1196 Discussion Thread | Erdős Problems

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The anticipated paper on this problem by Alexeev, Barreto, Li, Lichtman, Price, Shah, Tang, and Tao is now on arXiv!

natso26 — 14:35 on 04 May 2026
- I asked ChatGPT to draw a poster on the result and paper. Title:
  "The Glorious One plus Eight"
  
  The Glorious One plus Eight
  
  Let me know if there are serious errors in the poster, for instance by personal message.
  
  old-bielefelder — 11:22 on 12 May 2026
  - Are you sure the people's images are accurate? It's nonmathematical but seems important.
    
    natso26 — 12:44 on 12 May 2026
    - Of course, in this version they are not shown accurately.
      
      But before trying on this "detail", I want to make sure that the mathematical elements in the poster are fine. So, if we reach such a status I will look at more realisitc portraits.
      
      old-bielefelder — 13:32 on 12 May 2026
  - There are indeed some errors here.
    
    Major:
    - Bottom left: $𝑓 (𝐴) = \sum 𝑎 \in 𝐴 1 / 𝑎 ≪ 1 / (𝑛 l o g 𝑛)$ . It should be something like $𝑓 (𝐴) = \sum 𝑎 \in 𝐴 1 / (𝑎 l o g 𝑎)$ .
    - Bottom right: $ℙ (h i t 𝐴) ≪ 1 / (𝑛 l o g 𝑛)$ . What is proved is more like $ℙ (h i t 𝐴) \leq 1 + 𝑜 (1)$ . (Edit: should be $ℙ (h i t 𝐴) \leq 1$ .)
    - Downward Markov chain: it moves from $𝑛$ to $𝑛 / 𝑞$ , not from $𝑛$ to $𝑞$ . And $𝑞$ need not be a *proper* divisor.
    - Upward (adjoint) chain: the probability is not proportional to $Λ (𝑞)$ . (The adjoint formula is a bit more complicated than this.)
    
    Minor:
    - Divisibility poset: $Ω (𝑛) \geq 4$ ( $ℕ 24$ ). Should be $ℕ \geq 4$ .
    
    Credit to GPT-5.5 Thinking for discussion.
    
    natso26 — 17:34 on 17 May 2026
    - Thanks a lot. Good idea to ask "the" AI to find errors and weaknesses in the poster. The new version is now online, again
      
      here
      
      old-bielefelder — 18:56 on 17 May 2026
      - The probability of hitting $𝐴$ is at most $1$ rather than just at most $1 + 𝑜 (1)$ , since it is a probability. The $𝑜 (1)$ comes in since $𝑓 (𝐴)$ is the probability of hitting $𝐴$ plus $𝑜 (1)$ .
        
        WillSawin — 11:30 on 18 May 2026
        
        Yes, thanks, I (not GPT) have confused myself a bit there!
        
        natso26 — 14:29 on 19 May 2026
A possible relative-capacity application of the coprime-modulus variant

I have been trying to understand whether the new von Mangoldt flow / cut-capacity method gives sharp constants not only for the full divisibility poset, but also for natural local arithmetic subuniverses.

Let me write, provisionally, $C a p (𝑆) : = l i m s u p 𝑥 \to \infty s u p 𝐴 \subset 𝑆 \cap [𝑥, \infty) 𝐴 p r i m i t i v e \sum 𝑎 \in 𝐴 1 𝑎 l o g 𝑎 .$ Thus $C a p (ℕ) = 1$ in the sense of #1196.

Fix an odd prime $ℓ$ . Let $𝑈 ℓ$ be the set of positive integers $𝑁$ such that $ℓ$ divides each of $𝜑 (𝑁), 𝜆 (𝑁), 𝜓 (𝑁), 𝐽 𝑡 (𝑁) (𝑡 \geq 1), 𝜎 𝑡 (𝑁) (𝑡 \geq 0),$ where $𝜓$ denotes the Dedekind psi function.

The non- $𝜎$ conditions seem to be soft from the capacity point of view. For instance, failure of the $𝜑, 𝜆, 𝐽 𝑡$ gates forces avoidance of primes $𝑝 \equiv 1 (m o d ℓ)$ , except for the irrelevant escape $𝑣 ℓ (𝑁) \geq 2$ , and failure of the Dedekind- $𝜓$ gate similarly forces avoidance of primes $𝑝 \equiv - 1 (m o d ℓ)$ . Such residue-class avoidance should have zero relative capacity by the coprime-to-modulus variant.

The interesting part is therefore the local $𝜎 𝑡$ condition. Put $𝑚 = ℓ - 1$ . For $𝑝 \neq ℓ$ and $𝑒 \geq 0$ , define $𝑍 ℓ (𝑝, 𝑒) : = {𝑡 \in ℤ / 𝑚 ℤ : 1 + 𝑝 𝑡 + \dots + 𝑝 𝑒 𝑡 \equiv 0 (m o d ℓ)},$ where the class $𝑡 = 0$ is to be interpreted as the positive class $𝑡 \equiv 0 (m o d 𝑚)$ . This also handles $𝜎 0$ , since for $𝑝 \neq ℓ$ the condition at $𝑡 = 0$ is just $𝑒 + 1 \equiv 0 (m o d ℓ)$ . The prime $𝑝 = ℓ$ does not help with this positive class, since $1 + ℓ 𝑡 + \dots + ℓ 𝑒 𝑡 \equiv 1 (m o d ℓ) (𝑡 \geq 1) .$ The soft $𝜎$ -classes are those $𝑡$ for which $𝑥 𝑡 = - 1$ is solvable in $𝔽 \times ℓ$ , i.e. $g c d (𝑡, 𝑚) ∣ 𝑚 / 2 .$ For such a class, an exponent-one prime in a suitable nonempty residue class already kills $𝜎 𝑡$ , so failure should again have zero capacity. Thus the genuinely finite gates are $H a r d ℓ : = {𝑡 \in ℤ / 𝑚 ℤ : g c d (𝑡, 𝑚) ∤ 𝑚 / 2} .$ This leads to the candidate formula $C a p (𝑈 ℓ) = ℙ (H a r d ℓ \subset ⋃ 𝑝 \neq ℓ 𝑍 ℓ (𝑝, 𝑉 𝑝)),$ where the $𝑉 𝑝$ are independent geometric valuations $ℙ (𝑉 𝑝 = 𝑒) = 1 - 1 / 𝑝 𝑝 𝑒 (𝑒 \geq 0) .$ For example, when $ℓ = 3$ , one has $H a r d 3 = {0}$ , so this gives $C a p (𝑈 3) = 1 - \prod 𝑝 \neq 3 (1 - 1 𝑝 2 + 𝑝 + 1) .$ Similarly, for $ℓ = 5$ , $C a p (𝑈 5) = 1 - \prod 𝑝 \neq 5 (1 - 1 𝑝 4 + 𝑝 3 + 𝑝 2 + 𝑝 + 1) .$ The first genuinely coupled case is $ℓ = 7$ , where $H a r d 7 = {0, 2, 4} .$ The one-gate collapse seems to occur exactly when $ℓ - 1$ is a power of $2$ .

The proof route I have in mind is:

1. classify the local gates for $𝜑, 𝜆, 𝜓, 𝐽 𝑡, 𝜎 𝑡$ ;
2. discard the residue-class avoidance failures and the soft $𝜎$ -failures as zero-capacity sets;
3. use the coprime-to-fixed-modulus variant to evaluate finite exact valuation atoms, namely $C a p {𝑛 : 𝑣 𝑝 (𝑛) = 𝑒 𝑝 f o r 𝑝 \in 𝑃} = \prod 𝑝 \in 𝑃 1 - 1 / 𝑝 𝑝 𝑒 𝑝$ for each finite set of primes $𝑃$ , up to the harmless fixed logarithmic shift from dividing out $\prod 𝑝 \in 𝑃 𝑝 𝑒 𝑝$ ;
4. pass from finite $𝑃$ to all primes using the convergent $𝑝 - 2$ -type tails for the hard gates.

My main question is about step 3. Is the exact finite-valuation atom statement really an immediate corollary of the coprime-to-fixed-modulus/cut-capacity inequality, after dividing out the fixed prime powers, or is there some obstruction from the fact that an exact valuation atom is not divisor-closed under the downward flow?

rptx — 18:29 on 10 May 2026
A feature of this solution that seems worth emphasizing is that it is not merely a Markov-chain trick, but a dual certificate for the divisibility poset.

The cleanest way I now understand the proof is as follows. Put a weighted directed edge $𝑛 𝑞 \to 𝑛, 𝑤 (𝑛 𝑞, 𝑛) = Λ (𝑞) 𝑛 𝑞 l o g 2 (𝑛 𝑞) .$ Then the von Mangoldt identity gives the exact outflow $O u t (𝑛) = 1 𝑛 l o g 𝑛,$ while Mertens' theorem gives the asymptotic inflow $I n (𝑛) = \sum 𝑞 Λ (𝑞) 𝑛 𝑞 l o g 2 (𝑛 𝑞) = 1 𝑛 l o g 𝑛 + 𝑂 (1 𝑛 l o g 2 𝑛) .$ Thus the flow is almost divergence-free on the tail $𝑛 \geq 𝑥$ , with total divergence $\sum 𝑛 \geq 𝑥 | d i v (𝑛) | ≪ 1 l o g 𝑥 .$ Now take $Ω = {𝑑 \geq 𝑥 : 𝑑 ∣ 𝑎 f o r s o m e 𝑎 \in 𝐴} .$ The primitivity of $𝐴$ is used in exactly one decisive place: every edge entering $𝑎 \in 𝐴$ comes from outside $Ω$ , because if $𝑎 𝑞 \in Ω$ , then $𝑎 ∣ 𝑎 𝑞 ∣ 𝑏$ for some $𝑏 \in 𝐴$ , forcing $𝑎 = 𝑏$ , impossible for $𝑞 > 1$ . Hence the inflow into $Ω$ sees $\sum 𝑎 \in 𝐴 1 𝑎 l o g 𝑎$ up to the $𝑂 (1 / l o g 𝑥)$ error. On the other hand, every edge leaving $Ω$ must cross below $𝑥$ , and the same Mertens computation gives boundary outflow $1 + 𝑂 (1 l o g 𝑥) .$ The discrete divergence theorem then immediately gives $\sum 𝑎 \in 𝐴 1 𝑎 l o g 𝑎 \leq 1 + 𝑂 (1 l o g 𝑥) .$ This formulation makes the role of $Λ$ especially transparent. It is doing two independent jobs at once: locally, it gives the exact divisor identity $\sum 𝑞 ∣ 𝑛 Λ (𝑞) = l o g 𝑛,$ and globally, it gives the correct Mertens normalization at the boundary. That combination is precisely what removes the extra constants appearing in earlier approaches.

In this form, the proof feels like a weighted LYM/Sperner certificate for the divisibility lattice: construct a nearly divergence-free flow whose chains are divisibility chains, observe that an antichain can be crossed only once, and compute the boundary mass. It would be very interesting to know how far this flow-certificate viewpoint extends, especially to restricted prime supports, $𝑘$ -almost-prime layers, and Banks--Martin type extremal questions.

mbr63 — 16:13 on 27 Apr 2026
We have built what one might call the von Mangoldt downward process $𝑛 \mapsto 𝑛 / 𝑞$ (with transition probability $Λ (𝑞) / l o g 𝑛$ ), the von Mangoldt measure $𝜈$ , and the von Mangoldt upward process $𝑛 \mapsto 𝑞 𝑛$ (with transition probability $Λ (𝑞) 𝜈 (𝑛 𝑞) / 𝜈 (𝑛) l o g (𝑛 𝑞)$ , all powered by the basic identity $\sum 𝑞 | 𝑛 Λ (𝑞) = l o g 𝑛$ . But this process can "jump" over primes due to the fact that the support of $Λ$ contains prime powers in addition to just primes.

But suppose we take the fundamental theorem of arithmetic identity $\sum 𝑝 | 𝑛 𝑣 𝑝 (𝑛) l o g 𝑝 = l o g 𝑛$ instead, where $𝑣 𝑝 (𝑛)$ is the number of times $𝑝$ divides $𝑛$ . This then gives a "downward prime process" in which $𝑛$ transitions to $𝑛 / 𝑝$ with probability $𝑣 𝑝 (𝑛) l o g 𝑝 / l o g 𝑛$ , and then should give a "prime measure" $˜ 𝜈$ that is invariant under this process and asymptotic to $1 / 𝑛 l o g 𝑛$ , as well as an "upward prime process" in which $𝑛$ transitions to $𝑛 𝑝$ with probability $𝑣 𝑝 (𝑛 𝑝) l o g 𝑝 ˜ 𝜈 (𝑛 𝑝) / l o g (𝑛 𝑝) ˜ 𝜈 (𝑛)$ . This should be very similar to the von Mangoldt processes, but now the inequality $\sum 𝑎 \in 𝐴 ˜ 𝜈 (𝑎) \leq 1$ should be an identity in the case $𝐴$ is the set of products of $𝑘$ primes (possibly repeated), because this process does not jump over primes. (This is similar to the modification of $𝜈$ proposed by Will, but has a slight difference, in that when repeated primes occur in the factorization, Will often factors them out consecutively in the downward process, whereas in this process they will be factored out at random times.)

Given how nice the formula for $𝜈$ was - basically the Laplace transform of $1 / 𝜁$ - perhaps one can hope that there is a similarly clean formula for $˜ 𝜈$ ? Among other things, this should lead to an improvement of the $1 / l o g 𝑥$ type error term (indeed, one might now hope to not just improve the constant, but get better asymptotic decay of the error term than just one power of the logarithm).

A variant question: suppose we restrict the von Mangoldt downward process $𝑛 \mapsto 𝑛 / 𝑞$ to squarefree numbers (note that the property of being squarefree is preserved by the flow). What is the new canonical measure $𝜈 𝑠 𝑓$ attached to this restricted process? My initial guess is that it will have to do with the Laplace transform of $𝜁 (2 𝑠) / 𝜁 (𝑠)$ .

TerenceTao — 01:07 on 16 Apr 2026
- For the prime process $𝑛 \to 𝑛 / 𝑝$ with probability $𝑣 𝑝 (𝑛) l o g 𝑝 / l o g 𝑛$ , let us set
  $𝑎 (𝑛) : = \prod 𝑝 1 𝑣 𝑝 (𝑛)!, 𝑃 (𝑠) : = \sum 𝑝 𝑝 - 𝑠,$ the prime zeta function. Then I claim $˜ 𝜈 (1) : = 1$ and for $𝑛 > 1$ , $˜ 𝜈 (𝑛) : = 𝑎 (𝑛) \int \infty 1 𝑛 - 𝑠 (𝑒 - 𝑃 (𝑠))' d 𝑠 I B P = 𝑎 (𝑛) l o g 𝑛 \int \infty 1 𝑛 - 𝑠 𝑒 - 𝑃 (𝑠) d 𝑠 = 𝑎 (𝑛) 𝑛 \int \infty 0 e x p (- 𝑡 - 𝑃 (1 + 𝑡 / l o g 𝑛)) d 𝑡$
  is invariant. Indeed, noting that $𝑣 𝑝 (𝑛 𝑝) 𝑎 (𝑛 𝑝) = 𝑎 (𝑛)$ , we see that $\sum 𝑝 𝑣 𝑝 (𝑛 𝑝) l o g 𝑝 l o g (𝑛 𝑝) ˜ 𝜈 (𝑛 𝑝) = 𝑎 (𝑛) \int \infty 1 𝑛 - 𝑠 𝑒 - 𝑃 (𝑠) \sum 𝑝 𝑝 - 𝑠 l o g (𝑝) d 𝑠 = 𝑎 (𝑛) \int \infty 1 𝑛 - 𝑠 (𝑒 - 𝑃 (𝑠))' d 𝑠 = ˜ 𝜈 (𝑛) .$
  
  Moreover confirming your identity comment, if $Ω (𝑛) : = \sum 𝑝 𝑣 𝑝 (𝑛)$ , then $\sum Ω (𝑛) = 𝑘 𝑎 (𝑛) 𝑛 - 𝑠 = 𝑃 (𝑠) 𝑘 𝑘!,$ so $\sum Ω (𝑛) = 𝑘 ˜ 𝜈 (𝑛) = \int \infty 1 𝑃 (𝑠) 𝑘 𝑘! (𝑒 - 𝑃 (𝑠))' d 𝑠 = \int \infty 0 𝑒 - 𝑢 𝑢 𝑘 𝑘! d 𝑢 = 1 .$ So, indeed, for the set of integers with exactly $𝑘$ prime factors, counted with multiplicity, the total $˜ 𝜈$ -mass is exactly $1$ . Near $𝑠 = 1$ one has $𝑃 (1 + 𝑢) = l o g (1 / 𝑢) + 𝑂 (1)$ , so $˜ 𝜈 (𝑛) ≍ 1 𝑛 l o g 𝑛 \prod 𝑝 𝑣 𝑝 (𝑛)! .$
  
  Now for the squarefree-restricted von Mangoldt process, define for squarefree $𝑛$ , $Φ 𝑛 (𝑠) : = \prod 𝑝 ∤ 𝑛 (1 - 𝑝 - 𝑠) = 1 𝜁 (𝑠) \prod 𝑝 ∣ 𝑛 (1 - 𝑝 - 𝑠) - 1,$ and set $𝜈 𝑠 𝑓 (1) : = 1$ and for $𝑛 > 1$ with $𝜇 (𝑛) 2 = 1$ , $𝜈 𝑠 𝑓 (𝑛) : = \int \infty 1 𝑛 - 𝑠 (Φ 𝑛 (𝑠))' d 𝑠 I B P = l o g 𝑛 \int \infty 1 𝑛 - 𝑠 Φ 𝑛 (𝑠) d 𝑠 .$ Then, again, I claim that $𝜈 𝑠 𝑓$ is invariant for the restricted chain. Indeed, for $𝑝 ∤ 𝑛$ ,
  $Φ 𝑛 𝑝 (𝑠) = Φ 𝑛 (𝑠) 1 - 𝑝 - 𝑠 a n d Φ' 𝑛 (𝑠) Φ 𝑛 (𝑠) = \sum 𝑝 ∤ 𝑛 𝑝 - 𝑠 l o g (𝑝) 1 - 𝑝 - 𝑠,$ so $\sum 𝑝 ∤ 𝑛 l o g 𝑝 l o g (𝑛 𝑝) 𝜈 𝑠 𝑓 (𝑛 𝑝) = \int \infty 1 𝑛 - 𝑠 Φ 𝑛 (𝑠) \sum 𝑝 ∤ 𝑛 𝑝 - 𝑠 l o g (𝑝) 1 - 𝑝 - 𝑠 d 𝑠 = \int \infty 1 𝑛 - 𝑠 Φ' 𝑛 (𝑠) d 𝑠 = 𝜈 𝑠 𝑓 (𝑛) .$ Equivalently, $𝜈 𝑠 𝑓 (𝑛) = 1 𝑛 \int \infty 0 𝑒 - 𝑡 𝜁 (1 + 𝑡 / l o g 𝑛) \prod 𝑝 ∣ 𝑛 (1 - 𝑝 - 1 - 𝑡 / l o g 𝑛) - 1 d 𝑡 .$
  
  Kevin Barreto — 04:24 on 16 Apr 2026
  - Thanks! So there does seem to be something special about the original von Mangoldt process - the associated invariant measure $𝜈$ is extremely smooth (in the Archimedean sense), being asymptotic to $1 / 𝑛 l o g 𝑛$ , while all the variants of this measure pick up arithmetic factors such as $1 \prod 𝑝 𝑣 𝑝 (𝑛)!$ . A little surprising to me that removing individual primes instead of prime powers makes it *less* likely to have prime multiplicity, but I'll chalk it up to one of the numerous probability paradoxes that arise when one tries to compare various weighted expectations. But these factors mean that one cannot immediately solve #1196 by using these processes instead of the von Mangoldt one, as the invariant measure is no longer asymptotic to $1 / 𝑛 l o g 𝑛$ . So in some sense the AI was "lucky" in finding the one approach that actually worked; it would be interesting to publish the traces to see if there was a lot of brute force involved in trying nearby approaches which didn't quite work.
    
    (Side remark: the near-invariant measure I constructed for the Collatz iteration also had a rather nasty arithmetic factor involving the 3-adic structure of $𝑛$ , and which was the main reason that paper was so complicated. Again, it is a minor miracle that the invariant measure for the von Mangoldt process is completely insensitive to arithmetic structure.)
    
    On the other hand, it's possible that the identity $\sum Ω (𝑛) = 𝑘 ˜ 𝜈 (𝑛) = 1$ could be an alternate starting point to recover the results of [GLW24], if one can get enough understanding of how to compensate for the $1 \prod 𝑝 𝑣 𝑝 (𝑛)!$ factor.
    
    TerenceTao — 04:59 on 16 Apr 2026
    - These processes have many invariant measures. For the downward process, you can start with an arbitrary measure supported on large n and apply the process many times. Some features of the initial measure will be damped, but not all of them. (For example, the invariant measure for squarefrees is also an invariant measure for the original downward process, since the original process sends squarefrees to sqaurefrees). However, there are probably not so many invariant measures that have an elegant explicit formula.
      
      These processes have many invariant measures. For the downward process, you can start with an arbitrary measure supported on large n and apply the process many times. Some features of the initial measure will be damped, but not all of them. (For example, the invariant measure for squarefrees is also an invariant measure for the original downward process, since the original process sends squarefrees to sqaurefrees). However, there are probably not so many invariant measures that have an elegant explicit formula.
      
      The measure Kevin found where prime multiplicity is less likely is not a consequence of the changed downward probabilities, but rather a consequence of the way Kevin needed to modify the distribution to get an explicit formula.
      
      I suspect there is another measure obtained the same way you suggested originally, by applying $𝑘$ downward iterations to the $1 / (𝑛 l o g 𝑛)$ measure and taking the limit as $𝑘$ goes to $\infty$ . Furthermore, such a measure probably looks similar to the $𝜈'$ I wrote down: That is, it is approximately $1 𝑛 l o g 𝑛$ and has lower-order terms that are larger for $𝑛$ divisible by small primes. But there may be no elegant formula for such a measure.
      
      Such a measure might not be so useful in getting an improved bound, since what matters for the bound is the least value of $𝑛 l o g 𝑛 𝜈 (𝑛)$ for some $𝑛 \geq 𝑞$ . (There are other variants, but in none of them is dependence on arithmetic properties of $𝑛$ your friend). It's possible that one can compensate for this by a more intricate argument but given that the problem is almost a linear programming problem that is dual to the problem of finding a measure $𝜈$ , there might not be so much room to compensate for failures of $𝜈$ .
      
      A variant that might be worth considering, which definitely won't have an elegant formula, but might still be possible to analyze somehow, is the downward process where you divide by $𝑝$ with probably $l o g 𝑝 l o g s q f (𝑛)$ where $s q f$ denotes the squarefree part. This should make $𝑛$ with small prime factors less likely, but might be smoother overall and have a better lower bound for $𝑛 l o g 𝑛 𝜈 (𝑛)$ .
      
      The reason the von Mangoldt approach works so well is not exactly luck: The key feature is that in the recurrence formula $𝜈 (𝑛) = \sum 𝑞 Λ (𝑞) l o g (𝑛 𝑞) 𝜈 (𝑛 𝑞)$ , as long as the $𝜈$ on the right hand side doesn't depend on arithmetic properties of $𝑛$ , nothing else depends on arithmetic properties of $𝑛$ , so that it is self-consistent for $𝜈$ to be independent of arithmetic properties of $𝑛$ . However, given the overall performance of AI on similar problems which is not yet strong enough that I would expect it to always find such things, I do suspect in this case it was found with some luck.
      
      WillSawin — 11:28 on 16 Apr 2026
      - It might have been found by ansatz. In my initial conversation with GPT-5.4 Thinking trying to reverse engineer the proof, one requirement is that
        $𝑅 𝑤 (𝑚) : = l o g 𝑚 \sum 𝑞 𝑤 (𝑞) 𝑞 l o g (𝑚 𝑞) 𝐶 (𝑚 𝑞) \approx 1,$
        with downward divisor chain
        $𝑃 (𝑛, 𝑛 / 𝑞) = 𝑤 (𝑞) 𝐶 (𝑛), 𝐶 (𝑛) : = \sum 𝑞 ∣ 𝑛 𝑤 (𝑞) .$
        
        Now from this, GPT strongly gravitated towards setting $𝐶 (𝑚 𝑞) = l o g (𝑚 𝑞)$ , because otherwise "the row sum has no reason to simplify". And the von Mangoldt $𝑤 (𝑞) = Λ (𝑞)$ is pretty much the right thing for $\sum 𝑞 ∣ 𝑛 𝑤 (𝑞) = 𝐶 (𝑛)$ .
        
        Admittedly these reasons are rather ad hoc. But it could be similar to how the choice was found.
        
        natso26 — 13:25 on 16 Apr 2026
      - Good points: none of these downward flows are uniquely ergodic, as different boundary conditions at infinity generate different invariant measures. This explains the "probability paradox" I had sensed but was unable to rigorously pin down previously.
        
        For large typical $𝑛$ , all large prime factors should appear just once, and small prime factors appear rarely, so the prime downward process and von Mangoldt downward process are nearly identical; in particular $1 / (𝑛 l o g 𝑛)$ should be nearly invariant for both (up to relative errors of $𝑂 (1 / l o g 𝑛)$ , which is acceptable due to how quickly $𝑛$ shrinks under such processes). So there should indeed be an invariant measure for the downward process that does not have the $\prod 𝑝 1 / 𝑣 𝑝 (𝑛)!$ factor and is genuinely asymptotic to $1 / 𝑛 l o g 𝑛$ . But perhaps it will have a far messier formula than the one for Kevin's measure; one may need to analyze it by other means than explicit formulae.
        
        Good point about how the von Mangoldt process requires knowledge about the anatomy of $𝑞$ , but not about $𝑛$ , allowing for arithmetic independence to be self-consistent with the von Mangoldt recurrence. This can be contrasted with the prime recurrence
        $𝜈 (𝑛) = \sum 𝑝 (𝜈 𝑝 (𝑛) + 1) l o g 𝑝 l o g (𝑛 𝑝) 𝜈 (𝑛 𝑝)$
        where now there is a $𝜈 𝑝 (𝑛)$ factor that involves the anatomy of $𝑛$ . But this factor is lower order; this sum should be dominated by large primes $𝑝$ rather than small primes $𝑝$ (in fact primes of the form $𝑝 = 𝑛 𝑂 (1)$ give the bulk of the contribution), and then $𝜈 𝑝 (𝑛)$ is almost certainly zero.
        
        A side note: another process that implicitly appears in the literature is that of removing a prime factor $𝑝$ from $𝑛$ uniformly amongst all such factors not counting multiplicity, i.e., with a probability of $1 / 𝜔 (𝑛)$ . This process is tied to the "Ramare identity"
        $\sum 𝑛 𝑓 (𝑛) = \sum 𝑝 \sum 𝑛 𝑓 (𝑛 𝑝) 𝜔 (𝑛 𝑝) + 𝑓 (1)$
        which turns out to be useful in the analysis of multiplicative functions on short intervals (particularly after modifying the process to restrict $𝑝$ to some suitable range of interest, and terminating when $𝑛$ is not divisible by any primes in that range), though this process is unlikely to be relevant for #1196 type problems (it does not come close to preserving $1 / 𝑛 l o g 𝑛$ ). It seems that Markov divisibility processes are secretly present in quite a lot of the analytic number theory literature!
        
        EDIT: Perhaps one can first work with the function field model to explore these various alternate processes?
        
        TerenceTao — 14:44 on 16 Apr 2026
    - Here is the link to the 5.4 Pro chat that solved the problem if you're interested in the summarised chain of thought.
      
      Liam Price — 15:55 on 16 Apr 2026
      - Interesting chain of thought. It seems to have latched early on to the idea of using a Sperner/LYM type approach in which antichains are controlled by a suitable random chain with a desired hitting probability - which is essentially what the final proof is - but halfway through the chain of thought it seemed to abandon that idea and try quite different directions, in particular starting to search for counterexamples at one point. Then it returned to the Sperner/LYM type approach, but using weights closer to the Mertens measure Jared mentioned than the measure $𝜈$ . Oddly, at no point in the published chain of thought does the von Mangoldt function make an appearance, so it sheds no light on how the LLM landed on that particular process. All in all, it's quite a chaotic internal thought process, with many dead ends, though to be fair many human mathematicians's thought processes during the initial stages of problem solving would likely also be similarly chaotic.
        
        TerenceTao — 16:59 on 16 Apr 2026
        
        [Post deleted]
        
        Unknown — 17:13 on 16 Apr 2026
        
        Shame, this would otherwise be extremely useful analysis for Mathematicians. There is a massive gap here between what the labs are doing and understand versus what mathematicians are doing and understand. These CoTs could really help bridge that gap better for everyone's benefit. The core concern is distillation attacks, however, so getting them to improve this might be difficult. It's not a valid moral concern, but certainly a rational business one.
        
        qrdl — 17:51 on 16 Apr 2026
        
        I used GPT-5.4 Thinking last week (17 April) to conduct a provenance audit of the 55-page reasoning chain for the 80 minute proof of GPT-5.4 Pro of 13 April. It is a very interesting technique that drills further and further down into the reasoning chain. I am not a mathematician (just a computer scientist interested in LLMs), but I think the detailed analysis should be of interest to you: https://chatgpt.com/share/69e208a2-decc-83eb-be4d-796a0a2160fd . This thread starts with the original proof and then continues with a sequence of provenance prompts and responses. All the prompts were written by GPT-5.4 Thinking.
        
        uwepleban — 22:52 on 23 Apr 2026
      - I have been discussing reasoning chains, summarization, etc. with GPT-5.4T recently, and asked it just now to look at the issue of the summarization having “dropped the rabbit” (GPT-5.4T’s formulation), and it suggested the following:
        
        'There is also a plausible compromise short of full raw-trace release. OpenAI’s monitorability work says follow-up questioning can surface previously unverbalized reasoning, especially when the model retains access to its earlier hidden chain of thought. For theorem discovery, that suggests a post-solution interrogation phase could sometimes recover the missing mathematical pivot — “Where did the von Mangoldt choice come from?” — even when the first summary failed to mention it. That still would not make the result fully faithful, but it would be much more useful than today’s prettified one-paragraph digest. [4][2]
        [2] OpenAI, “Reasoning models,” OpenAI API Docs, n.d., https://developers.openai.com/api/docs/guides/reasoning
        [4] OpenAI, “Evaluating chain-of-thought monitorability,” OpenAI, 2025-12-18 (>90d), https://openai.com/index/evaluating-chain-of-thought-monitorability/ '
        
        If you are interested in additional details, here is my conversation with the model: https://chatgpt.com/share/69e1f007-3704-83eb-8c83-16565be09758
        You need to scroll down to the end, since the earlier portion is an unrelated follow-up on a Pulse notification concerning proof-auditable models.
        
        uwepleban — 08:40 on 17 Apr 2026
      - I continued your conversation with 5.4 Pro with a request for a provenance audit of its solution. The detailed prompt was created by GPT-5.4 Thinking. The audit result may have information that you may find helpful.
        Link: https://chatgpt.com/share/69e208a2-decc-83eb-be4d-796a0a2160fd -- this link has been superseded by
        [1] Erdös Problem 1196: GPT-5.4 Pro provenance audit of its own solution trace — https://chatgpt.com/share/69e208a2-decc-83eb-be4d-796a0a2160fd
        
        uwepleban — 10:22 on 17 Apr 2026
        
        That log was somewhat helpful, but still inconclusive at the most critical components of the problem solving process, which remain frustratingly opaque. My tentative theory is that these models are still quite weak at developing strategy and constructing novel coherent narratives (as opposed to explaining existing, human-generated, narratives, for which they are now rather good at); this may also be related to the tendency of AI-generated proofs (such as this one) to dwell at length on rather routine components of an argument, while not stressing the most original and important aspects of a proof.
        
        For me, there are two points in the thought process that would be particularly illuminating to locate. The first, obviously, is where the idea of using the von Mangoldt process first emerged. But dual to this is why this idea was not later abandoned, like so many other ideas generated in the chain of thought. Problem solving is not just about coming up with the one right idea amongst a sea of bad ideas; it is also about the filtering that isolates that one viable idea from all the unviable ideas. This filtering seems to be largely absent in the published chain of thought; and yet the AI was somehow able to perform the remaining technical steps needed to convert the viable idea into an actual proof.
        
        Perhaps one has to compensate for the effect of "survivor bias": amongst the Erdos problems alone, thousands of instances of these models are being thrown at these problems, but it is pretty much only the successful (or partially successful) instances which are being reported here, and one possible explanation for the lack of coherent strategic narrative in this instance is that there simply isn't one: it could instead be more of a numbers game of "throwing things repeatedly at the wall and seeing what sticks". Perhaps one experiment which would be helpful would be to run additional instances of these models on the same problem #1196 (turning off internet access to avoid contamination) and seeing how they perform, and whether their chains of thought are noticeably different from this one. (If someone does wish to perform such an experiment, it would be appreciated (and scientifically valuable) if this were announced in advance of the experiment being performed, with a commitment to report results regardless of whether they are positive or negative, to avoid the aforementioned survivor bias.)
        
        TerenceTao — 15:10 on 17 Apr 2026
        
        Arb Research has kindly shared with me ten separate runs of GPT 5.4 Pro on this problem #1196 (with a request not to use internet search). From a quick reading, it appears that 8 of them claimed successes, with the other 2 rating the claim as plausible. Interestingly, several of the successful runs actually obtained the sharper formula $\sum 𝑛 \leq 𝐴 𝜈 (𝑛) \leq 1$ that was also derived here, with $𝜈$ essentially the Mellin transform of $1 / 𝜁 (𝑠)$ . Almost all of the runs latched on to the approach of constructing a random chain with a good hitting probability (many runs referred to this as the "Lubell method", after the Lubell of the LYM inequality).
        
        Another notable fact is that none of the runs highlighted the von Mangoldt process that was a prominent feature of the original run (and none of them mention flow networks either). Runs 4 and 7 have an interesting alternate construction of the upward divisibility chain in terms of exponential clocks in the prime factorization indices that actually looks rather tractable to work with; I will need to study this construction further when I have more time.
        
        Basically it seems that for this particular type of problem there are several natural ways to proceed that make the problem actually quite tractable; the literature had managed to focus on a somewhat suboptimal approach in which the opening move was to transfer the problem to a continuous setting, but the AI runs consistently stayed in the discrete world and managed to utilize various existing tools from discrete mathematics (mostly centering around methods relating to the LYM inequality) to reach a solution.
        
        TerenceTao — 18:22 on 17 Apr 2026
        
        I got some more information from Gavin Leech (Arb Research).
        Here are the run times for the ten independent runs:
        
        run-1: 65m 11s
        run-2: 58m 55s
        run-3: 57m 4s
        run-4: 52m 10s
        run-5: 53m 23s
        run-6: 47m 23s
        run-7: 67m 27s
        run-8: 2m 11s (Gavin: unsure what happened here, run could have bugged out)
        run-9: 37m 36s
        run-10: 62m 37s
        
        Runs 1 and 8 were those, where no proof was found.
        
        Here are the byte-lengths of the outputs:
        
        1 6826 *
        2 5477
        3 4318
        4 4502
        5 4060
        
        6 4920
        7 4933
        8 8780 *
        9 5919
        10 5279
        
        Funnily (or typically) the runs without proofs
        are the longest ones (in byte-count)
        
        old-bielefelder — 08:47 on 18 Apr 2026
        
        Thanks for the analysis! This makes sense to me: the method seems "robust" enough that there are many slight workable variations towards the solution (which is a good sign for a "discovery").
        
        Also in my experience it seems rare that GPT could have "stumbled" upon a completely new approach without any motivation - it seems intrinsically difficult to do so (from a computational complexity perspective). So either it can't come up with anything new or, if it *does* come up with something new, that's likely not by sheer luck. This has always been what I assume to be the case theoretically (and it contrasts with some narrative that AI could outperform humans at the highest levels simply by trying out random things). But so far before this we haven't seen clear evidence yet - mostly because the average performance level of AI is not yet high enough. Hopefully there will be more cases of this nature in the future!
        
        If so, the future of mathematics seems very bright.
        
        natso26 — 09:39 on 18 Apr 2026
        
        Can you elaborate on your point about continuous vs. discrete? I don't see at all where the transfer to the continuous setting you mention occurs in the literature. In the paper GLW24 there is a transfer to the continuous setting where the sum of $1 𝑛 l o g 𝑛$ over the primitive set of numbers with exactly $𝑘$ prime factors is estimated by a $𝑘$ -fold integral. However, the estimate they obtain on this sum is better than the estimate that one can obtain by Markov chain methods (which is perhaps unsurprising as these Markov chain methods apply for an arbitrary primitive set). So this transfer to the continuous setting does not currently seem suboptimal.
        
        On the other hand, the argument from the original paper of Erdős is not continuous at all and proceeds by partitioning the set of numbers at most $𝑁$ for large $𝑁$ into subsets corresponding to each member of a finite primitive set. Later arguments on an arbitrary primitive set, such as Lichtman's, build on this one and also seem discrete to me.
        
        As Jared has pointed out, this argument can be interpreted using a "Markov chain" which passes from $𝑛$ to $𝑛 / 𝑝 𝑛$ with probability $1$ where $𝑝 𝑛$ is the largest prime factor of $𝑛$ , which while deterministic in the downwards direction has an interesting invariant measure and upwards version. This can be viewed as a deterministic approximation to the Markov chain that led to a solution. In some sense, the flaw in this approach is that it is too discrete, working with partitions of sets of numbers when it should work with functions on numbers (either the transition probabilities of the Markov chain, or the invariant measure, or the flows...).
        
        WillSawin — 12:26 on 18 Apr 2026
        
        I guess what I meant was more "Archimedean" vs. "arithmetic" rather than "continuous" vs. "discrete". In the previous arguments, the main data being tracked for a given number $𝑛$ were, on the one hand, its Archimedean size (as measured by say $l o g 𝑛$ or $l o g l o g 𝑛$ ), as well as its largest prime factor $𝑃 (𝑛)$ , which is arithmetic, but is only a single numerical statistic. So this type of data can be tracked either by a continuous analysis (using some sort of Mertens weight like $\prod 𝑝 \leq 𝑃 (1 - 1 𝑝)$ to track the influence of the largest prime factor) or a discrete analysis. But the von Mangoldt type methods require one to track all the prime factors, not just the largest one (this is particularly clear in the Arb research versions of the argument that rely on Arratia-type distributions of natural numbers with nice distributions of prime factor valuations). Somehow it is using "all" of the adelic structure present in the problem, whereas the previous approaches are relying more (though not 100%) on the Archimedean structure only.
        
        TerenceTao — 15:18 on 18 Apr 2026
        
        When you say "all" of the adelic structure, this is exactly my thoughts in different language. In the Erdos Primitive Set Conjecture, I introduced the weaker notion of lexicographically-primitive (i.e. L-primitive) sets, which capture just the information of the largest prime $𝑃 (𝑛)$ . With this definition, Erdos' orginal 1935 proof holds for L-primitive sets. The key difference in [L23] is to study the *second* largest prime factor (c.f. the key Lemma 3.1) by associating a family of L-primitive sets to the initially given primitive set.
        
        For a long time, I had the idea to iterate this $𝑘$ times to obtain a nested sequence of L-primitive sets for the initial primitive set, to capture the largest $𝑘$ primes. Perhaps this would correspond to recovering Terry's notion of "all adelic structure" in the limit as $𝑘 \to \infty$ .
        
        However, the corresponding translation to the Archimedian setting produced a $𝑘$ dimensional integral optimization problem (which [L23] trivially solves when $𝑘 = 1$ ), which seems quite complicated. That's why it was so satisfying to see this "Gordian knot" being cut by the arithmetic approach.
        
        P.S. Incidentally, I'd been periodically asking Noam Brown at OpenAI, along with Freddie Manners more recently at Deepmind, and there had been some mixed results for small $𝑘$ . However, the integral optimization in general seems quite complicated.
        
        JaredDukerLichtman — 21:05 on 18 Apr 2026
        
        I had GPT-5.4 Thinking dig deeper into the analysis of the GPT-5.4 Pro reasoning trace and used your post as a basis for driving the analysis. Here is its summary, followed by links to [1] Erdös Problem 1196: GPT-5.4 Pro provenance audit of its own solution trace and [2] Companion prompt-analysis thread. In [1], the audit thread directly follows the solution to the Erdös Problem 1196. In [2], you need to scroll past an initial discussion of reasoning in general to get to the relevant portion.
        
        GPT-5.4T’s Summary
        I think your two harder questions were exactly the right ones, so I tried to probe them directly by running a same-thread post-solution interrogation of the original GPT-5.4 Pro conversation. I’ll link both the model’s self-audit thread and the companion analysis thread separately.
        My main conclusion is that the interrogation did not recover the exact microscopic eureka moment. In particular, the precise point where the von Mangoldt route first appeared, and the precise point where it became obviously proof-closing, both remain partly opaque. So the original complaint still stands in an important sense: the visible reasoning summary is too lossy at the most mathematically interesting step.
        That said, the follow-up audit did recover something useful about the selection dynamics, which gets at your second question. The best nutshell answer was:
        The winning branch was retained not because it had fewer defects, but because its defects were more repairable.
        What emerged was not a story of one branch being obviously superior from the outset. Rather, the surviving branch kept turning uncertainties into explicit local checks that then passed: normalization, monotone divisibility / one-hit structure, disjoint last-step events, predecessor recurrence, and manageable source mass. The losing branches, by contrast, kept re-encountering the same native obstruction after attempted repairs. The clearest case was the prime-only least-prime-factor / Buchstab-style branch, which remained haunted by the same shift defect even after several refinements.
        So my present read is that the search was not a coherent top-down strategic narrative in the human sense. But neither did it look like pure unguided wall-throwing. The best fit was something like local opportunistic exploration with recurring structural motifs: repeated attempts to realize the right occupation scale, exploit one-hit behavior along divisibility chains, replace approximate recursions by exact local closure, and control the boundary/source term. In that sense there was some strategic spine, but it seems to have been expressed mainly through iterated local repair and viability filtering, not through one stable master plan.
        So, at least in this instance, I would answer your two questions this way:
        1. Why was the von Mangoldt branch retained?
        Because it kept shedding structural debt: each apparent defect was converted into a local check that passed, whereas rival branches kept carrying one unresolved core defect.
        2. Was there coherent strategy?
        Not in the strong “top-down proof plan” sense. The recoverable record looks more like opportunistic search plus local verification/repair, with some recurring strategic motifs but no single clean narrative.
        This still leaves your survivor-bias point entirely intact. The case is informative, but it is still a successful instance analyzed after the fact. So I agree that the scientifically cleaner next step would be repeated, contamination-controlled runs on #1196 announced in advance, with a commitment to report failures as well as successes.
        
        Citations
        [1] Erdös Problem 1196: GPT-5.4 Pro provenance audit of its own solution trace — https://chatgpt.com/share/69e208a2-decc-83eb-be4d-796a0a2160fd .
        [2] Companion prompt-analysis thread — https://chatgpt.com/share/69e29938-dc3c-83eb-ab12-d21274b81f8b
        
        uwepleban — 21:02 on 17 Apr 2026
I got Gemini to do a literature review on the GPT-generated proofs provided by Arb Research, and connections to existing probabilistic constructions of prime factorizations, particularly focusing on the work of Arratia. I summarize some of the points of that review here.

First, we recall the zeta distribution $𝑍 𝑠$ , which for any $𝑠 > 1$ is a probability distribution on the natural numbers with law

$𝐏 (𝑍 𝑠 = 𝑛) = 1 𝜁 (𝑠) 𝑛 𝑠 .$

Crucially, because of the Euler identity $𝜁 (𝑠) = \prod 𝑝 (1 - 𝑝 - 𝑠) - 1$ (as well as the fundamental theorem of arithmetic identity $1 𝑛 𝑠 = \prod 𝑝 𝑝 - 𝑣 𝑝 (𝑛) 𝑠$ ), $𝑍 𝑠$ has a very clean prime factorization
$𝑍 𝑠 = \prod 𝑝 𝑝 𝑒 𝑝, 𝑠$
where the $𝑒 𝑝, 𝑠$ are independent geometric random variables with mean $𝑝 𝑠$ , thus
$𝐏 (𝑒 𝑝, 𝑠 = 𝑘) = 𝑝 - 𝑘 𝑠 (1 - 𝑝 - 𝑠) .$

For $𝑠$ close to one, $𝑍 𝑠$ tends to obey the size law $l o g 𝑍 𝑠 = 𝑂 (1 𝑠 - 1)$ . So we expect $𝑍 𝑠$ to go to infinity as $𝑠$ tends to $1$ . But it turns out we can make this intuition far more precise and useful, coupling all these individual zeta random variables together into a single "zeta process" with nice properties.

A geometric random variable of mean $𝑝 𝑠$ can also be thought of as the longest consecutive success time of a sequence of independent events, each having a success probability of $𝑝 - 𝑠$ . One example of such an event is when an exponential variable of rate $l o g 𝑝$ exceeds $𝑠$ . Thus, if for each prime $𝑝$ we create an infinite sequence $𝐸 𝑝, 1, 𝐸 𝑝, 2, \dots$ of exponential clocks of rate $l o g 𝑝$ , i.e., independent random variables with the exponential distribution
$𝐏 (𝐸 𝑝, 𝑘 \geq 𝑠) = 𝑝 - 𝑠$
then we can couple all the zeta variables $𝑍 𝑠$ together by defining the $𝑒 𝑝, 𝑠$ to be the last time $𝐸 𝑝, 𝑘$ stays consistently above $𝑠$ ,
$𝑒 𝑝, 𝑠 = m a x {𝑘 : 𝐸 𝑝, 1, \dots, 𝐸 𝑝, 𝑘 \geq 𝑠}$
or equivalently the first time $𝐸 𝑝, 𝑘$ dips below $𝑠$ , minus $1$ ,
$𝑒 𝑝, 𝑠 = m i n {𝑘 : 𝐸 𝑝, 𝑘 < 𝑠} - 1,$
and then setting $𝑍 𝑠 : = \prod 𝑝 𝑝 𝑒 𝑝, 𝑠$ . By doing so, the $𝑒 𝑝, 𝑠$ are now non-increasing in $𝑠$ , and hence the zeta variables have formed a continuous divisibility chain: $𝑍 𝑠 2 | 𝑍 𝑠 1$ when $1 < 𝑠 2 < 𝑠 1$ . Thus, as $𝑠$ decreases down to $1$ , $𝑍 𝑠$ will mostly stay constant, but experience upward "jumps" every so often when it gets multiplied by a copy of itself. Or: as $𝑠$ increases up to $1$ , $𝑍 𝑠$ is mostly constant, but occasionally "jumps" downward to a factor of itself, caused by one of the $𝑒 𝑝, 𝑘$ dropping to some lower value.

How often does this drop occur? Let $𝑠 > 1$ , and $𝑑 𝑠$ be infinitesimal. If $𝑍 𝑠 = 𝑛$ for some natural number $𝑛$ , then there will be a drop between $𝑍 𝑠$ and $𝑍 𝑠 + 𝑑 𝑠$ if, for some prime $𝑝$ , one of the exponential random variables $𝐸 𝑝, 𝑖$ , $1 \leq 𝑖 \leq 𝑒 𝑝, 𝑠$ is not just greater than or equal to $𝑠$ , but also less than $𝑠 + 𝑑 𝑠$ . Conditionally on the first statement, the second statement occurs with probability $- 𝑑 𝑑 𝑠 𝑝 - 𝑠 𝑑 𝑠 𝑝 - 𝑠 = l o g 𝑝 𝑑 𝑠$ up to higher order terms, for a given $𝑝$ and $𝑖$ . Thus, up to higher order terms, the probability of a jump is
$\sum 𝑝 𝑒 𝑝, 𝑠 \sum 𝑖 = 1 l o g 𝑝 𝑑 𝑠 = l o g 𝑛 𝑑 𝑠 .$
Thus, if one increases $𝑠$ from a starting point with $𝑍 𝑠 = 𝑛$ , there is an exponential clock with rate $l o g 𝑛$ to time when a drop occurs. The same analysis shows that the value $𝑍 𝑠'$ that $𝑍 𝑠$ drops to is determined by the downwards von Mangoldt process, i.e., it will drop to $𝑛 / 𝑞$ with probability $Λ (𝑞) / l o g 𝑛$ . So this process is a "Poissonization" of the von Mangoldt process, where we have reindexed using continuous time with exponential clock spacings, rather than indexed by the natural numbers. (AI search suggests that this type of Poissonization trick goes back to this 1968 paper of Athreya and Karlin.) The upwards jump process can also be analyzed in a similar fashion (comparing $𝑍 𝑠$ to $𝑍 𝑠 - 𝑑 𝑠$ ), but the formulae are a little bit messier (in particular the transition probabilities now depend on $𝑠$ in an arithmetic fashion).

So we have now generated an infinite divisibility chain process. How often is any given natural number $𝑛$ hit? Any such hit will create a (unique) transition point where $𝑍 𝑠 = 𝑛$ and $𝑍 𝑠 + 𝑑 𝑠 \neq 𝑛$ for some infinitesimal $𝑑 𝑠$ . By the above analysis, the probability that this occurs for a given $𝑠, 𝑑 𝑠$ is approximately $1 𝜁 (𝑠) 𝑛 𝑠 l o g 𝑛 𝑑 𝑠$ , and so the hitting probability is
$\int \infty 1 1 𝜁 (𝑠) 𝑛 𝑠 l o g 𝑛 𝑑 𝑠$
which is one of our formulae for $𝜈 (𝑛)$ ! So, because any divisibility chain hits a primitive set $𝐴$ at most once, we again get a proof of
$\sum 𝑛 \in 𝐴 𝜈 (𝑛) \leq 1$
which when combined with routine asymptotics for $𝜈$ gives a solution to #1169.

At each $𝑠 > 1$ , the drop in $𝑍 𝑠$ should occur in time of the order of $1 l o g 𝑍 𝑠$ , which is in turn of the order of $𝑠 - 1$ . So we should expect about one such drop for every dyadic shell $[1 + 𝑒 - 𝑚 - 1, 1 + 𝑒 - 𝑚]$ of $𝑠 > 1$ on the average. This is broadly consistent with the Hardy-Ramanujan law $𝜔 (𝑛) \sim l o g l o g 𝑛$ given the heuristic $l o g 𝑍 𝑠 \sim 1 𝑠 - 1$ . It should also "explain" the Erdos-Kac law via some sort of martingale central limit theorem, but I haven't worked out the details of this.

The work of Arratia, Barbour, and others analyzed individual zeta distributions $𝑍 𝑠$ for fixed choices of $𝑠$ to achieve various couplings between the prime factorization of large random natural numbers at fixed scales, and the Poisson-Dirichlet process, giving yet another proof of Billingsley's theorem; I would imagine that this approach would also recover other basic results in the anatomy of integers such Dickman's theorem on the distribution of smooth numbers. The novelty here uncovered by the AI proofs (together with subsequent human analysis) is that these zeta distributions $𝑍 𝑠$ can be coupled together in a natural way to generate an infinite divisibility process, which turns out to be particularly pleasant to work with when Poissonized into a continuous process rather than a discrete one.

The description of this "zeta process" in terms of the independent clock variables $𝐸 𝑝, 𝑘$ makes it quite tractable for all sorts of probabilistic computations. It may be worthwhile exploring other results or problems in the anatomy of integers in terms of this process, and specifically in terms of these clock variables.

EDIT: Here is a quiver diagram reflecting my current mental "map" between all these concepts.

TerenceTao — 18:36 on 18 Apr 2026
- As another application of the aforementioned zeta process, there is a conceptually straightforward proof of a weak form of Billingsley's theorem, in which one uses the Dirichlet density in place of the natural density, i.e., one gets the Poisson-Dirichlet process for the normalized prime factors $𝑃 𝑛 : = (l o g 𝑝 1 l o g 𝑛, \dots, l o g 𝑝 𝑘 l o g 𝑛)$ if $𝑛$ is drawn from the zeta distribution $𝑍 1 + 𝜎$ in the limit $𝜎 ↓ 0$ rather than from the natural (or logarithmic) measure on $[1, 𝑥]$ as $𝑥 \to \infty$ . A sketch of the argument is as follows. Pick a large parameter $𝑇$ , let $𝜎 > 0$ be sufficiently small depending on $𝑇$ , and consider the upwards process from $𝑍 1 + 𝜎$ to $𝑍 1 + 𝑇 𝜎$ . This will involve a stochastic number of transitions governed by the downwards von Mangoldt process, with the number of transitions generally close to $l o g 𝑇$ . Every time one has a transition from $𝑍 𝑠$ to $𝑍 𝑠 + 𝑑 𝑠$ in which $𝑛 = 𝑍 𝑠$ is dropped to $𝑛 / 𝑞 = 𝑍 𝑠 + 𝑑 𝑠$ , the normalized prime distributions $𝑃 𝑍 𝑠 + 𝑑 𝑠, 𝑃 𝑠$ are basically related by the relation
  $𝑃 𝑠 = (1 - l o g 𝑞 l o g 𝑛) \cdot 𝑃 𝑠 + 𝑑 𝑠 \cup {l o g 𝑞 l o g 𝑛};$
  this isn't quite right because of the possibility of $𝑞$ being a prime power rather than prime, but for $𝑠$ close to $1$ this possibility is negligible. From Mertens' theorem, $l o g 𝑞 l o g 𝑛$ is very close to uniformly distributed in $[0, 1]$ if one uses the von Mangoldt distribution. So this recursion is basically describing the stick breaking process, and after $≍ l o g 𝑇$ applications of this process (so that $𝑃 1 + 𝜎$ is described iteratively in terms of $𝑃 1 + 𝑇 𝜎$ ) one should start becoming very close to the Poisson-Dirichlet process if one starts sending $𝑇$ slowly to infinity (with the contribution of $𝑃 1 + 𝑇 𝜎$ being almost entirely localized to a $𝑂 (1 / 𝑇)$ -neighborhood of the origin and ultimately negligible in this limit).
  
  A similar argument shows the Dirichlet density version of Dickman's theorem, that if $𝑛$ is drawn from $𝑍 𝑠$ then the probability that $𝑛$ is $𝑛 1 / 𝑢$ -smooth converges to $𝜌 (𝑢)$ in the limit $𝑠 ↓ 1$ , where the Dickman function $𝜌$ is defined through the delay-integral equation $𝜌 (𝑢) = 1 𝑢 \int 10 𝜌 (𝑢 - 𝑡) 𝑑 𝑡$ for $𝑢 > 1$ , with $𝜌 (𝑢) = 1$ for $𝑢 \leq 1$ .
  
  One can also get the Dirichlet density version of the Hardy-Ramanujan law. The weakest form of this is $𝐄 𝜔 (𝑍 1 + 𝜎) \approx l o g 1 𝜎$ , which one can get by observing that as one transitions from $𝑍 1 + 𝜎$ to $𝑍 1 + 𝜎 + 𝑑 𝜎$ , there is a $l o g 𝑍 1 + 𝜎 𝑑 𝜎$ chance of $𝜔 (𝑍 1 + 𝜎)$ decrementing by one, leading to the equation
  $𝑑 𝑑 𝜎 𝐄 𝜔 (𝑍 1 + 𝜎) \approx - 𝐄 l o g 𝑍 1 + 𝜎 .$
  The right-hand side can be computed directly using the zeta distribution to be approximately $- 1 𝜎$ , giving the claim. A more complicated argument of this type can also control the second moment $𝐄 (𝜔 (𝑍 1 + 𝜎) - l o g l o g 𝑍 1 + 𝜎) 2$ to be able to conclude that $𝜔 (𝑍 1 + 𝜎) \sim l o g l o g 𝑍 1 + 𝜎$ statistically in the limit $𝜎 ↓ 0$ . Probably one can get the Erdos-Kac law for Dirichlet density by similar reasoning.
  
  If one could couple all these facts with the distribution of $l o g 𝑍 1 + 𝜎$ and get some sort of local limit theorem then one could upgrade Dirichlet density to logarithmic or natural density. The calculations get somewhat involved though. EDIT: from the Hardy-Littlewood Tauberian theorem one can at least get from Dirichlet density to logarithmic density for free.
  
  TerenceTao — 15:38 on 19 Apr 2026
- Very interesting — the proposed proof of [1217] appears to use a related coupled zeta-distribution / divisibility-chain viewpoint as well.
  
  Quanyu Tang — 03:56 on 20 Apr 2026
  - Indeed, it looks like one can use the zeta process to also prove [1217], adapting the solution already provided. Write $𝑐 = l i m s u p 𝑥 \to \infty 1 l o g l o g 𝑥 \sum 𝑎 𝑛 < 𝑥 1 𝑎 𝑛 l o g 𝑎 𝑛$ , then for a sequence of $𝑥$ going to infinity we have
    $1 l o g l o g 𝑥 \sum 𝑎 𝑛 < 𝑥 1 𝑎 𝑛 l o g 𝑎 𝑛 = 𝑐 + 𝑜 (1)$
    which we can rewrite using the asymptotics of $𝜈$ as
    $\sum 𝑎 \in 𝐴 \cap [1, 𝑥] 𝜈 (𝑎) = (𝑐 + 𝑜 (1)) l o g l o g 𝑥 .$
    By linearity of expectation, if we take $𝑑 1 | 𝑑 2 | 𝑑 3 | \dots$ to be the random divisibility chain arising from the jumps in the zeta process, we thus have
    $𝐄 \sum 𝑗 1 𝐴 \cap [1, 𝑥] (𝑑 𝑗) = (𝑐 + 𝑜 (1)) l o g l o g 𝑥$
    so the event $\sum 𝑗 1 𝐴 \cap [1, 𝑥] (𝑑 𝑗) \geq (𝑐 + 𝑜 (1)) l o g l o g 𝑥$ occurs with positive probability. This already gives the "single scale" version of 1217. To get the full asymptotic version, we would like the event $𝐸 𝑥$ defined by
    $\sum 𝑗 1 𝐴 \cap [1, 𝑥] (𝑑 𝑗) \geq (𝑐 - 𝜀) l o g l o g 𝑥$
    to occur with probability $≫ 𝑐, 𝜀 1$ , which implies by elementary measure theoretic arguments (apply Fatou's lemma to $1 - 1 𝐸 𝑥$ ) that the $𝐸 𝑥$ occur for infinitely many $𝑥$ with positive probability.
    
    We first observe that we can cut off the process $𝑍 𝑠$ to $𝑠 - 1 ≫ 𝑐, 𝜀 1 l o g 𝑥$ without significantly impacting the hitting probability $𝜈$ in the range $[1, 𝑥]$ by more than an epsilon type factor, so we can work in the regime $𝑠 \geq 𝑠 0 : = 1 + 𝑐 0 l o g 𝑥$ for some small constant $𝑐 0 = 𝑐 0 (𝑐, 𝜀) > 0$ .
    
    It will suffice to obtain some uniform integrability, and specifically a second moment bound
    $(𝐄 (\sum 𝑗 1 𝐴 \cap [1, 𝑥] (𝑑 𝑗)) 2) 1 / 2 ≪ 𝑐, 𝜀 l o g l o g 𝑥$
    will suffice. If we let $𝑁 𝑠$ denote the number of jumps of the zeta process after $𝑠$ , it suffices to show that
    $(𝐄 𝑁 2 𝑠 0) 1 / 2 ≪ l o g 1 𝑠 0 - 1 ≍ 𝑐, 𝜀 l o g l o g 𝑥 .$
    For infinitesimal $𝑑 𝑠$ , $𝑁 𝑠$ will equal $𝑁 𝑠 + 𝑑 𝑠 + 1$ with probability $l o g 𝑍 𝑠 𝑑 𝑠$ and $𝑁 𝑠$ otherwise if we condition on $𝑍 𝑠$ , so we have
    $𝜕 𝑠 𝐄 𝑁 2 𝑠 = - 2 𝐄 𝑁 𝑠 l o g 𝑍 𝑠 .$
    From the zeta distribution one can check that $𝐄 (l o g 𝑍 𝑠) 2 ≪ 1 (𝑠 - 1) 2$ , so by Cauchy-Schwarz
    $𝜕 𝑠 𝐄 𝑁 2 𝑠 ≪ (𝐄 𝑁 2 𝑠) 1 / 2 1 𝑠 - 1$
    or
    $𝜕 𝑠 (𝐄 𝑁 2 𝑠) 1 / 2 ≪ 1 𝑠 - 1$
    and the claim follows from integration.
    
    The solution provided in the comments in [1217] uses a slight variant of the zeta process (also constructed in a number of the Arb solutions here), in which the $𝑒 𝑝, 𝑠$ jump periodically in $1 𝑠 - 1$ with spacing determined by an exponential random variable; this leads to a slightly denser hitting probability than $𝜈$ (it no longer skips over primes) but there are now some arithmetic irregularities that make it slightly messier to work with. But it is essentially the same process asymptotically.
    
    TerenceTao — 20:25 on 21 Apr 2026

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