What these results do—and do not—say

The accelerated Collatz map is used here as a demanding calibration for finite presentation dynamics:

$$T(n)=\begin{cases} n/2,&n\equiv0\pmod2,\\ (3n+1)/2,&n\equiv1\pmod2. \end{cases}$$

The page proves exact finite-trace statements. It does not prove the Collatz conjecture, finiteness of paradoxical traces, or any claimed improvement to the recorded exponential candidate bound.

FPRD-D38 · guarded-affine presentation

One trace has four exact readouts

For a chronological parity word$w\in\{0,1\}^k$ with$s=|w|_1$, iteration has the affine form

$$T^k(n)=\frac{3^s n+c_w}{2^k}.$$

Since $3^s$ is a unit modulo$2^k$, the starting values realizing that word form exactly one guard cylinder

$$n\equiv r_w\pmod{2^k}, \qquad r_w\equiv-c_w3^{-s}\pmod{2^k}.$$

For example,$7\mapsto11\mapsto17\mapsto26\mapsto13$has word $1110$ and

$$T^4(n)=\frac{27n+19}{16}, \qquad n\equiv7\pmod{16}.$$

The orbit list, word, affine action, and guard are presentations of the same finite event, but they preserve different information. At a fixed layer $(K,S)$, the additional readout$D=2^K-3^S$ and$\Pi(w)=c_w-Dr_w$ measures descent at the least admissible representative. This definition concerns finite traces, not complete orbits.

FPRD-T138 · exact affine sewing

A cut transports every continuation guard

Let a prefix $u$ of length$k$ have weight$p$, guard$r_u$, and endpoint$y_u=T^k(r_u)$. Every start in its cylinder is $r_u+2^k\mu$, and the prefix sends it to$y_u+3^p\mu$.

The proof is one modular substitution:$y_u+3^p\mu\equiv q\pmod{2^\ell}$. Multiplication by the odd unit$3^{-p}$ gives one lift class. The endpoint records one realized input; the seam map records the prefix's action on every hypothetical continuation guard.

FPRD-T139 · exact interface demand

The complete seam profile needs linearly many bits

For the all-odd prefixes$u_s=1^s$, the horizon profile is

$$F_{u_s}^{(\ell)}(q)=3^{-s}(q+1)-1\pmod{2^\ell}.$$

For $\ell\ge3$, the order$\operatorname{ord}_{2^\ell}(3)=2^{\ell-2}$gives that many distinct slopes and hence distinct complete seam maps. Naming the exact map therefore needs at least$\ell-2$ bits. Its slope and intercept can be stored in$O(\ell)$ bits, so the exact profile demand is $\Theta(\ell)$.

This counts exact continuation functions. A recognizer may merge prefixes whose full functions differ, so this is not an automata or language-recognition lower bound. Within a total trace budget$K$, this family supplies only$\ell\le\lfloor\log_2(K+1)\rfloor+2$.

FPRD-T140 · exact recovery calculus

Recovery separates into disorder and seam lift

Fix a contracting layer$(K,S)$, so$D=2^K-3^S>0$. Split a word$w=uz$ after$j$ symbols. If the suffix has length $L=K-j$ and weight$q$, its unique maximal affine constant is

$$C^+_{L,q}=2^{L-q}(3^q-2^q).$$

Write$\delta_z=C^+_{L,q}-c_z\ge0$for loss from that scalar maximizer, and let$\mu_z$ be the least nonnegative seam lift satisfying$\mu_z\equiv3^{-p}(r_z-y_u)\pmod{2^L}$. With$B_u=3^qy_u+C^+_{L,q}-2^Lr_u$, direct substitution gives

The prefix budget must pay two nonnegative costs: local loss from the best fixed-weight word and modular misalignment at the cut. The scalar maximizer does not by itself guarantee a feasible continuation of the required length and weight.

FPRD-T141 · one-bit slot conservation

One presentation step injects child slots into parent slots

The prefix budget obeys the exact one-bit recurrence

$$2B_{ub}=B_u-\varepsilon_bD-bG_{L,q},$$

where$\varepsilon_b=b\mathbin{\mathrm{xor}}(y_u\bmod2)$and the nonnegative odd-branch toll is$G_{L,q}=3^{q-1}(2^{L-q}-1)$when that branch is feasible. For$S_u^{\mathrm{slot}}=\{m:0\le m<2^L,\ Dm\le B_u\}$, send child slot $\nu$ to$\varepsilon_b+2\nu$.

Equality can occur. The theorem is an additive conservation law, not a universal one-step contraction factor. Its abstract lift-slot count is also different from the lowering-capped count below.

FPRD-T142 · marginal no-gain theorem

More unconditional bands cannot improve the best envelope

Let $t_1\ge\cdots\ge t_L\ge0$be nested guard thresholds and let$A_i\ge0$ be unconditional bounds on increasing disorder sublevels. The refined envelope is

$$M_L=t_L+\sum_{i=2}^L\min(t_{i-1}-t_i,A_i)+A_1.$$

The induction has two cases. Put$\Delta=t_{L-1}-t_L$. If$\Delta\le A_L$, the last correction equals $\Delta$and $M_L=M_{L-1}$. If$\Delta>A_L$, discarding the earlier nonnegative corrections gives$M_L\ge t_L+A_L$. Together with the base case, this proves the claim.

The theorem closes only envelope arguments built from unconditional marginals. It does not rule out a genuinely joint guard–disorder estimate. A finite exhaustive audit over small integer thresholds was rerun as a falsification check; the displayed induction, not that enumeration, proves the theorem.

FPRD-T143 · capped continuation potential

Capacity and word supply meet at a bottleneck

Let $N_u^{\mathrm{cap}}$bound admissible guard extensions, let$W_u=\binom{L}{q}$ count remaining fixed-weight words, and define

$$m_u=\min(N_u^{\mathrm{cap}},W_u), \qquad P_h=\sum_{|u|=h}m_u.$$

Child capacity is subadditive, while Pascal's identity makes word supply additive. Therefore$P_{h+1}\le P_h$; at full depth the cap equals the exact acceptance indicator. Locally, if$\sigma_u=N_u^{\mathrm{cap}}-N_{u0}^{\mathrm{cap}}-N_{u1}^{\mathrm{cap}}$and $\ell_u=m_u-m_{u0}-m_{u1}$, then

$$2\ell_u=\sigma_u +|N_{u0}^{\mathrm{cap}}-W_{u0}| +|N_{u1}^{\mathrm{cap}}-W_{u1}| -|N_u^{\mathrm{cap}}-W_u|.$$

This identity locates exact loss where arithmetic capacity and combinatorial continuation supply fail to align. It follows from$2\min(x,y)=x+y-|x-y|$.

Monotonicity is not multiplicative contraction. Capacity-one paths can preserve their only slot, and$N^{\mathrm{cap}}$ must not be identified with $N^{\mathrm{slot}}$. The theorem has a written all-length proof but no independent all-length executable checker.

The remaining frontier is a correlation problem

Exact seams, recovery budgets, and monotone potentials organize finite traces. They do not determine which available guard slot receives which fixed-weight suffix or how much disorder that forced suffix carries. The unresolved question is therefore joint:

$$r_w\le R, \qquad \delta_w\le B-Dr_w.$$

FPRD-C03 records the honest alternatives: prove enough loss for the actual forced continuation, or construct an exact accepted-trace counterfamily. Larger marginal tables do not answer that question.

Read the counterfamilies, reproduced finite boundary, and open joint problem →

Evidence, provenance, and review status

The uniform results come from the preserved guarded-affine proof package and its pinned repository record. The affine, seam, and recovery checker previously reran 370,252 split instances; the budget checker reran 77,022 prefix nodes. A new finite audit exhaustively checked 532,700 small nested-threshold instances. These computations are regression and falsification evidence, not replacements for the written all-parameter proofs.

Read the complete source guide, failure ledger, and executable-evidence boundaries →