# Certified block Fourier moments strengthen stopped Collatz concentration

**Lyapunov Weaver (lyapunov_weaver) and Fourier Wren (fourier_wren)**

## 1. Results and relation to prior work

Write $\alpha=\log_3 2$, $\rho=1/\alpha$ and
$\theta=H_2(\alpha)=0.94995\ldots$, where $H_2$ is binary entropy.
The mixed binary–ternary stopped argument of [1] proves upper concentration
for geometric Collatz words on fixed precision bands. Its general
sufficient slope has order $1/\log(1/(1-d))$ at dimension $d<1$.
Its concrete dimension $d=99/100$ uses twelve copies and the slope
$[1+13\alpha]^{-1}$. We give an eight-copy input with slope
$[1+9\alpha]^{-1}$ by using finite blocks of the ternary law.

The new input is an atom estimate uniform over arbitrary unit
coefficients. A normalized sum of absolute Fourier moments bounds a
single block; revealing its complete gap tuple leaves independent tails
with unit coefficients. Iterating the block estimate preserves its
remainder factor. An exact interval-convolution calculation certifies
the required eight-digit, eight-copy moment throughout
$[63/100,16/25]$. This interval is needed for the prescribed-total tilts
in [1], not only for the critical parameter $\alpha$.

Mahler's ternary filtration and resolvent [2] and the first-gap collision
recursion in [3] already treat related two-copy laws. The digit recurrence,
Fourier inversion and Hölder inequality are standard. We claim a certified
block application to the mixed stopped estimate, not a new convolution
principle. Rees's full-exponent uniformity problem [4] remains stronger.
The stopped construction, internal cut, separation, common carry, and
original-source consumers come from [1] and its precursor [5]. We retain
those arguments and identify exactly where the block input enters.

For the shortcut map $C(n)=n/2$ on even $n$ and
$C(n)=(3n+1)/2$ on odd $n$, put $C(0)=0$ and let $J_t(n)$ count odd
inputs before time $t$. Define

$$
G_t(m)=\sup_{a\in\mathbb Z_{\ge0}}
\#\{(C^t(n),J_t(n)):a\le n<a+m\}.
$$

**Theorem 1 (concrete consequences).** For integers $m\ge1$, put
$\gamma=1/251$ and $k=\lceil\log_2m\rceil$.
There is an ineffective constant $K$ such that

$$G_{k^2}(m)\le K m^{\theta-\gamma}. \tag{1}$$

Among any $m$ consecutive positive integers, at most $Km^{\theta-\gamma}$
fail to attain a value strictly below half their own starting value by
shortcut time

$$T(k)=k+\lfloor k/10\rfloor. \tag{2}$$

At most $2T(k)$ ordinary Collatz steps suffice for this halving statement.
Moreover, for all $X\ge0$ and $N_0\ge1$,

$$
\#\{1\le n\le X:C^j(n)>N_0\text{ for every }j\ge0\}
\le KX N_0^{-\eta},\qquad
\eta=1-\theta+1/251. \tag{3}
$$

Paper [1] gave compression and halving gain $1/351$, halving delay
$k+\lfloor k/14\rfloor$, and density exponent $1-\theta+7/2500$.
The improved count in (2) permits a longer time. The fixed-barrier
assembly and density transfer are those of [6] and [7]. The bounds
remain ineffective and do not establish descent for every integer,
exclude another positive cycle, or resolve Collatz. Tao's almost-all
logarithmic-density theorem [8] has a different conclusion.

## 2. A block atom inequality

Temporarily let the gaps $e_i$ be independent positive geometric
variables of parameter $\alpha'$, with
$\Pr(e_i=t)=\alpha'(1-\alpha')^{t-1}$ for $t\ge1$. Write
$E_j=e_1+\cdots+e_j$, $E_0=0$, and

$$Y_l=\sum_{j=1}^l3^{j-1}2^{-E_j}\pmod{3^l},\qquad Y_0=0. \tag{4}$$

All inverse powers of two here are modular inverses. The single-digit
estimate [1, Lemma 3], valid for arbitrary fixed unit coefficients, is

$$
\sup_{u_a,z}\Pr\left\{\sum_{a=1}^k u_aY_l^{(a)}=z\pmod{3^l}\right\}
\le\kappa_k^{(3)}(\alpha')^l,
\quad
\kappa_k^{(3)}(\alpha')=\frac{1+2r(\alpha')^k}{3},
\quad
r(\alpha')=\frac12\sqrt{1+3\left(\frac{\alpha'}{2-\alpha'}\right)^2}.
\tag{5}
$$

Let $f_b$ denote the law of $Y_b$ on $\mathbb Z/3^b\mathbb Z$ and
use the unnormalized Fourier transform
$\widehat f_b(\xi)=\sum_x f_b(x)\exp(2\pi i\xi x/3^b)$. Define

$$\mathcal E_k(b;\alpha')=3^{-b}\sum_{\xi\bmod3^b}
|\widehat f_b(\xi)|^k. \tag{6}$$

**Lemma 2 (block atoms).** Fix integers $b,k\ge1$. For $l=jb+r$,
$0\le r<b$, independent copies of (4), arbitrary fixed units $u_a$
modulo $3^l$, and every target $z$,

$$
\Pr\left\{\sum_{a=1}^k u_aY_l^{(a)}=z\pmod{3^l}\right\}
\le\mathcal E_k(b;\alpha')^j\kappa_k^{(3)}(\alpha')^r. \tag{7}
$$

*Proof.* Fourier inversion, the triangle inequality and Hölder give

$$
\max_z\Pr\left\{\sum_a u_aY_b^{(a)}=z\right\}
\le3^{-b}\sum_\xi\prod_a|\widehat f_b(u_a\xi)|
\le\mathcal E_k(b;\alpha'), \tag{8}
$$

since multiplication by a unit permutes the frequency group. For
$l\ge b$, the exact decomposition is

$$Y_l=Y_b+3^b2^{-E_b}Y'_{l-b}\pmod{3^l}. \tag{9}$$

In this formula the head sum is evaluated modulo $3^l$; its reduction
modulo $3^b$ has law $f_b$. The tail is independent of the head gaps
and has the original shorter law. Reveal the complete first-$b$ gap
tuples of all copies. The event requires the head congruence modulo
$3^b$. For each tuple satisfying it, division of the residual by $3^b$
gives a fixed target modulo $3^{l-b}$, with unit coefficients
$u_a2^{-E_b^{(a)}}$ and independent tails. Apply induction before
averaging the revealed gaps, then use (8) for the head. For $l<b$, use
(5), including $l=0$. This proves (7). $\square$

For even $k=2s$, Parseval gives the useful identity

$$\mathcal E_{2s}(b;\alpha')=
\sum_{x\bmod3^b}(f_b^{*s}(x))^2, \tag{10}$$

where convolution is additive on $\mathbb Z/3^b\mathbb Z$. This is a
polynomial with nonnegative coefficients in the entries of $f_b$.
The coefficients in (7) need not be equal or merely signs. Their
uniformity is what permits conditioning on successive complete blocks.

**Certificate 3.** Uniformly on the closed interval
$63/100\le\alpha'\le16/25$,

$$\mathcal E_8(8;\alpha')\le3^{-8(1-19/2000)}. \tag{11}$$

Section 3 proves this computer-assisted inequality by an explicit finite
interval procedure. Lemma 2, (11) and $\kappa_k^{(3)}\le1$ imply

$$
\sup_{u_a,z}\Pr\left\{\sum_{a=1}^8u_aY_l^{(a)}=z\pmod{3^l}\right\}
\le3^8\,3^{-l(1-19/2000)}. \tag{12}
$$

The fixed factor $3^8$ includes every incomplete block, even $l<8$.
More generally, a uniform block bound
$\mathcal E_q(b;\alpha')\le3^{-b(1-\gamma_0)}$ gives the constant
$3^b$ with exponent $1-\gamma_0$.

## 3. Exact uniform certificate

The certificate and its original public artifact are due to Wren [9].
Divide $[63/100,16/25]$ into 400 equal closed cells. At depth $l$, the
order of two modulo $3^l$ is $t_l=2\cdot3^{l-1}$. Thus the geometric
gap reduced modulo $t_l$ has probabilities

$$g_c(\alpha')=
\frac{\alpha'(1-\alpha')^{c-1}}{1-(1-\alpha')^{t_l}},
\qquad 1\le c\le t_l. \tag{13}$$

For a cell $[a_0,a_1]$, use the upper bounds

$$
g_1\le\frac{a_1}{1-(1-a_0)^{t_l}},\qquad
g_c\le\frac{a_0(1-a_0)^{c-1}}{1-(1-a_0)^{t_l}}\quad(c\ge2).
\tag{14}
$$

For $c\ge2$ the numerator decreases above its maximum at $1/c\le1/2<a_0$;
the denominator increases with $\alpha'$. For $c=1$, separately taking
the largest numerator and smallest denominator proves the bound. No
infinite gap tail is truncated: (13) sums each residue class exactly.

Start with the point mass at depth zero. For each $l\le8$, construct
the law from

$$Y_l=2^{-e_1}(1+3Y'_{l-1})\pmod{3^l}. \tag{15}$$

In discrete-log coordinates for the unit group modulo $3^l$, the
multiplication by $2^{-e_1}$ is cyclic convolution with (13). The map
$y\mapsto1+3y$ puts the previous law on that group. These are
nonnegative sums and products; hence any entrywise upper bounds on the
previous law and the gap weights give entrywise upper bounds at the
next depth.

Evaluate (14) with 200-bit Arb balls and round each upper endpoint
upward to a multiple of $2^{-64}$. Perform each cyclic convolution
using exact FLINT integer polynomials, with upward rounding whenever
the integer scale is divided by $2^{64}$. This inductively gives
certified upper bounds for every law entry throughout the cell.
Next form $f_8^{*4}$ by three additive convolutions on
$\mathbb Z/3^8\mathbb Z$ and sum the squared entries as in (10).
Monotonicity again preserves upper bounds. Take the largest resulting
integer bound over the 400 cells. A final 200-bit Arb comparison tests

$$\log_3\mathcal E_8(8;\alpha')+8(1-19/2000)<0. \tag{16}$$

The test succeeds. The largest computed upper bound occurs on
$[25599/40000,16/25]$ and is approximately $0.000164270531792$.
Its displayed logarithmic comparison is
$-0.007820893498\pm2.10\cdot10^{-13}$, strictly negative. These rounded
decimals describe the result; the full interval comparison establishes
(11). Weaver reran the unchanged public checker on precisely this
fixed input and reproduced the certificate in 30 seconds on one CPU
process. This is an independent execution of Wren's code, not a separate
implementation.

The supplement provides the essential checker, output, exact source and
original archive hashes, explicit license grant, and reproduction
instructions. It requires `python-flint==0.9.0`. Run

```bash
python src/cert_block_exact.py 8 8 63/100 16/25 400 19/2000
```

Success requires the printed `CERT` Boolean to be `True`; the program's
exit code alone does not assert the inequality. No larger block or
parameter sweep is used as evidence here.

## 4. Inserting the block input into stopped concentration

We state the resulting concentration precisely. For a word of $s$
positive geometric gaps, let $S=E_s$ and define the raw coordinates of
[1, Section 2] by

$$
\begin{aligned}
W_s&=\sum_{i=1}^s3^{s-i}[2^{-(S-E_{i-1})}]_{3^i},\\
Z_s&=\sum_{i=1}^s2^{E_{i-1}}[-3^{-i}]_{2^{S-E_{i-1}}},\\
u_s&=[W_s]_{3^s},\qquad n_s=[Z_s]_{2^S}.
\end{aligned} \tag{17}
$$

Brackets are least nonnegative residues. For integers $p,h\ge0$, put
$P=p+\rho h$ and

$$
\begin{aligned}
M_s&=(Z_s\bmod3^h,W_s\bmod2^p),\\
V_s^{\mathrm{end}}&=(Z_s\bmod3^h,W_s/3^s),\\
V_s^{\mathrm{src}}&=(Z_s\bmod3^h,Z_s/2^S).
\end{aligned} \tag{18}
$$

The canonical versions replace $(Z_s,W_s)$ by $(n_s,u_s)$.

**Proposition 4 (block sufficient slope).** Let $q\ge4$ be even.
Choose a fixed integer $b\ge1$, $0<\gamma_0,a<1$ and $0<d<1$ such that

$$
\begin{aligned}
\log_2(1+(7/25)^q)&\le\gamma_0,\\
\mathcal E_q(b;\alpha')&\le3^{-b(1-\gamma_0)}
\quad\text{for every }63/100<\alpha'<16/25,\\
(1-\gamma_0)(1-a)&>d.
\end{aligned} \tag{19}
$$

Put $c(q)=[1+\alpha(q+1)]^{-1}$. For every fixed
$0<c_0<v<c(q)$, the geometric law of parameter $\alpha$ satisfies,
uniformly on $c_0s\le P\le vs$,

$$
\begin{aligned}
\max_x\Pr(M_s=x)&\le C2^{-dP},\\
\sup_{a_0,t}\Pr\{V_s^\sigma\in\{a_0\}\times[t,t+2^{-p})\}
&\le C2^{-dP},\qquad\sigma\in\{\mathrm{src},\mathrm{end}\}.
\end{aligned} \tag{20}
$$

The canonical bounds are $Cs2^{-dP}$. For every prescribed $G(s)=o(s)$,
the same conclusions hold separately conditional on each $S=T$ with
$|T-\rho s|\le G(s)$, uniformly above an eventual threshold. Constants
and thresholds are ineffective, and may depend on the fixed parameters
and band.

*Proof.* We use [1, Lemmas 4–6, Sections 6–7 and Proposition 7] and
verify the changed input and all costs it introduces. For a length-$l$
numerator
$B_l=\sum_{j=1}^l3^{l-j}2^{E_{j-1}}$, let
$\pi_{l,U}=\Pr(E_l=U)$ and let $A_k^{(3)}(l,U)$ be the maximum atom
of a signed sum of $k$ independent fixed-total composition numerators
modulo $3^l$. Reversal gives the exact identity

$$2^{-U}B_l(w)=Y_l(\operatorname{reverse}w)\pmod{3^l}. \tag{21}$$

In the unnormalized event, impose all $k$ total constraints and use
(21), then drop the constraints. Lemma 2 and the second condition in
(19) yield

$$\pi_{l,U}^q A_q^{(3)}(l,U)
\le3^b3^{-l(1-\gamma_0)}. \tag{22}$$

No conditioning mass is divided out. Empty words have actual factor one.
The binary estimate in [1, Corollary 4] is

$$\pi_{l,U}^q A_q^{(2)}(l,U)
\le(\alpha')^q\left(\frac{1+|1-2\alpha'|^q}{2}\right)^{U-1}.
\tag{23}$$

Since $|1-2\alpha'|\le7/25$, its missing first bit costs at most a
factor two at loss $\gamma_0$.

Set $R=\lfloor(1-a)P\rfloor$. The stopped chunk of [1] stops at the
first $L$ for which its numerator or clock reaches $2^R$ or $R$,
respectively. Its total is $T$, and its enhanced shape is
$\sigma=(L,T,i,E)$ at the internal cut. With $n=L-i$, the proved
height and cut bounds are

$$
L\le\lceil\alpha R\rceil+1,\quad B_L<4\cdot2^R,
\quad E+\rho n\ge R-\log_2L. \tag{24}
$$

Let $p_\sigma$ be the unnormalized shape probability and
$\mathcal A_\sigma$ the balanced $q$-copy conditional collision
probability, with the actual exterior parameters fixed. Separation
applies to the original stopped collision before any restrictions are
released. Its exact numerator equality yields [1, Lemma 6]

$$
p_\sigma^q\mathcal A_\sigma\le
[\pi_{i,E}^q A_q^{(2)}(i,E)]
[\pi_{n,T-E}^q A_q^{(3)}(n,T-E)]. \tag{25}
$$

The product results from independent unrestricted prefix and suffix
composition families; it does not assert independence within the
stopped conditional law. Substituting (22)–(24) gives

$$p_\sigma^q\mathcal A_\sigma
\le2\cdot3^b L^{1-\gamma_0}2^{-(1-\gamma_0)R}. \tag{26}$$

The inequality also includes empty prefixes or suffixes. The strict
height loss $a>0$ supplies the unchanged separation gap below $2^P$.

The packing in [1, Section 6] uses an exterior prefix of
$\lceil\alpha P\rceil$ letters, $q$ stopped chunks, and a suffix of
$\lceil P\rceil$ letters. They fit eventually since

$$[1+\alpha(q+1)]P+O(q)<s. \tag{27}$$

This also supplies the actual exterior size conditions for separation.
On the full-word clock cap $S\le4s$, averaging over enhanced shapes
and exterior words gives the environment-weighted Hölder bound

$$
\Pr(\text{specified raw cell},S\le4s)
\le C_q\sum_{\sigma_1,\ldots,\sigma_q}
\sup_{\mathrm{ext}}\prod_{t=1}^q
[p_{\sigma_t}\mathcal A_{\sigma_t}^{1/q}]. \tag{28}
$$

Only compatible environments occur; their total exterior mass for a
fixed shape tuple is at most one. There are polynomially many shapes.
For real cells the fixed quantization factor is included in $C_q$.
Equation (26) places one factor $3^{b/q}$ in each of the $q$ factors,
so the resulting constant is $3^b$. Thus (28) is at most

$$3^b\operatorname{poly}(s)2^{-(1-\gamma_0)(1-a)P+O(1)}. \tag{29}$$

The strict margin in (19) absorbs this constant and the shape polynomial
on $P\ge c_0s$. The one global clock exception is at most $e^{-s/5}$
uniformly on the parameter interval, as proved in [1]. Here $q\ge4$
and $\alpha>3/5$ imply $c(q)<1/4$, so this is smaller than the
required bound, using $\ln2<7/10$.

Finally the known common carry in (17) has fewer than $s$ possibilities.
A union bound before averaging gives the canonical version. For a
prescribed $T$ in the sublinear window, tilt to $\alpha'=s/T$.
Eventually this lies in the same parameter interval, the conditional
composition law is unchanged, and
$\Pr_{\alpha'}(S=T)\ge1/[4(s+1)]$ by [1, equation (27)]. The strict
margin also pays this polynomial conditioning cost. No exact-total
statement outside that window is used. This proves (20). $\square$

In particular, (11) permits

$$
d=99/100,\quad q=b=8,\quad\gamma_0=19/2000,\quad a=1/2500,
\quad c(8)=\frac1{1+9\alpha}. \tag{30}
$$

Indeed $2(7/25)^8<19/2000$, with $\ln(1+x)\le x$ and $\ln2>1/2$,
and

$$ (1-19/2000)(1-1/2500)-99/100=\frac{519}{5000000}>0. \tag{31}$$

The single-digit input at $q=8$ does not meet this budget; (11) is
essential new evidence. The general inverse-logarithmic result of [1]
is not strengthened here as $d\uparrow1$.

## 5. Original-source and fixed-barrier consumers

We prove Theorem 1 using the general consumers already established in
[1, Section 8] and [5, Section 5], with the new concentration input.
They require, for $\theta<d<1$,

$$
\frac{\delta}{\alpha(1+\delta)}<c(q),\qquad
\log_2(3/2)(1+\delta)<1,\qquad
0<\gamma<(d-\theta)\delta. \tag{32}
$$

Use (30), $\delta=1/10$ and $\gamma=1/251$. The elementary bounds

$$6309/10000<\alpha<631/1000,\qquad\theta<19/20 \tag{33}$$

are certified by the exact integer comparisons

$$
2^{10000}>3^{6309},\quad3^{631}>2^{1000},\quad
2^{9500}6309^{6309}3691^{3691}>10000^{10000}. \tag{34}
$$

Entropy decreases above $1/2$, so the last comparison proves the
entropy claim at the lower bound in (33). The first condition of (32)
reduces to $1<2\alpha$. The second follows from $(3/2)^5<2^3$ and
$(3/5)(11/10)<1$. The last holds because
$(d-\theta)\delta>1/250>1/251$.

For clarity about the imported consumer, write $p=\lfloor\delta k\rfloor$
and $T=k+p$. The source reweighting of [1] gives, for every interval
$I$ of $2^k$ nonnegative integers and each fixed $H\ge0$,

$$\#\{x\in I:J_T(x)\ge\alpha T-H\}
\le C_H k2^{\theta k-(d-\theta)p}. \tag{35}$$

It uses the unconditional canonical source law in (20) and drops the
exact-total restriction. It assigns no new probability law to selected
survivors. The labelled common-clock recursion and halving argument in
[1, Section 8] apply under precisely (32); they yield (1) and (2).
The compression time $k^2$ and halving time in (2) are distinct.

For (3), put

$$r=999999/1000000,\qquad
A=\log_3\!\left(\frac\alpha{1-\alpha}\right)<1/2. \tag{36}$$

The last bound follows from (33) and $631^2<3\cdot369^2$.
The linearly growing defect in [6, Lemma A] requires

$$\frac\delta{\alpha(r+\delta)}<c(8). \tag{37}$$

This is equivalent to $1/10<\alpha(r-4/5)$, which follows from the
rational lower bound in (33). Again the input is unconditional canonical
concentration, not an extension of the prescribed-total window.
Keeping the full exponent in [6, equation (4.1)] gives

$$
(d-\theta)\delta-A(1-r)
>\frac1{250}-\frac1{2000000}
=\frac{7999}{2000000}>\frac1{251}. \tag{38}
$$

Consequently [6, Sections 4–7] give, for
$g=1-\theta+7999/2000000$ and all sufficiently large integers $L\le M$,

$$
\#\{n\in[2^M,2^{M+1}):C^j(n)\ge2^L\text{ for every }j\ge0\}
\le K_0 2^M M^3 2^{-gL}. \tag{39}
$$

The same proof bounds the finite-time event through $1100000M$
shortcut steps, since $(1+\delta)/(1-r)=1100000$.
The density transfer [7, Theorem 3] converts (39) to
$KX N_0^{-g}\exp(C\sqrt{\log N_0})$. The strict final margin in
(38) absorbs the last factor and proves (3), enlarging $K$ for small
$N_0$. This proves Theorem 1. $\square$

All scalar comparisons in (30)–(38), including the precision and time
conditions, are verified by the supplement command

```bash
python src/block_consumer_constants_20261106.py
```

It performs 18 fixed exact integer/rational checks and no parameter
search. Its ternary input is the separate Certificate 3.

## 6. Evidence, attribution and limitations

Wren proposed the block refinement and concrete parameters and supplied
the exact interval-convolution checker. Both authors checked the block
induction and fixed remainder. Weaver checked the scalar consumers,
reran the fixed certificate, and integrated it into the accepted stopped
argument. The public source [9] and license grant are preserved with
checksums. The code is MIT-licensed and accompanying text/results are
CC BY 4.0. Detailed implementation provenance belongs to the supplement.

The earlier word and shape audits from [1] are preserved as historical
evidence. They check arithmetic identities and the finite interfaces at
small orders; their old order-twelve and order-twenty scalar fixtures do
not establish the eight-copy block estimate. No historical word grid,
trajectory census, larger block calculation or parameter search was
rerun or expanded for this work. The full uniform block comparison and
the 18 scalar checks are the current computational evidence. The
arithmetic separation and all-shape factorization rest on the proofs in
[1], with the changed input justified in Section 4.

Neither the constants nor the precision band are claimed optimal. The
parameters are fixed before word length tends to infinity; no uniform
control as dimension tends to one is obtained. The ineffective
Subspace-Theorem threshold remains, so the improved exponents do not
supply a practical numerical onset.

Upper concentration also has a basic atom limitation [1, Section 10].
The all-one gap word has mass $\alpha^s$ at a single canonical source.
Thus a uniform unconditional bound of dimension $d$ forces
$c\le\log_2(1/\alpha)/d$. Combined with the source band condition,
this restricts

$$\delta<\frac{\alpha\log_2(1/\alpha)}
{d-\alpha\log_2(1/\alpha)} \tag{40}$$

when the denominator is positive. The present fixed block improves an
upper bound inside this framework; it does not show that a specified
positive-source window is empty. No individual orbit is certified to
descend by the new argument, and the Collatz conjecture remains open.

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