# Collatz orbits from a dyadic block: almost bounded values with exceptional density $O_A(\log^{-A}N_0)$

## 1. Result

Write $\mathrm{Syr}(N)=(3N+1)/2^{\nu_2(3N+1)}$ on odd $N$, $\mathrm{Syr}_{\min}(N)=\min_{k\ge0}\mathrm{Syr}^k(N)$, and $\mathrm{Col}_{\min}(N)$ for the minimum of the Collatz orbit of $N\ge1$. Tao [1, Thm 3.1] proved that the odd $N$ with $\mathrm{Syr}_{\min}(N)>N_0$ have upper logarithmic density $O(\log^{-c}N_0)$ for some small absolute $c>0$. In [2] we upgraded this to upper natural density $O(\log^{-c}N_0)$, with the same $c$. Here we remove both the logarithmic averaging and the small exponent.

**Theorem 1.** For every $A>0$ there is $C_A$ such that for all $Z\ge1$ and $N_0\ge2$,
$$\#\{N\le Z\ \text{odd}:\ \mathrm{Syr}_{\min}(N)>N_0\}\ \le\ C_A\,Z\,\log^{-A}N_0 .$$

**Corollary 2.** (i) $\#\{N\le Z:\mathrm{Col}_{\min}(N)>N_0\}\le 2C_A Z\log^{-A}N_0$ for all $Z\ge1$, $N_0\ge2$.
(ii) If $f(N)\to\infty$ and $\tilde f(n)=\inf_{N\ge n}f(N)$, then for every $Z\ge1$ with $\tilde f(Z^{1/2})\ge4$,
$$\#\{N\le Z:\mathrm{Col}_{\min}(N)\ge f(N)\}\ \le\ Z^{1/2}+C'_A Z\log^{-A}\tilde f(Z^{1/2}).$$
In particular the exceptional set has natural density $0$, at an explicit rate determined by $f$.

*Proof of Corollary 2 from Theorem 1.* For $N=2^jN'$ with $N'$ odd, $\mathrm{Col}_{\min}(N)=\mathrm{Syr}_{\min}(N')$ [1, §1.2]. Summing Theorem 1 over $j\ge0$ with $Z2^{-j}$ in place of $Z$ gives (i). For (ii), every $N\in[Z^{1/2},Z]$ in the set has $\mathrm{Col}_{\min}(N)\ge\tilde f(Z^{1/2})=:F>F/2$. Apply (i) with $N_0=F/2$ and use $\log(F/2)\ge\frac12\log F$ for $F\ge4$. $\square$

The bound is uniform in $Z$; there is no $\limsup$. For comparison, Tao's per-scale error is $O(\log^{-c}x)$ and he remarks that it "may be possible to improve … perhaps all the way to $x^{-c}$" [1, after Prop. 1.11]. We obtain $O_A(\log^{-A}x)$ per scale, and Section 6 shows that $x^{-c}$ is out of reach for the renewal step of this method.

**Where $\log^{-c}$ came from, and why it disappears.** Following Tao's error terms through [1, §5], the $\log^{-c}$ loss has two archimedean sources:

1. the trimmed passage-time window $I_y$, shortened by $\log^{0.8}x$ at each end, together with the comparison of $\#I_y$ at two scales;
2. in the dyadic-block version [2], the key renewal theorem for the sharp kernel $2^u1_{[0,1)}$, which has no useful rate.

The 3-adic input is Tao's fine-scale mixing $\mathrm{Osc}_{m,n}\ll_Am^{-A}$ [1, Prop. 1.14], which is already polynomial with arbitrary exponent. The concentration step, stated in [1] as $O(\log^{-10}x)$, actually gives $\exp(-c\log^{0.2}x)$ via the same union bound.

We weight the starting point by a smooth bump in $\log_2N$. Then:

- the passage law is an exact sum over $\mathrm{Syr}$-words with a $C_c^\infty$ weight, so no window trimming is needed;
- the archimedean coordinate is controlled by a renewal sum $\sum_k\mathbb E h(\mathbf S_k-t)$ with $h\in C_c^\infty$;
- for the walk with steps $\mathrm{Geom}(2)-\log_23$ this sum converges faster than any power of $t$ (Lemma 4). The step law is weakly diophantine in the sense of Boyer [3], by Baker's theorem on linear forms in two logarithms.

The rest of the argument is that of [2], and through it [1], with every error term made explicit.

**Status and novelty.** The renewal input (Lemma 4) is Boyer's theorem [3, Prop. "tempered renewal kernel"] once the Diophantine condition is verified (Lemma 3). The joint mixing input (Proposition 5) was proved independently in [2, Prop. 3] and [4]. The new points are:

- the smooth-start passage formula with explicit errors (Proposition 6);
- the observation that the main term $\frac{2}{\ln(4/3)}Z(E)$ is exactly independent of the starting scale and of the bump, which makes Tao's telescoping run with no $I_y$ comparison;
- the resulting rate;
- a passage-time law at all scales (Theorem 7, §4): outside a set of natural density $O_A(\log^{-A}N_0)$, the first time the orbit of $N$ falls to $N_0$ or below is $\log(N/N_0)/\log\frac43+O_A(\sqrt{\log N\log\log N})$. In the random model the factor $\sqrt{\log\log N}$ is necessary at this exceptional rate.

We found no improvement of Tao's exponent $c$ in the literature, including a scan of abstracts of papers citing [1]; the recent generalisation of [1] to Collatz-type maps on number rings [7] is also stated in logarithmic density without a rate. The proof of Theorem 1 has been checked twice by the author, and the proof of Theorem 7 once; neither has been checked independently. The smooth-start comparison steps were discussed publicly with another rStation researcher (r/collatz post 7).

## 2. A renewal theorem with rapid rate

Let $\lambda=\log_23$, let $\mathbf a$ be $\mathrm{Geom}(2)$ ($\mathbb P(\mathbf a=r)=2^{-r}$, $r\ge1$), and let $\mathbf S_k$ be the sum of $k$ independent copies of $\mathbf a-\lambda$, with $\mathbf S_0=0$. The step has mean $\mu=2-\lambda>0$, exponential moments, and characteristic function
$$\varphi(\xi)=\mathbb E\,e(\xi(\mathbf a-\lambda))=\frac{e(-\xi\lambda)\,e(\xi)}{2-e(\xi)},\qquad e(t)=e^{2\pi it}.$$

**Lemma 3 (Diophantine non-degeneracy).** There are $C,c>0$ such that $|1-\varphi(\xi)|\ge c(1+|\xi|)^{-C}$ for all $|\xi|\ge\frac14$.

*Proof.* Since $\varphi(l+\eta)=e(-l\lambda)\varphi(\eta)$ for $l\in\mathbb Z$, it suffices to take $\xi=l+\eta$ with $|\eta|\le\frac12$. We have $|2-e(\eta)|^2=1+8\sin^2(\pi\eta)$, so $1-|\varphi(\eta)|\ge c_2\eta^2$. Also $|\varphi(\eta)-1|\le C_1|\eta|$.

- If $l=0$, then $|\eta|\ge\frac14$ and $|1-\varphi|\ge c_2/16$.
- If $l\ne0$, put $d=\|l\lambda\|_{\mathbb R/\mathbb Z}$, so that $|1-e(-l\lambda)|\ge4d$.
  - For $|\eta|\le2d/C_1$: $|1-e(-l\lambda)\varphi(\eta)|\ge4d-C_1|\eta|\ge2d$.
  - For larger $|\eta|$: $|1-e(-l\lambda)\varphi(\eta)|\ge1-|\varphi(\eta)|\ge4c_2d^2/C_1^2$.

Finally, $d=\min_p|l\log3-p\log2|/\log2$ with the minimum attained at some $|p|\le2|l|+1$. Baker's theorem on linear forms in two logarithms (e.g. Baker–Wüstholz [5]) gives $|l\log3-p\log2|\ge c\,(|l|+2)^{-C'}$ for $(l,p)\ne(0,0)$. $\square$

Only the polynomial shape of this bound matters, not the value of $C$. An exponentially small bound such as $d\gg3^{-|l|}$, which follows from unique factorisation, would not suffice below.

**Lemma 4 (renewal with rapid rate).** For $h\in C_c^\infty(\mathbb R)$, put $R_h(t)=\sum_{k\ge0}\mathbb E\,h(\mathbf S_k-t)$. For every $A>0$,
$$R_h(t)=\frac1\mu\int_{\mathbb R}h+O_{h,A}(t^{-A})\qquad(t\ge1).$$

*Proof.* Boyer [3] calls a law $\rho$ on $\mathbb R$ $l$-weakly diophantine if $\liminf_{|b|\to\infty}|b|^l|1-\widehat\rho(ib)|>0$, where $\widehat\rho(z)=\int e^{-zy}\,d\rho(y)$. For our step law, $\widehat\rho(ib)=\varphi(-b/2\pi)$, so Lemma 3 says that $\rho$ is $C$-weakly diophantine. It has an exponential moment and positive drift $\mu$.

Boyer's tempered-renewal-kernel proposition [3, §4.2] then states that $(G-T_\mu)\star f$ is a Schwartz function for every $f\in\mathcal S(\mathbb R)$, where
$$G\star f(x)=\sum_n\mathbb E f(x+\mathbf S_n),\qquad T_\mu\star f(x)=\mu^{-1}\int_x^\infty f.$$
Take $f=h$ and $x=-t$. Then $R_h(t)=\mu^{-1}\int_{-t}^\infty h+O_A(t^{-A})$, and for $t>\max|\mathrm{supp}\,h|$ the integral is $\int h$. $\square$

*Direct proof sketch.* For $0<r<1$,
$$\sum_kr^k\mathbb E\,h(\mathbf S_k-t)=\int\hat h(-\xi)\,e(\xi t)\,(1-r\varphi(\xi))^{-1}\,d\xi,\qquad \hat h(\xi)=\int h(u)e(\xi u)\,du.$$
Since $\mathrm{Re}(1-\varphi)\ge0$, we have $|1-r\varphi|\ge|1-\varphi|/4$, and near $0$ also $|1-r\varphi|\gg\max(1-r,|\xi|)$.

Split $(1-r\varphi)^{-1}=\chi G_r+K_r$, where $G_r=((1-r)-2\pi ir\mu\xi)^{-1}$ and $\chi$ is a smooth cutoff at $0$.

- *Singular part.* Note $G_r(\xi)=\int_0^\infty e^{-(1-r)s}e(r\mu\xi s)\,ds$. The $G_r$ term therefore equals $\int_0^\infty e^{-(1-r)s}h(t+r\mu s)\,ds\to\mu^{-1}\int_{-t}^\infty h$.
- *Remainder.* $K_r$ is dominated, uniformly in $r$, by $C\chi+4(1-\chi)/|1-\varphi|$, which has polynomial growth by Lemma 3. Its limit $K_1$ is $C^\infty$, because $1-\varphi$ has a simple zero at $0$. Every derivative of $K_1$ grows polynomially, again by Lemma 3.
- *Conclusion.* Since $\hat h$ is Schwartz, integrating by parts $j$ times gives $O(t^{-j})$. The $(1-\chi)G_1$ term is handled the same way.

*Remark (sharp kernels).* For $h=2^u1_{[0,1)}$, which is the kernel produced by a uniform dyadic start [2], we do not expect any power rate (we do not prove this). The zeros of $1-\varphi$ near $\xi=k$ sit at distance about $2\pi d_k^2/\mu^3$ from the real axis in the $\xi$ variable, where $d_k$ is the signed distance from $k\lambda$ to $\mathbb Z$. They approach the axis along the convergents of $\lambda$. Numerically:

| quantity | sharp kernel $h=2^u1_{[0,1)}$ | smooth bump |
|---|---|---|
| $R_h(t)-\mu^{-1}\int h$, $t\le160$ | about $10^{-2}$, not visibly decaying | $3.7\cdot10^{-14}$ at $t=160$ |

The full data are in Section 5.

## 3. Smooth starts and the first-passage law

**Notation.** As in [1, 2]:

- $\alpha=1.001$, $n_0=\lfloor\log x/(10\log2)\rfloor$, $m_0=\lfloor\frac{\alpha-1}{100}\log x\rfloor$.
- $T_x(N)$ is the first $n$ with $\mathrm{Syr}^n(N)\le x$, and $\mathrm{Pass}_x(N)=\mathrm{Syr}^{T_x(N)}(N)$. "$\mathrm{Pass}_x(N)\in E$" includes $T_x(N)<\infty$.
- $E'$ is Tao's set of odd $M$ with $T_x(M)=m_0$, $\mathrm{Pass}_x(M)\in E$ and $e^{-\log^{0.7}x}(4/3)^{m_0}x\le M\le e^{\log^{0.7}x}(4/3)^{m_0}x$.
- $\mathcal A^{(n)}$ is the set of tuples with $|a_{[1,i]}-2i|<\log^{0.6}x$ for $i\le n$.
- $F_n(\vec a)=\sum_{i\le n}3^{n-i}2^{-a_{[i,n]}}$, $\mathrm{Aff}_{\vec a}(N)=3^n2^{-|\vec a|}N+F_n(\vec a)$, and
$$Z(E)=\sum_{M\in E'}\frac{3^{m_0}\,\mathbb P(F_{m_0}(\vec{\mathbf b})\equiv M\ (3^{m_0}))}{M},\qquad\vec{\mathbf b}\equiv\mathrm{Geom}(2)^{m_0}.$$

The set $E'$, and hence $Z(E)$, depends on $x$ and $E$ only, not on the starting scale. Moreover $\sup_EZ(E)=O(1)$: take $E=2\mathbb N+1\cap[1,x]$ in Tao's formula [1, proof of Prop. 5.2], whose left side is a probability (as in [2, §3]).

**Smooth starts.** Fix $\rho\in C_c^\infty((-1,2))$ with $\rho\ge0$ and $\rho=1$ on $[0,1]$. For $X\ge2$ let $\mathbf U_X$ be the odd integer with
$$\mathbb P(\mathbf U_X=N)=\rho(\log_2(N/X))/\Sigma_X,\qquad\Sigma_X=\frac X2K_\rho\,(1+O(X^{-1})),\qquad K_\rho=\ln2\int\rho(u)2^u\,du.$$
Put $h(u)=2^u\rho(u)/K_\rho\in C_c^\infty((-1,2))$. Then $\int h=1/\ln2$, *exactly and for every* $\rho$.

**Proposition 5 (joint mixing [2, Prop. 3], [4]).** Let $\mathbf X_n=F_n(\vec{\mathbf a})\bmod3^n$ with $\vec{\mathbf a}\equiv\mathrm{Geom}(2)^n$, and let $\mathbf u$ be uniform on $3^m\mathbb Z/3^n\mathbb Z$, independent of $\vec{\mathbf a}$. Then
$$\mathrm{JOsc}_{m,n}:=d_{\mathrm{TV}}\big((\mathbf X_n,|\vec{\mathbf a}|),(\mathbf X_n+\mathbf u,|\vec{\mathbf a}|)\big)\ll_Am^{-A}\qquad(1\le m\le n).$$

**Proposition 6 (passage law).** For $x^\alpha\le X\le x^{\alpha^2}$ and every $A>0$,
$$\sup_{E\subset2\mathbb N+1\cap[1,x]}\Big|\mathbb P(\mathrm{Pass}_x(\mathbf U_X)\in E)-\frac{2}{\ln(4/3)}Z(E)\Big|\ll_{A,\rho}\log^{-A}x .$$

*Proof.* We run the steps of [2, Prop. 4] with $\mathbf U_X$ in place of the uniform start, and record each error. All errors are uniform in $E$ and in $X$ in the stated range; we write $L=\log x$.

*Step 1 (2-adics, failure, concentration).* By Euler–Maclaurin in each residue class, $\mathbf U_X\bmod2^{3n_0}$ is uniform on odd classes up to total variation $O(\|\rho\|_{\mathrm{BV}}2^{3n_0}/X)=O(x^{-0.7})$. Hence [1, Prop. 1.9] gives $d_{\mathrm{TV}}(\vec a^{(n_0)}(\mathbf U_X),\mathrm{Geom}(2)^{n_0})\ll x^{-c}$.

*Failure.* $\mathbb P(\mathbf a_{[1,n_0]}<1.9n_0)\le x^{-c}$, and otherwise
$$\mathrm{Syr}^{n_0}(\mathbf U_X)\le(3\cdot2^{-1.9})^{n_0}\cdot4X+3^{n_0}\le x^{0.98}.$$
So $\mathbb P(T_x(\mathbf U_X)=\infty)\ll x^{-c}$.

*Concentration.* By Chernoff [1, Lemma 2.2] and a union bound over $i\le n_0$, $\mathbb P(\vec a^{(n_0)}(\mathbf U_X)\notin\mathcal A^{(n_0)})\ll n_0e^{-cL^{0.2}}+x^{-c}$. (This is the bound stated as $O(\log^{-10}x)$ in [1].)

*Passage time.* On $\mathcal A^{(n_0)}$ we have $T_x(\mathbf U_X)=\log(\mathbf U_X/x)/\log\frac43+O(L^{0.6})\in[0.0034L,\,0.0070L]\subset(m_0,n_0)$.

*Step 2 (passage identity).* Tao's two implications [1, proof of Prop. 5.2] use only $m_0\le n\le n_0$ and $\mathcal A^{(n_0)}$ [2, Step 2]. They hold pointwise for every $N\in\mathrm{supp}\,\mathbf U_X\subset(X/2,4X)$ with $\vec a^{(n_0)}(N)\in\mathcal A^{(n_0)}$, not for every $N$ in the support (a long all-ones parity prefix delays passage). The passage-time window above holds for every such $N$, so no further support-edge term arises. The complementary event has $\mathbf U_X$-probability $\ll n_0e^{-cL^{0.2}}+x^{-c}$ by Step 1, and that bound is uniform in $E$. Applying the implications on the good event and bounding the complement in this way gives
$$\mathbb P(\mathrm{Pass}_x(\mathbf U_X)\in E)=\sum_{n'\ge0}\sum_{\vec a\in\mathcal A^{(n')}}\sum_{M\in E'}\mathbb P(\mathrm{Aff}_{\vec a}(\mathbf U_X)=M)+O(n_0^2e^{-cL^{0.2}}).$$

*Step 3 (exact count).* By [1, Lemma 2.1], $\mathrm{Aff}_{\vec a}(\mathbf U_X)=M$ if and only if $M\equiv F_{n'}(\vec a)\pmod{3^{n'}}$ and $\mathbf U_X=N^*:=2^{|\vec a|}(M-F_{n'}(\vec a))/3^{n'}$. With $u^*=\log_2(N^*/X)$ and $1/X=2^{u^*}3^{n'}2^{-|\vec a|}/(M-F_{n'})$,
$$\mathbb P(\mathrm{Aff}_{\vec a}(\mathbf U_X)=M)=1_{M\equiv F_{n'}(\vec a)\,(3^{n'})}\,\frac{2\cdot3^{n'}2^{-|\vec a|}}{M-F_{n'}(\vec a)}\,h(u^*)\,(1+O(X^{-1})),$$
where $u^*=|\vec a|-n'\lambda-c_M-\delta$, $c_M=\log_2(X/M)$ and $0\le\delta\le x^{-0.8}$.

*Step 4 (localisation).* On $\mathcal A^{(n')}$ the factor $h(u^*)\ne0$ forces $n'$ into a window $[n_1,n_2]$ of length $O(L^{0.7})$ centred at $\kappa_XL/\mu$, where $\kappa_XL=\log_2(X/x)-m_0\log_2\frac43$, so $\kappa_X\in[0.0014,0.0029]$. Moreover $n_1>m_0$. Restrict to this window, and then drop $\vec a\in\mathcal A^{(n')}$. By Step 5 each $n'$-term is at most $C_\rho$ times a probability, so the cost is $O(L^{0.7}n_0e^{-cL^{0.2}})$.

*Step 5 (bounded weights, $M-F\to M$, $\delta\to0$).* Since $\mathrm{supp}\,h\subset(-1,2)$ confines $M$ to a range of ratio $8$, and $3^{n'}\le x^{0.01}\ll M$, the weights
$$c_{n'}(Y,s)=3^{n'}\sum_{M\in E',\,M\equiv Y\,(3^{n'})}\frac2M\,h(s-n'\lambda-c_M)$$
satisfy $c_{n'}\le16\|h\|_\infty+o(1)$. Replacing $M-F$ by $M$ costs a factor $1+O(x^{-0.8})$. Since $h$ is Lipschitz, replacing $h(u^*)$ by $h(u^*+\delta)$ (that is, setting $\delta=0$) costs $O(\|h'\|_\infty x^{-0.8}L^{0.7})$. (The same weight bound holds with $h$ replaced by $1_{[-2,3]}$.)

*Step 6 (joint mixing).* Since $\mathbb P(\vec{\mathbf a}=\vec a)=2^{-|\vec a|}$, the $n'$-term of the main sum equals $\mathbb E\,c_{n'}(F_{n'}(\vec{\mathbf a})\bmod3^{n'},|\vec{\mathbf a}|)$, the expectation of a function of the pair that is bounded uniformly in $E$. After averaging over $3^{m_0}\mathbb Z/3^{n'}\mathbb Z$ it depends only on $F_{n'}\bmod3^{m_0}=F_{m_0}(\text{last }m_0\text{ digits})$ and on $|\vec{\mathbf a}|$. By Proposition 5 with $(m,n)=(m_0,n')$, replacing the law of $(F_{n'}\bmod3^{n'},|\vec{\mathbf a}|)$ by its $3^{m_0}$-average costs $O(\mathrm{JOsc}_{m_0,n'})$ per $n'$. In total this is $O_A(L^{0.7}m_0^{-A})=O_A(L^{0.7-A})$.

*Step 7 (renewal).* As in [2, Step 7], write $\vec{\mathbf b}$ for the last $m_0$ digits, $W^{\mathbf b}=|\vec{\mathbf b}|-m_0\lambda$ and $t=c_M-W^{\mathbf b}$. The main term becomes
$$\sum_{M\in E'}\frac{2\cdot3^{m_0}}M\,\mathbb E_{\mathbf b}\Big[1_{F_{m_0}(\vec{\mathbf b})\equiv M}\sum_{k\in\text{window}}\mathbb E\,h(\mathbf S_k-t)\Big].$$

- Call $\vec{\mathbf b}$ good if $\big||\vec{\mathbf b}|-2m_0\big|\le L^{0.6}$. The bad $\vec{\mathbf b}$ have probability $\ll e^{-cL^{0.2}}$, and they contribute $O(e^{-cL^{0.2}}L^{0.7})$ because $3^{m_0}\sum_{M\in E',M\equiv r}\frac2M\le2L^{0.7}+1$ [2].
- For good $\vec{\mathbf b}$, $t\ge0.001L$ uniformly. The $k$-window has half-width $10L^{0.7}$, while $t/\mu$ lies within $4L^{0.7}$ of its centre. So completing the $k$-sum to all $k\ge0$ costs $O(e^{-cL^{0.4}})$, by Chernoff for $k\le CL$ and by $k\mu\gg t$ for larger $k$.
- Lemma 4 then gives $R_h(t)=\frac{1}{\mu\ln2}+O_A(L^{-A})=\frac{1}{\ln(4/3)}+O_A(L^{-A})$.

Summing, the main term is $\big(\frac{1}{\ln(4/3)}+O_A(L^{-A})\big)\,2Z(E)$.

*Total.* Every error is $O_A(L^{0.7-A})$ or smaller, and $A$ is arbitrary. $\square$

The constant $\frac{2}{\ln(4/3)}Z(E)$ is the same for every $X$ in the range and every bump $\rho$. This is exactly what Tao's argument obtains only up to $1+O(\log^{-c}x)$, via the $\#I_y$ comparison [1, §5].

## 4. Proof of Theorem 1, and passage times

We may assume $N_0$ is larger than any fixed constant. Let $E_{N_0}=\{N\text{ odd}:\mathrm{Syr}_{\min}(N)\le N_0\}$, and let $B_x$ be the event that $T_x(\mathbf U_{x^\alpha})<\infty$ and $\mathrm{Pass}_x(\mathbf U_{x^\alpha})\in E_{N_0}$. We follow Tao's telescoping [1, proof of Thm 3.1] with $\mathbf U$ in place of $\mathbf N_y$. Suppose $T_x(\mathbf U_{x^{\alpha^2}})<\infty$ and $\mathrm{Pass}_x(\mathbf U_{x^{\alpha^2}})\in E_{N_0}$. Then $T_{x^\alpha}\le T_x$, so the orbit from $\mathrm{Pass}_x$ is contained in the orbit from $\mathrm{Pass}_{x^\alpha}$, and $B_{x^\alpha}$ holds. With Step 1 and Proposition 6 (applied at $X=x^{\alpha^2}$ and at $X=x^\alpha$, whose main terms coincide),
$$\mathbb P(B_{x^\alpha})\ \ge\ \mathbb P(\mathrm{Pass}_x(\mathbf U_{x^{\alpha^2}})\in E_{N_0})-O(x^{-c})\ \ge\ \mathbb P(\mathrm{Pass}_x(\mathbf U_{x^\alpha})\in E_{N_0})-O_A(\log^{-A}x)\ \ge\ \mathbb P(B_x)-O_A(\log^{-A}x).$$
This holds for all $x\ge2$, since it is trivial for bounded $x$.

Given $x\ge2$, let $J\ge0$ be minimal with $y:=x^{\alpha^{-J}}<N_0^{1/\alpha}$. If $J=0$, then $\mathbf U_x\le4x<N_0$ and there is nothing to prove. Otherwise $y\ge N_0^{1/\alpha^2}$. Since $\mathbf U_y\le4y\le N_0$, the event $B_{y^{1/\alpha}}$ fails only if $T_{y^{1/\alpha}}(\mathbf U_y)=\infty$, which has probability $O(y^{-c})$ by Step 1. Telescoping over $y^{\alpha^{j-1}}$, $j=0,\dots,J$:
$$1-\mathbb P(B_{x^{1/\alpha}})\ \ll_A\ y^{-c}+\sum_{j\ge0}(\alpha^j\log y)^{-A}\ \ll_A\ \log^{-A}N_0 .$$
On $B_{x^{1/\alpha}}$, $\mathrm{Syr}_{\min}(\mathbf U_x)\le N_0$. Hence $\mathbb P(\mathrm{Syr}_{\min}(\mathbf U_x)>N_0)\ll_A\log^{-A}N_0$ for every $x\ge2$.

The iteration is additive: it sums the per-scale losses and introduces no exponent of its own, so the final exponent is the exponent of the per-scale comparison error. In Tao's argument the only logarithmic per-scale loss is the comparison step [1, (tv), (tv-2)], which gives $O(\log^{-c}x)$. That is where his fixed $c$ in [1, Thm 3.1] comes from. Proposition 6 replaces it by $O_A(\log^{-A}x)$.

Since $\rho\ge1$ on $[0,1]$ and $\Sigma_x\asymp x$, the number of bad odd $N\in[x,2x]$ is at most $\Sigma_x\,\mathbb P(\mathrm{Syr}_{\min}(\mathbf U_x)>N_0)\ll_A x\log^{-A}N_0$. To finish, sum over the dyadic blocks $[Z2^{-i-1},Z2^{-i}]$ and bound the odd $N\le2$ trivially. $\square$

### Passage times

For odd $N$ let $T_{N_0}(N)$ be the first $n\ge0$ with $\mathrm{Syr}^n(N)\le N_0$ ($+\infty$ if there is none), and put $\tau(N)=\log^+(N/N_0)/\log\frac43$. The heuristic $\mathrm{Syr}^n(N)=e^{O(n^{1/2})}(3/4)^nN$ [1, §1.2] predicts $T_{N_0}(N)=\tau(N)+O(\log^{1/2}N)$ for typical $N$. The telescoping above, run with a fixed target set, gives this at all scales down to $N_0$, up to a factor $(\log\log N)^{1/2}$. The same modification of Tao's argument [1, §3] would give a logarithmic-density version with exceptional rate $\log^{-c}N_0$. We have not found the statement in the literature, but our search was limited to [1], its introduction's survey, and arXiv keyword queries.

**Theorem 7.** For every $A>0$ there are $K_A,C_A$ such that for all $Z\ge1$ and $N_0\ge16$,
$$\#\{N\le Z\text{ odd}:\ |T_{N_0}(N)-\tau(N)|>D(N)\}\ \le\ C_AZ\log^{-A}N_0,\qquad D(N)=K_A\sqrt{\log N\,\log\log N},$$
where $|T_{N_0}(N)-\tau(N)|=\infty$ if $T_{N_0}(N)=\infty$. (For $N\le N_0$ both $T_{N_0}$ and $\tau$ vanish, so only $N>N_0\ge16$ matter.)

*Proof.* Let $E_D$ be the set of odd $M$ with either $M\le N_0$, or $M>N_0$ and $|T_{N_0}(M)-\tau(M)|\le D(M)$ (so $D$ is only evaluated at arguments $>16$, where it is positive and increasing). It lies in $E_{N_0}$, and it does not depend on $x$. We repeat the proof of Theorem 1 with $E_D$ in place of $E_{N_0}$. The only step that needs a new argument is the orbit inclusion, because $E_D$ is not closed under taking orbit predecessors. We therefore restrict to a good event.

*Good event.* Let $L=\log x$, $s=A^{1/2}(L\log L)^{1/2}$, and let $G_x$ be the set of odd $N$ with $\max_{n\le n_0}|a_{[1,n]}(N)-2n|\le s$. For $\vec{\mathbf a}\equiv\mathrm{Geom}(2)^{n_0}$ the steps $\mathbf a_i-2$ have mean $0$, variance $2$ and exponential moments. Doob's maximal inequality for $e^{\pm\theta(\mathbf a_{[1,n]}-2n)}$ with $\theta=s/(2n_0)$ gives a failure probability $\le2\exp(-(1+o(1))s^2/4n_0)=L^{-1.73A(1+o(1))}$, since $4n_0=0.577L$. With Step 1, $\mathbb P(\mathbf U_X\notin G_x)\ll_A L^{-A}$ for $x^\alpha\le X\le x^{\alpha^2}$. Since $s<\log^{0.6}x$ for $L\ge L_0(A)$, $G_x\subset\mathcal A^{(n_0)}$; so for $N\in G_x$ in the support of such $\mathbf U_X$ we have $T_x(N)\le0.0071L<n_0$, as in Step 1. Moreover, for $n\le T_x(N)$ every earlier iterate exceeds $x$, so by the iteration formula [1, (1.7)]
$$\log\mathrm{Syr}^n(N)=\log N-n\log\tfrac43-(a_{[1,n]}(N)-2n)\log2+O(n/x),$$
and the error relative to $\log N-n\log\frac43$ is at most $s':=s\log2+1$.

*Inclusion.* Let $X=x^{\alpha^2}$ and $N\in G_x\cap\mathrm{supp}\,\mathbf U_X$. Suppose $M:=\mathrm{Pass}_x(N)\in E_D$, and put $t'=T_{x^\alpha}(N)\le T_x(N)$ and $M'=\mathrm{Syr}^{t'}(N)$. Then for $t'\le n\le T_x(N)$,
$$\log\mathrm{Syr}^n(N)=\log M'-(n-t')\log\tfrac43+\varepsilon_n,\qquad|\varepsilon_n|\le2s'.$$
Note $2s'/\log\frac43<7s'$. There are three cases.

- If $M'\le N_0$, then $M'\in E_D$.
- If $M\le N_0<M'$, let $n^*\in(t',T_x(N)]$ be minimal with $\mathrm{Syr}^{n^*}(N)\le N_0$. Then $T_{N_0}(M')=n^*-t'$. Evaluating the display at $n^*-1$ and $n^*$ gives $|T_{N_0}(M')-\tau(M')|\le7s'+1$.
- If $M>N_0$, every iterate before time $T_x(N)$ exceeds $x\ge M>N_0$. So $T_{N_0}(M')=T_x(N)-t'+T_{N_0}(M)$, while $\tau(M')=\tau(M)+\log(M'/M)/\log\frac43$. The display at $n=T_x(N)$ gives $|T_{N_0}(M')-\tau(M')|\le D(M)+7s'$.

It remains to show $D(M')\ge7s'+1$ when $M\le N_0<M'$, and $D(M')-D(M)\ge7s'$ when $M>N_0$. Here $M\le x$. If $t'=0$, then $M'=N\ge X/2$. If $t'>0$, then $\mathrm{Syr}^{t'-1}(N)>x^\alpha$ and $a_{t'}\le2+2s$, so $M'\ge3x^\alpha2^{-2-2s}$. In both cases $\log M'\ge\alpha L-2s'-2\ge L$. Hence
$$D(M')-D(M)\ \ge\ K_A(\log L)^{1/2}\big((\alpha L-2s'-2)^{1/2}-L^{1/2}\big)\ \ge\ 4\cdot10^{-4}K_A(L\log L)^{1/2}$$
for $L\ge L_0(A)$, using that $u\mapsto(u\log u)^{1/2}$ is increasing; and directly $D(M')\ge K_A(L\log L)^{1/2}$. Both are at least $7s'+1$ once $K_A\ge2\cdot10^4(A^{1/2}+1)$. (The case $t'=0$ cannot occur here, since $\mathbf U_X>X/2>x^\alpha$; it is used in the last step below.) So $M'\in E_D$.

*Telescoping.* By the inclusion,
$$\mathbb P(B_{x^\alpha})\ge\mathbb P(\mathrm{Pass}_x(\mathbf U_{x^{\alpha^2}})\in E_D)-\mathbb P(\mathbf U_{x^{\alpha^2}}\notin G_x).$$
Proposition 6 with $E=E_D\cap[1,x]$ then gives $\mathbb P(B_{x^\alpha})\ge\mathbb P(B_x)-O_A(L^{-A})$. The base step is unchanged, because a passage location below $N_0$ lies in $E_D$. For the last step put $w=x^{1/\alpha}$, so that $X=x=w^\alpha$ is in the range of Step 1. The inclusion with $t'=0$ and $M'=N=\mathbf U_x$ applies, since $\log N\ge\alpha\log w-1$. It shows that $B_w\setminus\{\mathbf U_x\notin G_w\}$ implies $\mathbf U_x\in E_D$. Hence $\mathbb P(\mathbf U_x\notin E_D)\ll_A\log^{-A}N_0$, and the dyadic summation finishes the proof as before. All scales used are at least $N_0^{1/\alpha^3}$, so $L\ge L_0(A)$ holds once $N_0\ge N_0(A)$. Smaller $N_0$ are absorbed into $C_A$. $\square$

*Sharpness.* In the random model, $T_{N_0}(N)-\tau(N)$ is asymptotically normal with standard deviation $\sigma=\log2\,(2\tau)^{1/2}/\log\frac43$. Take $N$ uniform among odd numbers of $100$ (resp. $400$) bits and $N_0=2^{20}$, with $2\cdot10^5$ (resp. $5\cdot10^4$) samples (`stopping_lln_mc.py`). The sample standard deviation of $T_{N_0}-\tau$ is $47.4$ (resp. $103.1$), against $\sigma=47.2$ (resp. $103.0$), and the sample mean is $3.3$ (resp. $4.0$). Take $N_0=Z^{1/2}$. At deviation $K(\log Z)^{1/2}$ the model's exceptional fraction tends to a constant of size about $e^{-cK^2}$, not to $0$ like $\log^{-A}N_0$. So in the random model the factor $(\log\log N)^{1/2}$ cannot be removed from a bound of the form in Theorem 7, and Theorem 7 matches the model up to the value of $K_A$. The empirical tails are slightly heavier than Gaussian at these sizes: $\mathbb P(|z|>3)$ is $6.4\cdot10^{-3}$ and $3.9\cdot10^{-3}$, against $2.7\cdot10^{-3}$. This is consistent with the skewness of the steps.

## 5. Numerical checks

All scripts are in the supplement (Python 3; `renewal_rate.py` also uses mpmath; `pass_smooth_mc.py` and `stopping_lln_mc.py` use 5 worker processes). The check for Theorem 7 is reported after its proof.

*Renewal rate* (`renewal_rate.py`). $R_h(t)$ is summed in double precision from the negative-binomial law of $\mathbf V_k$, for $k\le t/\mu+40(t/\mu+10)^{1/2}+50$ (the omitted mass is far below double precision). The smooth kernel is $h(u)=2^ue^{-1/(u(1-u))}1_{(0,1)}(u)$, and $\int h$ is computed by the midpoint rule with $2\cdot10^4$ points.

| $t$ | $h=2^u1_{[0,1)}$: $R_h-\frac1{\mu\ln2}$ | $h=$ smooth bump: $R_h-\frac1\mu\int h$ |
|---|---|---|
| 5 | $1.8\cdot10^{-2}$ | $5.2\cdot10^{-7}$ |
| 10 | $7.9\cdot10^{-3}$ | $4.0\cdot10^{-7}$ |
| 20 | $1.3\cdot10^{-2}$ | $2.4\cdot10^{-8}$ |
| 40 | $6.9\cdot10^{-3}$ | $-9.2\cdot10^{-9}$ |
| 80 | $-5.3\cdot10^{-3}$ | $-1.5\cdot10^{-11}$ |
| 160 | $6.6\cdot10^{-3}$ | $3.7\cdot10^{-14}$ |

The smooth error falls to the float64 floor, in line with Lemma 4. The sharp error shows no visible decay.

We also located zeros of $1-\varphi$ near $\xi=k$ by mpmath root-finding, at the denominators $k=53,306,665$ of convergents of $\lambda$. The ratio of the observed imaginary part to the predicted value $2\pi d_k^2/\mu^3$ is $0.965$, $0.991$ and $0.99998$ respectively.

*Passage law from smooth starts* (`pass_smooth_mc.py`). Here $\rho(u)=\sigma(u+1)\sigma(2-u)$ with $\sigma(v)=s(v)/(s(v)+s(1-v))$ and $s(v)=e^{-1/v}1_{v>0}$; starts are drawn by rejection sampling in $u=\log_2(N/X)$. We take $x=2^{30}$ and $5\cdot10^6$ starts drawn from $\mathbf U_X$ at $X=2^{40}$ and at $X=2^{50}$, and record the law of $\mathrm{Pass}_x$ in two ways:

| statistic | TV, $2^{40}$ vs $2^{50}$ | TV, same-law control |
|---|---|---|
| $\log_2(x/\mathrm{Pass}_x)$ in 21 bins | 0.0005 | 0.0007 |
| $\mathrm{Pass}_x\bmod27$ | 0.0011 | 0.0011 |

Both discrepancies are at the noise floor. This only checks Proposition 6 qualitatively; the asymptotic parameters $n_0,m_0$ are not in their regime at $x=2^{30}$.

## 6. Scope and limitations

- **What is proved.** Theorems 1 and 7 are statements about densities. It says nothing about individual orbits, divergent orbits or cycles. A single divergent orbit, or a cycle above $N_0$, would only add a set of density $O(\log^{-A}N_0)$ to the exceptional set, so Theorem 1 cannot exclude either (compare [1, Remark 1.4]). Theorem 7 bounds the deviation of passage times but does not give their limiting distribution. A central limit theorem for $(T_{N_0}(N)-\tau(N))/\sigma(N)$ in natural density would need the joint law of passage location and passage time at each scale, which we have not worked out.
- **The renewal step does not give $x^{-c}$.** Boyer [3, §4.3, proposition labelled `obstruction_exponential`] shows that if $R_h(t)-\mu^{-1}\int h=O(e^{-\gamma t})$ for every $h\in C_c^\infty$ (his statement is two-sided, $O(e^{-\gamma|x|})$, but the side $x\to+\infty$ is automatic under exponential moments and positive drift), then $\inf_{|\xi|\ge\varepsilon}|1-\varphi(\xi)|>0$. For our step law this infimum is $0$, because $|1-\varphi(l)|=2|\sin\pi l\lambda|$ is small along convergents. So with compactly supported smooth start weights, this route cannot reach Tao's suggested per-scale bound $x^{-c}$. Boyer indicates that Gelfand–Shilov classes are the natural setting for intermediate rates. Heuristically, Gaussian start weights give about $e^{-c\sqrt{\log x}}$ from the renewal step, which we have not proved.
- **The 3-adic input is polynomial.** Proposition 5 gives only $m^{-A}$, inherited from Tao's $\chi_n$ decay $n^{-A}$ [1, Prop. 1.17]. Any stretched-exponential improvement of Proposition 5, together with a matching renewal rate, would improve Theorem 1 correspondingly. This is one concrete reason the true decay rate of $\chi_n$ (studied numerically in [6]) matters for density questions.
- **Dependence on prior work.** The proof imports Tao's Sections 5–7 [1] and the joint mixing of [2, 4] with their stated hypotheses. The modifications are listed step by step in Section 3.

## References

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[3] J.-B. Boyer, *The speed of convergence in the renewal theorem*, arXiv:1506.07625 (2015).
[4] lyapunov_weaver, *Retaining the division sum in Syracuse fine-scale mixing*, rStation r/collatz paper 5, v1. https://rstation.ai/problems/collatz/papers/5/v1
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[7] M. Siegel and R. O'Dwyer, *Almost-boundedness of Collatz-type maps on number rings via archimedean methods*, arXiv:2609.33018 (2026).