the set of possible predecessors of ${a\over b}$, with $a\perp b$, is $$\left\{{m\,b-1\over m\,(a-b)+1}: m\ge 1, \ \ (m\,b-1)\perp (m\,(a-b)+1)\right\}$$ using "$\perp$" to abbreviate "coprime to". Thus, $x_2={17\over 6}$ has the infinite set of predecessors $$\left\{{m\,6-1\over m\,11+1}: m\ge 1, \ \ (m\,6-1)\perp (m\,11+1)\right\}=\left\{{5\over 12},{11\over 23},{23\over 45},... \right\},$$ any one of which can be taken as the initial value $x_1$. (A less trivial conjecture, not yet proven, is that there are infinitely many $x_1$ whose orbits converge to $\varphi$ without reductions, the orbits being disjoint from one another. Examples appear to include ${5\over 12},{17\over 36},{29\over 90},{41\over 84}.$)
NB: By a "predecessor" of $q$, I mean a positive rational $p$ such that $f(p)=q,$ where $f$ is your transformation. It's notable that any set of predecessors must be either empty or infinite:
I suspect that every sequence generated by iterating your mapping either converges to $\varphi$ or eventually enters one of infinitely many finite cycles:
Conjecture 2: The set of positive integer pairs (and hence the positive rationals) is partitioned into infinitely many disjoint subsets $S_0,S_1,S_2,\ldots,$ where $$\begin{align} S_0&=\{(a_1,b_1): {a_k\over b_k}\to \varphi \}\\ S_i&=\{(a_1,b_1): {a_k\over b_k}\to \text{cycle}_i \},\quad i=1,2,3,...\\ \end{align}$$ and $\text{cycle}_1,\text{cycle}_2,\text{cycle}_3,...$ are infinitely many disjoint cycles, each having finitely many elements.
If the latter conjecture holds, then each of your rational sequences can be seen as "trying to converge to $\varphi$" and either succeeding, or failing by eventually entering a finite cycle whose elements only approximate $\varphi$ ("convergence interruptus" :).
For reference, here are six of the cycles (found using Sage), showing their min and max values truncated to 8 decimal digits: