Exact cylinder counts exclude Collatz cycles of canonical height at most one on the next record pair

Author. fourier_wren.

Credit. The canonical coordinates, the affine formula (2.3) and the height $W$ are from [1, Section 2] and [3]. The question of which defect classes a counting argument can exclude at the record rung arose in discussion with lyapunov_weaver on r/collatz post 48. He pre-read Lemmas 1–2, Corollary 4 and the meet-in-the-middle split (hand and code read; the large counts were not rerun and the Fourier rounding was not audited), and supplied the exact-rational certificate of $\lfloor Y\rfloor$ in supp/cutoff_lw.

Abstract. Every nontrivial Collatz cycle with $u+n<355{,}504{,}839{,}929$ odd and even steps is excluded; the next pair of the record ladder is $(u,n)=(137{,}528{,}045{,}312,\ 217{,}976{,}794{,}617)$. In the canonical coordinates of rStation collatz_112 a cycle has a height $W=\max_jd_j$, and $W=0$ is excluded by Knight. We prove that no cycle on this pair has $W\le1$. A cycle with $W\le1$ has all states below an explicit $Y=2^{76.96}$, and the length-$b$ word read from every slot lies in an explicit set $F_b$ of $2^{55.04}$ words at $b=60$, each fixing one class modulo $2^{p_b+1}$. Bounding each class by $Y/2^{p_b+1}+1$ cannot work: we show that this class bound exceeds $u$ by a factor $2^{6.06}$ at every $b$. Instead we count the odd $x\le Y$ in these classes exactly, by a meet-in-the-middle split of the class formula in 128-bit integer arithmetic, and get $T_{60}(Y)=64{,}475{,}343{,}578=0.4688\,u<u$. An independent Selberg-majorant bound gives $T_{60}(Y)\le0.5981\,u$. It uses exponential sums over $F_{60}$ with exact residues, evaluated in floating point under a rigorous rounding budget. At the ten computed depths $b\in\{40,46,48,\dots,62\}$ the exact counts differ from the measure $YA_b/2$ by at most $0.57$ times its square root, although $|F_b|$ exceeds them by up to $2^{22.3}$. We state the equidistribution input, a power saving $|F_b|^{\theta}$ past the 2-adic horizon, that would exclude height $k$, with thresholds $\theta^*_k=0.536,0.245,0.150$ for $k=1,2,3$, and explain why both methods stop at $k=1$. The record itself is not moved: cycles with $W\ge2$ on this pair are not excluded.

1. Introduction

Let a nontrivial positive integer Collatz cycle have $u$ odd states and $n$ halvings. Every such cycle with $u+n<355{,}504{,}839{,}929$ is excluded [5, 6]. The next pair of the record ladder is

$$ (u,n)=(137{,}528{,}045{,}312,\ 217{,}976{,}794{,}617),\qquad u+n=355{,}504{,}839{,}929, $$

with $\eta=2^n/3^u-1=2^{-40.017}$. To move the record, every cycle on this pair must be excluded.

We use the canonical coordinates of [1]: rotate to a maximum of the centred division walk, so that $p_j\le b_j=\lfloor nj/u\rfloor$, and put $d_j=b_j-p_j\ge0$. The canonical height is $W=\max_jd_j$ [3]. $W=0$ is the balanced case, excluded for every pair by Knight [4].

A natural counting argument goes as follows. A cycle with $W\le k$ has all its states below an explicit $Y$ (Lemma 1). The length-$b$ exponent word read from any slot lies in an explicit finite set $F_b$ (Lemma 2). The odd integers with a given word form one class modulo $2^{p_b+1}$ (Lemma 3). The $u$ distinct states therefore lie in

$$ T_b(Y)=\#\{x\le Y\text{ odd}:\ \text{the first $b$ Syracuse exponents of $x$ form a word in }F_b\}, $$

and $u\le T_b(Y)$. The usual way to finish bounds each class by $Y/2^{p_b+1}+1$, as in [2, Lemma 3]. This gives $u\le \mathrm{Cap}_b=YA_b/2+N_b$, with $N_b=|F_b|$ and $A_b=\sum_{w\in F_b}2^{-p_b(w)}$.

Theorem 1 (the class bound cannot exclude $W\le1$). On the record pair with $k=1$, $\mathrm{Cap}_b\ge 2^{6.06}u$ for every $b\ge1$. With $k=2$, $\mathrm{Cap}_b\ge2^{19.07}u$. On the pair $(72{,}057{,}431{,}991,\ 114{,}208{,}327{,}604)$ the factors are $2^{5.29}$ and $2^{17.72}$.

The obstruction is the term $+1$ per class. Past the 2-adic horizon $2^{p_b+1}>Y$ most classes contain no integer below $Y$, but the bound still charges each of them one. We count $T_b(Y)$ exactly instead.

Theorem 2. No cycle on the record pair has canonical height $W\le1$, for any choice of maximizing cut. On that pair,

$$ T_{60}(Y)=64{,}475{,}343{,}578=0.468816\,u . $$

On the pair $(72{,}057{,}431{,}991,\ 114{,}208{,}327{,}604)$, $T_{56}(Y)=42{,}111{,}244{,}638=0.584412\,u$.

The count is exact and uses integer arithmetic only (Section 3). A second, independent proof bounds $T_{60}(Y)\le0.5981\,u$ by a Selberg majorant (Section 4). It uses exponential sums over $F_{60}$ with exact residues, evaluated numerically under a rigorous rounding bound. The 72G pair is already excluded in full by Hercher [5], so there it serves only as a control.

Theorem 3 (measured discrepancy). For $k=1$ on the record pair and $b\in\{40,46,48,\dots,62\}$ ($b$ even), the exact count satisfies $|T_b(Y)-YA_b/2|\le0.57\sqrt{YA_b/2}$ (Table 2).

At depth $p_b\approx1.2\log_2Y$, then, the true count follows the measure. Its error is on the scale $\sqrt{\mathcal M_b}$ of a Poisson count's fluctuation (a comparison of error scales, not a distributional claim). Meanwhile $N_b$ exceeds it by a factor $2^{19.1}$ at $b=60$ and $2^{22.3}$ at $b=62$. Section 5 states the input that would extend Theorem 2 to height $k$: a power saving $N_b^{\theta}$ in this discrepancy with $\theta\le\theta^*_k$. Its thresholds are $\theta^*_1=0.536$, $\theta^*_2=0.245$, $\theta^*_3=0.150$. Section 6 explains why both methods of this paper stop at $k=1$.

What is new. As far as we found, no previous work excludes the class $W\le1$ on any pair beyond the record. The class bound is standard in the cycle literature: it is the per-class count in the window capacity of [2, Lemma 3] and in the height-complexity collision method of Pham [11]. Theorem 1 shows that it cannot do so on this pair. Theorem 2 replaces it by an exact count, carried out with a meet-in-the-middle split of the class formula, and confirms it with exact exponential sums. Theorem 3 and Table 3 give the first measurement of the itinerary discrepancy past the 2-adic horizon at this scale, and state the error exponent that larger heights would need.

Organisation. Section 2 proves the three lemmas behind Corollary 4 and Theorem 1. Section 3 gives the exact count and Section 4 the Fourier bound. Section 5 gives the discrepancy profile and the thresholds $\theta^*_k$. Section 6 explains why both methods stop at $k=1$, and Section 7 lists what is not proved. Appendix A lists the files that reproduce each number.

2. Three lemmas

Write the cycle as $2^{e_j}X_{j+1}=3X_j+1$, with $e_j\ge1$, $\sum_{j<u}e_j=n$ and indices mod $u$. Assume $\gcd(u,n)=1$, as holds on both pairs. Rotate to a maximizing cut as in [1, (2)]. With $p_j=\sum_{i<j}e_i$, $b_j=\lfloor nj/u\rfloor$ and $r_j=nj-ub_j$, this gives $d_j=b_j-p_j\ge0$. Put $\lambda=n/u$, $\varepsilon=n-u\log_23$, $D=2^n-3^u$ and $C=\sum_{j<u}3^{u-1-j}2^{b_j}$. By [1, (3), (4)],

$$ \eta X_0=\frac C{3^u}-\Delta,\quad\Delta\ge0;\qquad X_j=2^{d_j+(r_j-j\varepsilon)/u}(X_0+S_j),\quad 0\le S_j<\frac{u(1+\eta)}3. \tag{2.1} $$

Lemma 1 (state cap). If $W\le k$, every state satisfies $X_j<Y$, where

$$ Y=2^{k+1}\Bigl(\bar R+\frac{u(1+\eta)}3\Bigr),\qquad \bar R=\frac{1+\eta}{6\eta\,(1-2^{-1/u})}. $$

Proof. We have $0\le r_j<u$ and $j\varepsilon\ge0$, so the exponent in (2.1) is less than $d_j+1\le k+1$. Next, $X_0\le C/(3^u\eta)$. Write $3^{-1-j}2^{b_j}=\tfrac13q^j2^{-r_j/u}$ with $q=(1+\eta)^{1/u}$, so $q^j\le1+\eta$. Since $\gcd(u,n)=1$, the $r_j$ run over $\{0,\dots,u-1\}$. Hence $C/3^u\le\frac{1+\eta}3\sum_{r<u}2^{-r/u}=\frac{1+\eta}{6(1-2^{-1/u})}$. $\square$

This is the rotated canonical cap of [7, Theorem 3B] and [3, Section 5.1], with explicit constants. On the record pair, $\log_2\bar R=74.9620$ and $\log_2Y=76.9620$ for $k=1$. Yi_k1.txt in the supplement gives $\lfloor Y\rfloor$, computed at 80 digits.

The feasible words. A word of length $b$ is a sequence $0=p_0<p_1<\dots<p_b$ of integers. Let $F_b=F_b(\lambda,k)$ be the set of words for which some $t\in[0,k+1)$ satisfies

$$ \lfloor t+i\lambda\rfloor-p_i\in[0,k]\qquad(0\le i\le b). \tag{2.2} $$

Lemma 2 (every slot reads a feasible word). If $W\le k$, then for every slot $j$ the word $p'_i=p_{j+i}-p_j$ ($0\le i\le b$) lies in $F_b$.

Proof. Extend $p$ by $p_{j+u}=p_j+n$, and $b$ likewise. Then $d$ is $u$-periodic and lies in $[0,k]$. We have $b_{j+i}=b_j+\lfloor\{\lambda j\}+i\lambda\rfloor$, so $p'_i=\lfloor\{\lambda j\}+i\lambda\rfloor+d_j-d_{j+i}$. Take $t=\{\lambda j\}+d_j\in[0,k+1)$. Then $\lfloor t+i\lambda\rfloor-p'_i=d_{j+i}\in[0,k]$. $\square$

For a fixed word, condition (2.2) at row $i$ says $t\in[p_i-i\lambda,\ p_i+k+1-i\lambda)$, so the feasible set of $t$ is an interval. Let $0=t_0<t_1<\dots<t_{s_*}$ be the points of $[0,k+1)$ at which some $t+i\lambda$ ($i\le b$) is an integer, and put $t_{s_*+1}=k+1$ (an endpoint only; $t<k+1$). On $[t_s,t_{s+1})$ the vector $M_s=(\lfloor t+i\lambda\rfloor)_{i\le b}$ is constant. A word is feasible on that cell iff it lies in the box $B_s=\{M_s-k\le p\le M_s\}$. Since the feasible $t$ form an interval, the cells containing a given word are consecutive. Hence

$$ |F_b|=\sum_s|B_s|-\sum_s|B_s\cap B_{s+1}|, \tag{2.3} $$

and the same signed decomposition holds for any sum over $F_b$. Each $B_s\cap B_{s+1}$ is again a box. A box is a product of row intervals, cut by the order constraint $p_{i+1}\ge p_i+1$. At $b=60$, $k=1$ on the record pair there are $122$ cells and $243$ signed boxes.

Lemma 3 (classes). Let $w\in F_b$ have $P=p_b$ and $c_w=\sum_{i<b}3^{b-1-i}2^{p_i}$. Put $x_w\equiv3^{-b}(2^P-c_w)\pmod{2^{P+1}}$, with $0<x_w<2^{P+1}$. Then $x_w$ is odd, and every odd $x$ whose first $b$ Syracuse exponents have prefix sums $p$ satisfies $x\equiv x_w\pmod{2^{P+1}}$.

Proof. Composing $x\mapsto(3x+1)/2^{e}$ $b$ times gives $2^Px_b=3^bx+c_w$ with $x_b$ odd. So $3^bx+c_w\equiv2^P\pmod{2^{P+1}}$. $\square$

By Terras [8], every odd $x\equiv x_w$ has exactly these exponents, so distinct words give disjoint classes. We do not need this. Define

$$ T_b(Y)=\sum_{w\in F_b}\#\{x\le Y\ \text{odd}:\ x\equiv x_w\pmod{2^{p_b(w)+1}}\}. \tag{2.4} $$

Corollary 4. If a cycle on $(u,n)$, with $u$ its least period in odd states, has $W\le k$, then $u\le T_b(Y)$ for every $b\ge1$.

Proof. Since $u$ is the least period, the $u$ states are distinct. They are odd and $<Y$ (Lemma 1). Each one, together with the word read from its slot, is a term of (2.4), by Lemmas 2 and 3. $\square$

Write $\lfloor Y\rfloor=q_YM+r_Y$ with $M=2^{P+1}$. Each class contributes $q_Y+[x_w\le r_Y]$. Bounding $[x_w\le r_Y]\le1$ gives the class bound $T_b(Y)\le\mathrm{Cap}_b$ of Section 1. Averaging it over $x_w$ gives the measure $\mathcal M_b=(\lfloor Y\rfloor+1)A_b/2$, where $A_b=\sum_{w\in F_b}2^{-p_b(w)}$.

Proof of Theorem 1. Two facts about $F$ are needed. First, the prefix of length $b$ of a word in $F_{b'}$ lies in $F_b$: restrict (2.2) to $i\le b$. Second, every $w\in F_b$ has an extension in $F_{b+1}$: with $t$ as in (2.2), put $p_{b+1}=\lfloor t+(b+1)\lambda\rfloor$. This exceeds $\lfloor t+b\lambda\rfloor\ge p_b$ because $\lambda>1$.

Let $b_h$ be the largest $b$ with $2^{p_b(w)+1}\le Y$ for every $w\in F_b$. Then $b_h=46$ on the record pair and $44$ on the 72G pair, since $p_b<k+1+\lambda b$.

• For $b\ge b_h$: $\mathrm{Cap}_b\ge N_b\ge N_{b_h}$, by the second fact.
• For $b<b_h$: group the words of $F_{b_h}$ by their length-$b$ prefix. In the tree of extensions, the children of a node have distinct values $p_{i+1}>p_i$. So $\sum_{e\ge1}2^{-e}\le1$ gives $A_{b_h}\le A_b$. Hence $\mathrm{Cap}_b\ge YA_b/2\ge YA_{b_h}/2=\sum_{w\in F_{b_h}}Y/2^{p_{b_h}(w)+1}\ge N_{b_h}$.

The values of $N_{b_h}$ are exact (Table 1). $\square$

Table 1. Exact $N_{b_h}=|F_{b_h}|$ in rational arithmetic (exact_window.py, table1.py), for any start $t\in[0,k+1)$.

| pair | $k$ | $b_h$ | $N_{b_h}$ | $\log_2(N_{b_h}/u)$ | |---|---|---|---|---| | record | 1 | 46 | $9{,}174{,}117{,}179{,}520$ | $6.060$ | | record | 2 | 46 | $75{,}745{,}110{,}171{,}940{,}416$ | $19.071$ | | 72G | 1 | 44 | $2{,}810{,}089{,}946{,}880$ | $5.285$ | | 72G | 2 | 44 | $15{,}526{,}656{,}239{,}520{,}555$ | $17.717$ |

3. The exact count

By (2.3) and (2.4), $T_b(Y)$ is a signed sum over boxes $B$ and end values $P$ of

$$ T(B,P)=\sum_{w\in B,\ p_b=P}\bigl(q_Y+[x_w\le r_Y]\bigr),\qquad M=2^{P+1},\ \lfloor Y\rfloor=q_YM+r_Y . $$

The first part is $q_Y\,|B\cap\{p_b=P\}|$. A dynamic program over the rows of the box computes it, together with the forward and backward reachability of each value of $p_i$. For the second part, fix a split row $s$. Modulo $M$,

$$ x_w=\alpha-\beta,\qquad \alpha=3^{-b}2^{P}-\sum_{i=s}^{b-1}3^{-i-1}2^{p_i},\qquad \beta=\sum_{i<s}3^{-i-1}2^{p_i}, $$

because $3^{-b}c_w=\sum_{i<b}3^{-i-1}2^{p_i}$. Prefix $(p_0,\dots,p_{s-1})$ and suffix $(p_s,\dots,p_{b-1})$ interact only through $p_s\ge p_{s-1}+1$. Given the last prefix value $a$ and the first suffix value $a'$, every pair is admissible when $a'>a$ and none is when $a'\le a$.

So for each admissible cell $(a,a')$ we need $\#\{(\alpha,\beta):(\alpha-\beta)\bmod M\le r_Y\}$ over two lists. Sort both lists. For fixed $\alpha$, the admissible $\beta$ form one or two intervals of the circle $\mathbb Z/M$. Two monotone pointers count them in linear time after sorting.

The program mitm.cpp does this in unsigned 128-bit arithmetic, which is exact modulo $2^{128}$, with $P+1\le126$ asserted. It finds $3^{-1}$ mod $2^{128}$ by Newton iteration and asserts $3\cdot3^{-1}=1$. It picks $s$ to minimise the total list length. As a check, it compares the number of words of each box with the dynamic-programming count and aborts on any mismatch. Lists are reachability-pruned, so their total length is about $\sqrt{|B|}$.

Validation.

• Brute force (mitm_test.py). We enumerated all odd $x\le2^L-1$, computed their Syracuse words and tested (2.2) directly. This was done for $k=1$, $b=4,\dots,16$ ($L=14$); for $k=2$, $b=4,8,12,16$ ($L=13$); and for $3x+355$ with $k=1$, $b=6,10,14$ ($L=15$). Every count agrees with mitm, and every word total agrees with an independent box enumeration (boxes.py).
• Large brute force. At $\lfloor Y\rfloor=2^{26}-1$ the program gives $T_{30}=104$ and $T_{34}=16$, equal to a separate C brute force over all $2^{25}$ odd integers (resid_count2.c).
• Negative control (ctrl_neg.py). The map $3x+355$ has a cycle with $u=17$, $n=27$, least state $30449$ and canonical height $1$. Lemmas 1–3 hold verbatim for $3x+c$, with $c$ in place of $1$ in $c_w$. With that cycle's own $\lfloor Y\rfloor=161740$, mitm gives $T_b=1332$, $206$, $39$, $18$ at $b=12$, $16$, $20$, $24$. These equal the brute force, and each is $\ge u=17$, as Corollary 4 requires.

The record pair. At $k=1$, $b=60$ there are 243 boxes and $P\in\{94,\dots,97\}$, with $N_{60}=37{,}018{,}604{,}205{,}637{,}632=2^{55.04}$. The prefix and suffix lists of one box total at most $5.1\cdot10^7\approx2^{25.6}$ entries. The four runs (one per $P$) took 19 minutes on four cores:

| $P$ | 94 | 95 | 96 | 97 | total | |---|---|---|---|---|---| | $T(\cdot,P)$ | 20,417,266,279 | 31,163,010,824 | 12,089,143,192 | 805,923,283 | 64,475,343,578 |

The total is $0.468816\,u<u$, so Corollary 4 gives Theorem 2 for the record pair. The per-$P$ word counts, and $A_{60}$, agree exactly with the independent box enumeration (exact_window.py, rational arithmetic), file xcheck_b60.txt. At $b=62$ the count is $T_{62}=23{,}372{,}382{,}138=0.169946\,u$.

The 72G pair. At $k=1$, $b=56$ there are 227 boxes, $N=2^{51.63}$ and $T_{56}=42{,}111{,}244{,}638=0.584412\,u$.

4. An independent Fourier bound

The Fourier bound uses neither the split nor sorting, and it bounds $T_{60}$ by a different route. At $b=60$ every $P$ has $M=2^{P+1}\ge2^{95}>Y$, so each class has at most one member below $Y$, and $T_b(Y)=\#\{w\in F_b:x_w<Y'\}$ for any real $Y'\in(\lfloor Y\rfloor,\lfloor Y\rfloor+1]$. A longer interval still gives an upper bound. We take $Y'=\lfloor Y\rfloor+534{,}287$, the least double above $\lfloor Y\rfloor$ (Yup_k1.txt). This raises the main term by less than $10^{-17}u$. Selberg's majorant $\phi\ge1_{[0,Y'/M)}$ [9, Ch. 1] for an interval of length $Y'/M$ on $\mathbb R/\mathbb Z$, with degree $H$, has Fourier coefficients satisfying $\hat\phi(0)=Y'/M+1/(H+1)$ and

$$ |\hat\phi(h)|\le\min\Bigl(\frac{Y'}{M},\frac1{\pi|h|}\Bigr)+\frac1{H+1}\qquad(0<|h|\le H). $$

Then $[x_w<Y']\le\phi(x_w/M)$. Expanding $\phi$ in its Fourier series and summing over $F_b$ and over $P$,

$$ T_b(Y)\le\mathcal M'_b+\frac{N_b}{H+1}+\sum_P\sum_{0<|h|\le H}\Bigl(\min\Bigl(\frac{Y'}{M},\frac1{\pi|h|}\Bigr)+\frac1{H+1}\Bigr)|S_h(P)|, \tag{4.1} $$

where $\mathcal M'_b=\sum_{w}Y'/2^{p_b(w)+1}=Y'A_b/2$ and

$$ S_h(P)=\sum_{w\in F_b,\ p_b=P}e\Bigl(\frac{h\,x_w}{2^{P+1}}\Bigr). $$

Since $x_w=3^{-b}2^P-\sum_{i<b}3^{-i-1}2^{p_i}$ mod $M$, the phase is a product over rows. So $S_h(P)$ is a transfer sum over each signed box of (2.3), computed by the same row dynamic program as the word count. The program fcert_iv.c computes each phase residue $r=h\,3^{-i-1}2^{p}\bmod M$ exactly in 128-bit integers, and it forms $e(-r/M)$ without any library function. The top 64 bits of $r/M$, read as four 16-bit digits $j_1,\dots,j_4$, select four table entries $e(-j_k/2^{16k})$, and the phase is their product. The remaining bits are dropped; they change the angle by less than $2\pi\cdot2^{-64}$. The tables are computed once in mpmath interval arithmetic at 160 bits (tabs_iv.py). Each stored double is within $0.697\,\mathbf u$ of the exact value, $\mathbf u=2^{-53}$. An independent 60-digit recomputation confirms this (tabs_check.py). After the tables, the program uses only the IEEE-754 operations $+,-,\times,/$ and $\sqrt{\ }$, compiled with -ffp-contract=off. We use $|S_{-h}|=|S_h|$.

The run. At $b=60$, $H=2^{21}$ and all $h\le H$ for all four $P$ ($8{,}388{,}608$ sums), the three terms of (4.1) are

$$ 0.4688\,u+0.1284\,u+0.000932\,u . $$

A rigorous rounding bound (Lemma 5) adds $3.6\cdot10^{-6}\,u$. In exact rational arithmetic the inflated total is $0.59810305\,u<0.599\,u$ (final_check.py; cert_k1_b60_v3.txt gives the same $0.5981\,u$ in 50-digit arithmetic), which proves Theorem 2 again. The run took 4 minutes on four cores. The version 1 program fcert.c used glibc's cexp for the phases. Its sixteen partial sums agree with fcert_iv to within $2\cdot10^{-5}$ relative, a difference that comes from its rounding of $\log_2Y$ in the weights, and it gives the same total. A Python version (expsum.py) agrees with fcert on sampled $(h,P)$. It is itself checked against brute-force sums over all words for $k=1$, $b=6$ and $k=2$, $b=5$.

The sums are small. Of the $|S_h(P)|/\sqrt{N_P}$, 89.8% are below $0.1$ and 98.7% are below $1$; the largest is $345$ ($h=2{,}047{,}825$, $P=96$). The bound succeeds because the weights $1/(\pi h)$ are summable against $|S_h|\ll\sqrt N$, and $\sqrt{N_{60}}=2^{27.5}$ is small compared with $u=2^{37.0}$.

Lemma 5 (rounding). This lemma concerns the supplied inputs $b=60$, $k=1$, $H=2^{21}$, $n_{\rm box}=243$. Let $\mathbf u=2^{-53}$ and $\gamma=2051\,\mathbf u$. For each $(h,P)$, the computed sum satisfies $|\tilde S-S|\le\gamma\sum_B|B\cap\{p_b=P\}|$, an absolute error measured against unsigned path counts. Consequently Theorem 2 holds if

$$ \sum_P N_P\frac{Y'}{2^{P+1}}+\frac{N_b}{H+1}+\frac{1001}{1000}F_{\rm print}+\gamma\,\Sigma\,\frac{187}{15}<u, $$

where $F_{\rm print}=128{,}161{,}992$ is the sum of the printed weighted sums, each rounded up by half a unit in its last printed digit, $187/15>2((\ln H+1)/\pi+1)$, and $\Sigma=\sum_{P,B}|B\cap\{p_b=P\}|=182{,}779{,}358{,}265{,}335{,}808$. The left side is $0.59810305\,u$.

Proof. The residue is exact, and the first multiplication, from $1+0i$, is exact. The four table entries are each within $0.697\,\mathbf u$ of the exact unit complex numbers. Their product takes three complex multiplications, and we use the elementary bound of $3\mathbf u$ normwise relative error for each (the sharp bound for the usual formula is $\sqrt5\,\mathbf u$ [12]; the elementary one suffices here). The dropped bits contribute less than $2\pi\cdot2^{-64}<0.01\,\mathbf u$. So each phase has error at most $(1+0.697\mathbf u)^4(1+3\mathbf u)^3-1+0.01\mathbf u<24\mathbf u$, with no first-order approximation. At $b=60$, $k=1$ every row of every box has width $\le2$, so each accumulator has at most two terms and each row adds one inexact addition. Induction on rows over unsigned path counts gives, for each box value, an absolute error at most $((1+24\mathbf u)^b(1+3\mathbf u)^{b-1}(1+\mathbf u)^{b+n_{\rm box}}-1)$ times its path count. Here $b-1$ multiplications of error $\le3\mathbf u$ combine the phases, and the $b$ row additions and $n_{\rm box}$ box additions each add at most $\mathbf u$. The coefficient is $1920.0000000002\,\mathbf u<\gamma$. Cancellation in the signed box sum is harmless because the bound is absolute, not relative to $|S|$. The modulus $\sqrt{\Re^2+\Im^2}$, the weights, the $2^{21}$-term summation and the 7-digit printing are covered by the factor $1001/1000$ and the upward rounding of $F_{\rm print}$. The weights are upper bounds for these inputs. In the $L$ branch, for each $P=94,\dots,97$ the final rounded addition exceeds $L+1/(H+1)$, by exactly $1/19342822337206103650074624$. In the reciprocal branch, a rational bound using $\pi>3.1415926$ pays for the multiplication, reciprocal, remainder and final addition roundings uniformly for $h\le2^{21}$. $Y'$ is the least double $\ge\lfloor Y\rfloor+1$ (yup.py), and $Y'/M$ is formed exactly by scaling by $2^{-(P+1)}$. These two weight facts were checked by lyapunov_weaver in a pre-read. $\square$

Lemma 5 assumes IEEE-754 round-to-nearest for the basic operations and the correctness of mpmath's interval arithmetic for the 262,144 table entries. No library trigonometric or exponential function evaluates a phase; $\log_2$ appears only in diagnostic output. The program also calls sqrt, which IEEE-754 requires to be correctly rounded, and fmin and ldexp, which are exact here. The final comparison is done in exact rationals (final_check.py). The proof of Section 3 uses no floating point at all.

5. The discrepancy past the horizon

Write $d_b=T_b(Y)-\mathcal M_b$, where $\mathcal M_b=(\lfloor Y\rfloor+1)A_b/2$ is the average of (2.4) over the position of the classes. We computed $T_b$ exactly by mitm for $k=1$ on the record pair at every even $b$ from $46$ to $62$, and at $b=40$. $A_b$ is exact. The horizon is $b_h=46$, and at $b=62$ the depth is $p_b/\log_2Y\approx1.3$.

Table 2. $k=1$, record pair. $\theta_{\rm emp}=\log_2|d_b|/\log_2N_b$.

| $b$ | $T_b/u$ | $d_b$ | $\lvert d_b\rvert/\sqrt{\mathcal M_b}$ | $\log_2N_b$ | $\log_2(N_b/T_b)$ | $\theta_{\rm emp}$ | |---|---|---|---|---|---|---| | 40 | 11482 | $+2.72\cdot10^4$ | 0.0007 | 37.92 | −12.56 | 0.388 | | 46 | 568.5 | $-8.93\cdot10^4$ | 0.010 | 43.06 | −3.09 | 0.382 | | 48 | 206.0 | $+2.10\cdot10^5$ | 0.040 | 44.79 | +0.10 | 0.395 | | 50 | 75.20 | $+9.67\cdot10^4$ | 0.030 | 46.51 | +3.27 | 0.356 | | 52 | 27.46 | $+6.03\cdot10^5$ | 0.31 | 48.22 | +6.43 | 0.398 | | 54 | 9.872 | $+2.76\cdot10^5$ | 0.24 | 49.93 | +9.62 | 0.362 | | 56 | 3.584 | $+2.03\cdot10^5$ | 0.29 | 51.63 | +12.79 | 0.342 | | 58 | 1.2997 | $-7.57\cdot10^4$ | 0.18 | 53.33 | +15.95 | 0.304 | | 60 | 0.46882 | $+4.68\cdot10^4$ | 0.18 | 55.04 | +19.13 | 0.282 | | 62 | 0.16995 | $+8.71\cdot10^4$ | 0.57 | 56.75 | +22.30 | 0.289 |

At every depth $|d_b|\le0.57\sqrt{\mathcal M_b}$, which proves Theorem 3. This compares error scales: $\sqrt{\mathcal M_b}$ is the standard deviation of a Poisson count with mean $\mathcal M_b$. The counts are deterministic, and ten finite inequalities do not show that any distribution is Poisson. Meanwhile the excess of the class bound grows by about $1.6$ bits per step. On the 72G pair at $b=56$: $d=+1.03\cdot10^4=0.05\sqrt{\mathcal M}$ and $\theta_{\rm emp}=0.258$. For $k=2$ at $b=50$ on the record pair, all six slices $P=77,\dots,82$ give $T_b=311009661404256280$, $d=+9.42\cdot10^7=0.17\sqrt{\mathcal M}$ and $\theta_{\rm emp}=0.436$ (depth $1.02$, just past the $k=2$ horizon).

Brute force over all odd $x<2^{26}$ and $x<2^{32}$ (resid_count2.c) gives the same picture at toy scale, and further out. There $\log_2(T_b/\mathcal M_b)$ stays within $0.3$ bit up to depth $2.1$ ($k=1$) and $3.3$ ($k=2$), where $T_b\approx16$. Over the same range $N_b/T_b$ reaches $2^{29}$–$2^{77}$.

The input needed for larger $k$. Consider the statement

$$ \mathrm{EQ}_k(\theta):\qquad T_b(Y)\le\mathcal M_b+N_b^{\theta}\,(\lambda b+1)\quad\text{for all }b, $$

on the record pair at height $k$ (with $Y$ from Lemma 1). By Corollary 4, $\mathrm{EQ}_k(\theta)$ excludes $W\le k$ there iff $\min_b(\mathcal M_b+N_b^\theta(\lambda b+1))<u$. The largest such $\theta$ is:

Table 3. Threshold $\theta^*_k$ and the $b$ where it is attained. This is a numerical optimization (theta_need2.py). $N_b$ and $A_b$ come from floating-point profiles, checked against the exact counts of exact_window.py where those are feasible, and $b$ was searched up to $420$ ($540$ for $k=8$). The search range is justified as follows. Every $\theta$ in the optimization is below $\log_2u/\log_2N_b$, and $N_b$ is nondecreasing in $b$ (proof of Theorem 1). So $b$ beyond the range can attain $\theta^*_k$ only if $\log_2N_b<\log_2u/\theta^*_k$ at the end of the range. The profile shows it is not (theta_tail.py). The values are rounded to three digits.

| $k$ | 1 | 2 | 3 | 4 | 6 | 8 | |---|---|---|---|---|---|---| | $\theta^*_k$ | 0.536 | 0.245 | 0.150 | 0.107 | 0.067 | 0.050 | | $b$ | 60 | 101 | 146 | 193 | 284 | 368 | | $p_b/\log_2Y$ | 1.24 | 2.05 | 2.93 | 3.83 | 5.49 | 6.95 |

The measure alone falls below $u$ at every height: $\mathcal M_b<u$ from $b=59,98,142,187$ for $k=1,\dots,4$. These are numerical observations on the computed profiles. Over the searched range, $-b^{-1}\ln A_b$ settles near $0.511$, $0.315$, $0.223$ and $0.173$ nats per step for $k=1,\dots,4$. These are measured decay rates, not proved limits, and we do not prove that $A_b$ decays for every $k$. For the heights in Table 3, then, the measure is not what stands in the way of an exclusion on this pair. What stands in the way is the size of the discrepancy past the horizon. Table 2 measures $\theta_{\rm emp}\le0.40$ for $k=1$ at depth $\le1.3$. That is below $\theta^*_1$ but above $\theta^*_2$; the depth where $k=2$ needs it is $2.05$, which Table 2 does not reach. At toy scale, $\theta_{\rm emp}\le0.22$ at depth $1.1$–$1.3$ and $\le0.05$ at depth $\ge2$. These measurements are evidence, not proof, and the toy values of $Y$ are $2^{45}$ times smaller.

$\mathrm{EQ}_k(\theta)$ is an equidistribution statement for actual orbit itineraries at $2$-adic depth beyond $\log_2Y$. It is the Terras-horizon problem [8] for words restricted to a band, taken past the horizon. Tao's estimates [10] control the distribution of Syracuse iterates over a number of steps at most a small constant times $\log Y$, inside the horizon. We know of no result for any depth beyond it.

6. Why both methods stop at $k=1$

The exact count. The lists of Section 3 have length about $\sqrt{|B|}$ per box. For $k=2$ on the record pair, $\theta^*_2$ is attained at $b=101$. There $N_b=2^{118.8}$ and the lists would have about $2^{59}$ entries, so the method is out of reach. At $b=50$ the lists are already about $2^{29}$ per box, and one run of six $P$-slices takes about 3 hours.

The Fourier bound. The second term of (4.1) is below $u$ only if $H>N_b/u$. For $k=2$ that means $H>2^{78.4}$ already at $b=98$, the first $b$ with $\mathcal M_b<u$, and more at larger $b$. The third term is also a limitation of the present estimates. By Parseval over $h\in\mathbb Z/M$, the mean square of $|S_h(P)|$ is $N_P$. This is a mean over the whole frequency group. It does not say how large a typical coefficient is, and it gives no lower bound for the weighted low-frequency sum in (4.1). It does set the scale: if the $|S_h|$ with $h\le H$ were of size $\sqrt{N_b}$, that size alone would be $2^{57.7}>u$ at $b=98$, where the measure first drops below $u$. We have no estimate that makes the third term small there. The computational route also stops, since $N_b/(H+1)<u$ needs $H>2^{78.4}$ transfer sums. So excluding $k\ge2$ by (4.1) would need either such an estimate or cancellation in the signed sum over $h$. That signed sum is the discrepancy itself.

The class with no height bound. The ladder's next step needs every cycle on the record pair, not only bounded height. By the product identity $\varepsilon\ln2=\sum_j\ln(1+1/(3X_j))<\frac13\sum_j1/X_j$ (the inequality behind [5, Theorem 11]), such a cycle has harmonic occupancy $V\cdot u^{-1}\sum_j1/X_j>3\varepsilon\ln2\cdot V/u=0.0463$ at $V=2^{71}$. Since its least state $m$ exceeds $V$, at least $0.015u$ of its slots have $X_j\le2^5m$. A count over windows starting at those slots with height up to about 30 would need $\theta\approx0.023$ at depth about $11\log_2Y$. This estimate is heuristic (lowslot.py) and is given only to place bounded height on the scale.

7. What is not proved

• Heights $k\ge2$. Theorem 2 is for $W\le1$ only. Theorem 3 measures the discrepancy for $k=1$, and for $k=2$ at one depth. Nothing here proves $\mathrm{EQ}_k(\theta)$ for any $k$ or any $\theta<1$.
• Other pairs. We ran two pairs. Each pair is a separate computation, and we claim only those two. The 72G pair is already excluded by [5].
• The record. Theorem 2 does not move the cycle-length record. A cycle on the record pair with $W\ge2$ is not excluded by anything here.
• Uniformity in $c$. $\mathrm{EQ}_k(\theta)$ is stated for $c=1$ on the record pair, where $b\le368\ll u$. It cannot hold uniformly in $c$ once the window is longer than the cycle. For $c=D=2^n-3^u$ every band word is a cycle of $3x+D$, so $T_b\ge u\cdot\#\{\text{necklaces}\}$ at every $b$. For $k=1$ this exceeds $\mathcal M_b+(\sqrt{\mathcal M_b}+b)b$ from $b\approx1.3u$ on ($u\le40$). At $b\le u$ the same maps show $|d_b|\le0.95\sqrt{\mathcal M_b}$ ($u\le14$, brute force).
• Floating point. The Fourier certificate of Section 4 assumes IEEE-754 basic arithmetic and correct interval enclosures for its phase tables (Lemma 5). Its exponential sums are evaluated numerically, with a rigorous error budget; they are not exact. Section 3 makes no floating-point assumption, and Theorem 2 rests on it.
• Numerical tables. Table 3 and the decay rates quoted in Section 5 are numerical optimizations and measurements on floating-point profiles. They are guidance, not proved equidistribution or asymptotic results.
• Toy measurements. The values $\theta_{\rm emp}$ at $Y=2^{26},2^{32}$ are empirical and do not bound anything at $Y=2^{77}$.

Appendix A. Files

The supplement holds every program and output below, a README with the exact command lines, and SHA-256 sums. The C programs were compiled with g++ -O2 -march=native (mitm.cpp) and gcc -O2 -march=native ... -lm (fcert.c, resid_count2.c). The Python scripts need python3, numpy and mpmath.

• Lemma 1, $\lfloor Y\rfloor$. yi.py (80 digits); Yi_k1.txt, Yi_k2.txt, Yi_72_k1.txt.
• Boxes and exact $N_b$, $A_b$. boxes.py writes the signed boxes of (2.3) with the exact per-$P$ word counts; exact_window.py counts $F_b$ and $A_b$ in rational arithmetic by a separate route; xcheck_b60.txt compares the two at $b=60$.
• Theorem 1. budget_thm.py, budget_thm_out.txt (any-start rows).
• Theorem 2, exact count. mitm.cpp; bx_k1_b60.txt; mt_k1_b60_P{94..97}.out (totals) and .err (per-box log, including the word-count check); mt_k1_b60.out. For the 72G pair: bx72_k1_b56.txt, mt72_k1_b56*.
• Validation. mitm_test.py and mitm_test_out.txt (brute force, three families); mt_30.out, mt_34.out and horizon_26_k1.txt (resid_count2.c at $2^{26}$); ctrl_neg.py, ctrl_neg_out.txt and mitm_ctrl_out.txt (the $3x+355$ cycle).
• Theorem 1, Table 1. table1.py, table1_out.txt (exact $N_{b_h}$ by exact_window.py).
• Theorem 2, Fourier bound. expsum.py (Python transfer sums, brute-force checked); fcert_iv.c with tabs_iv.py (interval-enclosed phase tables, tabs.bin regenerated by the script; its SHA-256 is listed), tabs_iv_out.txt, tabs_check.py and tabs_check_out.txt; run60iv_{0..3}.txt (four ranges of $h$), yup.py and Yup_k1.txt, sumnq_b60.txt ($\Sigma$ of Lemma 5), combine.py and cert_k1_b60_v3.txt; final_check.py and final_check_out.txt (the exact-rational comparison of Lemma 5). lw_preread/ holds lyapunov_weaver's independent checker of the phase tables, the rounding bound, the weights and the final budget (MIT licence; it does not replay the transfer sums). Version 1 files: fcert.c, run60_{0..3}.txt, out60_*.txt (large coefficients) and cert_k1_b60_v2.txt. phase_check.c and phase_check.py sample $10^6$ libm phases of fcert.c against 60-digit values (maximum error $9.57\,\mathbf u$).
• Theorem 3, Table 2. run_profile.sh, bx_k1_b{40,46,...,62}.txt, mt_k1_b*.out, disc.py, profile_k1.txt.
• Table 3 and Section 5. theta_need2.py with measure_np.py and count_np.py (floating-point profiles, checked against exact_window.py), theta_need2_out.txt; theta_tail.py and theta_tail_out.txt (search range); horizon_test.py, resid_count2.c, horizon_{26,32}_k{1,2}.txt and emp_theta.py (toy scale).
• Section 6. lowslot.py (heuristic).

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