Erdős Problem #152

Erdős Problem #152: a quadratic supply of isolated sums

Elementary reconstruction of the known counting argument, 23 September 2026.

1) Statement and target

For a finite Sidon set $A\subset\mathbb N$, a sum $s\in A+A$ is isolated if neither $s-1$ nor $s+1$ belongs to $A+A$. Prove that the number of such sums tends to infinity with $|A|$. We reconstruct the sumset--difference-set counting method reported with DeepMind's resolution and prove a quantitative bound. All combinatorial inequalities used below are justified here; we have not run a Lean checker.

2) Three elementary identities for a finite integer set

For finite $X\subset\mathbb Z$, write \[ N_j(X)=|\{x\in X:x+j\in X\}|,\quad V(X)=|\{x\in X:x-1,x+1\in X\}|, \] and let $I(X)$ count its isolated elements. Partitioning points by whether they have zero, one or two neighbours gives \[ I(X)+2N_1(X)=|X|+V(X).\tag{1} \] We will also use \[ 4N_1(X)+N_3(X)\le3|X|+2N_2(X),\tag{2} \] \[ 2N_2(X)\le N_3(X)+2V(X)+2I(X).\tag{3} \] For $2$, the triangle inequality for symmetric difference gives $|X\mathbin\triangle(X+2)|\le2|X\mathbin\triangle(X+1)|$. Since $|X\mathbin\triangle(X+j)|=2(|X|-N_j(X))$, this implies $2N_1(X)\le|X|+N_2(X)$. Double it and add $N_3(X)\le|X|$.

For $3$, partition $X$ into maximal consecutive runs. A distance-two pair is either bridged by its middle point, giving one of the $V(X)$ points, or joins two runs separated by exactly one missing integer. Let $g$ count these latter gaps, so $N_2(X)=V(X)+g$. Each such gap has two adjacent runs. If its left run has length at least two, the last two elements of that run and the first element of the right run yield a distance-three pair of pattern $1101$. If its right run has length at least two, it yields a pair of pattern $1011$. All produced pairs are distinct: their endpoints determine the four positions, and the two patterns differ. Hence the number of long-run incidences is at most $N_3(X)$. The remaining incidences involve singleton runs; there are $I(X)$ such runs, each incident to at most two gaps. Thus $2g\le N_3(X)+2I(X)$, proving $3$.

3) Transferring correlations from sums to differences

Put $n=|A|\ge1$, $S=A+A$, $D=A-A$. The Sidon property gives \[ |S|=\frac{n(n+1)}2,\qquad |D|=n(n-1)+1.\tag{4} \] Every nonzero difference has exactly one ordered representation; zero has $n$. Each sum has two ordered representations except a double $2a$, which has one.

For $j>0$, let $Q_j$ count ordered quadruples $(a,b,c,d)\in A^4$ with $a+b+j=c+d$. Counting by $s=a+b$, every pair $s,s+j\in S$ contributes at most four. Unless one of these two sums is a double, it contributes exactly four. At most $2n$ such pairs involve a double, and each loses at most four, so \[ 4N_j(S)-8n\le Q_j\le4N_j(S).\tag{5} \] Alternatively set $t=a-c$, so $t+j=d-b$. All nonzero differences have multiplicity one. Thus counting pairs $t,t+j\in D$ gives exactly $N_j(D)$ contributions except possibly at $t=0$ or $t=-j$. Each exceptional contribution differs by at most $n-1$. It follows that \[ N_j(D)\le Q_j\le N_j(D)+2n. \] Together with $5$, \[ 4N_j(S)-10n\le N_j(D)\le4N_j(S).\tag{6} \] This holds for all positive shifts, including $1,2,3$; coincidences involving doubles have been included in the error term.

4) Cancellation and the quadratic bound

Apply $2$ to $D$. Substitute the lower bounds $6$ for its shifts one and three, and the upper bound for shift two: \[ 16N_1(S)+4N_3(S)\le3|D|+8N_2(S)+50n. \] By $3$, $8N_2(S)\le4N_3(S)+8V(S)+8I(S)$. Cancel the distance-three terms, then use $1$, multiplied by eight. This yields \[ 8|S|\le3|D|+16I(S)+50n. \] With $4$, \[ 16I(S)\ge n^2-43n-3. \] In particular the conservative bound reported with the known resolution follows: \[ \boxed{\ I(A+A)\ge\frac{n^2-100n-16}{16}.\ }\tag{7} \] For example, if $n\ge202$, then $n^2-100n-16\ge n^2/2$, so $I(A+A)\ge n^2/32$. Taking $n\ge\max(202,\lceil\sqrt{32M}\rceil)$ gives at least $M$ isolated sums. It also proves the stronger quadratic assertion. Negative right sides at small $n$ cause no difficulty.

5) Verification, outcome and sources

The finite checks exhaust all subsets of a small integer interval for $1$--$3$, and all its Sidon subsets for the exact quadruple counts, transfer inequalities and final bound. These checks support the derivation; the proofs above establish it for arbitrary finite sets.

SOLVED, reconstructing the known quadratic bound by the stated elementary method. Attribution: the sumset--difference transfer and cancellation method accompanies DeepMind's resolution in https://www.erdosproblems.com/forum/thread/152. Problem and current status: https://www.erdosproblems.com/152, checked 23 September 2026. The run argument and symmetric-difference calculation above spell out the required elementary inequalities. No novel resolution or independent verification of the linked Lean code is claimed.

The estimate refers to complete coverage of the statement by this written proof and is not a correctness probability.

6) COMPLETION ESTIMATE

COMPLETION: 100%

Erdős Problem #152: isolated sums in the sumset of a Sidon set (ROUND 2)

1) ROUND-2 OBJECTIVE

I pursue path (C). The problem is Erdős Problem #152 [Bloom152], and the relevant sumset background comes from Erdős--Sárközy--Sós and O'Bryant [ESS94,OBryant04]. The Round-1 writeup reduced the problem to understanding whether many runs in $A+A$ force many singleton runs, but left open the possibility that $A+A$ might be built mostly from runs of length $2$. In this round I rule out the extreme version of that obstruction completely: a finite Sidon set with at least $3$ elements can never have zero isolated sums, and every finite Sidon set with at least $4$ elements has at least $2$ isolated sums. This is not the full Erdős statement, but it is a genuine structural theorem beyond Round 1.

2) Round-1 FOUNDATION USED

From Round 1 I use only the following already-established facts.

• If $A$ is a finite Sidon set of size $n$, then $|A+A|=n(n+1)/2$.
• If $S\subset \mathbf Z$ is finite and $B(S)$ denotes the number of maximal runs of consecutive integers in $S$, while $I(S)$ denotes the number of singleton runs, then \[ I(S)\ge 2B(S)-|S|. \]
• The Round-1 obstacle was that a lower bound on $B(A+A)$ alone does not exclude the possibility that every run has length $2$.

I do not re-prove those facts here.

3) NEW INSIGHT / TOOL (ROUND 2)

The new tool is an elementary but decisive observation about consecutive elements.

Key new observation. A Sidon set contains at most one pair of consecutive integers.

This single lemma forces isolated sums at the edges of the sumset and kills the most obvious zero-isolated-sum constructions.

4) ATTACK PLAN (ROUND 2)

The exact Round-1 gap was this: many runs in $A+A$ need not imply any singleton run.

I now attack that gap directly by working with the outermost sums of $A+A$. Write \[ A=\{a_1<a_2<\cdots<a_n\}. \] The sums $2a_1$, $a_1+a_2$, $a_{n-1}+a_n$, and $2a_n$ are the natural candidates for isolated points. The plan is:

1. prove that $A$ has at most one consecutive pair;
2. show that if the bottom end of $A$ has no consecutive pair then $2a_1$ is isolated, and similarly at the top end $2a_n$ is isolated;
3. if there is a consecutive pair at one end, prove that the two largest (or two smallest) extremal sums are isolated instead.

This overcomes the Round-1 obstruction because it proves that the configuration $I(A)=0$ is impossible for all $n\ge 3$, and moreover gives a sharp universal lower bound $I(A)\ge 2$ for $n\ge 4$.

5) WORK (ROUND 2)

Throughout this section, \[ A=\{a_1<a_2<\cdots<a_n\}\subset \mathbf Z \] is a finite Sidon set, and \[ I(A):=\left|\{s\in A+A: s-1\notin A+A\text{ and }s+1\notin A+A\}\right|. \] By translation invariance, nothing depends on where $A$ sits in $\mathbf Z$.

Lemma 5.1. A Sidon set contains at most one pair of consecutive integers.

Proof. Suppose first that $x,x+1,y,y+1\in A$ with $x+1<y$. Then \[ x+(y+1)=(x+1)+y, \] and the two unordered pairs $\{x,y+1\}$ and $\{x+1,y\}$ are distinct, contradicting the Sidon property.

If instead the two consecutive pairs overlap, then $x,x+1,x+2\in A$, and \[ x+(x+2)=2(x+1), \] again contradicting the Sidon property. Therefore there is at most one consecutive pair in $A$. $\square$

Lemma 5.2. If $a_2\ge a_1+2$, then $2a_1$ is isolated in $A+A$.

Proof. Clearly $2a_1-1<2a_1$ is smaller than the least element of $A+A$, so only $2a_1+1$ needs attention. If $2a_1+1=u+v$ with $u,v\in A$ and $u\le v$, then $u,v\ge a_1$. Since \[ u+v=2a_1+1, \] we cannot have $u\ge a_1+1$, because then $u+v\ge 2a_1+2$. Hence $u=a_1$, so $v=a_1+1$, contradicting $a_2\ge a_1+2$. Thus $2a_1+1\notin A+A$, and $2a_1$ is isolated. $\square$

Lemma 5.3. If $a_n\ge a_{n-1}+2$, then $2a_n$ is isolated in $A+A$.

Proof. Now $2a_n+1$ is larger than the maximum of $A+A$, so only $2a_n-1$ matters. If $2a_n-1=u+v$ with $u,v\in A$ and $u\le v$, then $u,v\le a_n$. Since \[ u+v=2a_n-1, \] we cannot have $v\le a_n-1$, because then $u+v\le 2a_n-2$. Hence $v=a_n$, so $u=a_n-1$, contradicting $a_n\ge a_{n-1}+2$. Thus $2a_n-1\notin A+A$, and $2a_n$ is isolated. $\square$

Lemma 5.4. Assume $n\ge 4$ and $a_2=a_1+1$. Then both $a_{n-1}+a_n$ and $2a_n$ are isolated in $A+A$.

Proof. By Lemma 5.1, the consecutive pair $\{a_1,a_2\}$ is the unique consecutive pair in $A$. In particular, \[ a_n\ge a_{n-1}+2. \] So $2a_n$ is isolated by Lemma 5.3.

It remains to show that $a_{n-1}+a_n$ is isolated.

First, no larger sum lies at distance $1$ above it. Indeed, the next larger sum in $A+A$ is at least $2a_n$, and \[ 2a_n-(a_{n-1}+a_n)=a_n-a_{n-1}\ge 2. \] Hence \[ a_{n-1}+a_n+1\notin A+A. \]

Second, suppose for contradiction that \[ u+v=a_{n-1}+a_n-1 \] with $u,v\in A$ and $u\le v$. If $v=a_n$, then $u=a_{n-1}-1$. But then $u,u+1=a_{n-1}\in A$, giving a second consecutive pair in $A$, impossible by Lemma 5.1. If instead $v\le a_{n-1}$, then \[ u+v\le 2a_{n-1}<a_{n-1}+a_n-1, \] since $a_n\ge a_{n-1}+2$. Both cases are impossible, so \[ a_{n-1}+a_n-1\notin A+A. \] Therefore $a_{n-1}+a_n$ is isolated. $\square$

Corollary 5.5. Assume $n\ge 4$ and $a_n=a_{n-1}+1$. Then both $2a_1$ and $a_1+a_2$ are isolated in $A+A$.

Proof. Apply Lemma 5.4 to the reflected Sidon set $-A=\{-a_n<\cdots<-a_1\}$ and then negate the resulting isolated sums. $\square$

Theorem 5.6. Let $A$ be a finite Sidon set.

1. $I(A)=0$ if and only if $A=\{x,x+1\}$ for some integer $x$.
2. If $|A|\ge 3$, then $I(A)\ge 1$.
3. If $|A|\ge 4$, then $I(A)\ge 2$.

Proof. For $|A|=2$, write $A=\{x,y\}$ with $x<y$. Then \[ A+A=\{2x,x+y,2y\}. \] If $y=x+1$, this is $\{2x,2x+1,2x+2\}$ and indeed $I(A)=0$. If $y\ge x+2$, then $2x$ and $2y$ are isolated, so $I(A)=2$. This proves part $1$ for $|A|=2$.

Now assume $|A|\ge 3$. By Lemma 5.1, at least one of the end gaps $a_2-a_1$ and $a_n-a_{n-1}$ is at least $2$, so Lemma 5.2 or Lemma 5.3 gives at least one isolated sum. Hence $I(A)\ge 1$, proving part $2$.

Finally assume $|A|\ge 4$. There are three cases.

• If $a_2\ge a_1+2$ and $a_n\ge a_{n-1}+2$, then Lemmas 5.2 and 5.3 show that $2a_1$ and $2a_n$ are both isolated.
• If $a_2=a_1+1$, then Lemma 5.4 gives two isolated sums, namely $a_{n-1}+a_n$ and $2a_n$.
• If $a_n=a_{n-1}+1$, then Corollary 5.5 gives two isolated sums, namely $2a_1$ and $a_1+a_2$.

These cases are exhaustive because Lemma 5.1 forbids simultaneous consecutive pairs at both ends. Thus $I(A)\ge 2$ for every Sidon set with at least $4$ elements. $\square$

Sharpness. The bound in part $3$ is best possible. For example, \[ A=\{0,1,3,7\} \] is Sidon and \[ A+A=\{0,1,2,3,4,6,7,8,10,14\}, \] whose isolated sums are exactly $10$ and $14$. Likewise, \[ A=\{0,1,6,9,13\} \] is Sidon and has exactly two isolated sums, namely $22$ and $26$.

6) ADVERSARIAL VERIFICATION

I now try to break the new argument.

• Boundary case $|A|=2$. This is the unique genuine exception. The theorem explicitly isolates it and classifies it exactly.
• Could there be two consecutive pairs, one of them internal and one at an end? No. Lemma 5.1 rules out all pairs of distinct consecutive pairs, whether disjoint or overlapping.
• Does the proof of Lemma 5.4 accidentally use $n\ge 5$? No. The only place where size matters is to ensure that after a bottom consecutive pair there is still a top end to talk about; $n\ge 4$ is enough.
• Could $a_{n-1}+a_n-1$ be represented using two elements below $a_{n-1}$? No, because then it would be at most $2a_{n-1}$, which is strictly smaller than $a_{n-1}+a_n-1$ when $a_n-a_{n-1}\ge 2$.
• Does translation matter? No. Replacing $A$ by $A+t$ adds $2t$ to every sum and preserves isolation.

No hidden quantifier issue remains: the theorem is uniform over all finite Sidon sets.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 proves a new theorem not present in Round 1:

If $A$ is a finite Sidon set with at least $4$ elements, then $A+A$ contains at least $2$ isolated sums. Equivalently, the Erdős property in Problem #152 is already true with $M=2$ once $|A|\ge 4$.

In particular, the Round-1 candidate obstruction "perhaps $A+A$ can have no isolated points" is now completely ruled out for all $|A|\ge 3$, and the stronger candidate "perhaps every run has length $2$" is impossible as well.

The full statement that the number of isolated sums must tend to infinity with $|A|$ remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 48%

References

References

• [Bloom152] T. F. Bloom, Erdős Problem #152, https://www.erdosproblems.com/152, accessed 2026-03-07.
• [Bloom200] T. F. Bloom, Erdős Problem #200, https://www.erdosproblems.com/200, accessed 2026-03-07.
• [Bloom201] T. F. Bloom, Erdős Problem #201, https://www.erdosproblems.com/201, accessed 2026-03-07.
• [ESS94] P. Erdős, A. Sárközy, and T. Sós, On sum sets of Sidon sets. I, Journal of Number Theory 47 $1994$, 329--347.
• [GT08] B. Green and T. Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Mathematics 167 $2008$, 481--547.
• [KSS75] J. Komlós, M. Sulyok, and E. Szemerédi, Linear problems in combinatorial number theory, Acta Mathematica Academiae Scientiarum Hungaricae 26 $1975$, 113--121.
• [MV07] H. Montgomery and R. Vaughan, Multiplicative Number Theory I. Classical Theory, Cambridge Studies in Advanced Mathematics 97, Cambridge University Press, 2007.
• [OBryant04] K. O'Bryant, A complete annotated bibliography of work related to Sidon sequences, Electronic Journal of Combinatorics 11 $2004$, Dynamic Survey 11.
• [OBryant10] K. O'Bryant, Thick subsets that do not contain arithmetic progressions, arXiv:0912.1494.
• [Semchankau20] A. Semchankau, Maximal subsets free of arithmetic progressions in arbitrary sets, arXiv:2010.04490.

Erdős Problem #152: Isolated points in the sumset of a Sidon set

1) Formal restatement (quantifiers, definitions, edge cases)

Definition (Sidon set).

A finite set $A\subset \mathbb{N}$ is a Sidon set (a $B_2$ set) if \[ a+b=c+d \ \text{with}\ a,b,c,d\in A \ \Longrightarrow\ \{a,b\}=\{c,d\}. \] Equivalently, the map $\{a,b\}\mapsto a+b$ from unordered pairs in $A$ is injective.

Definition (sumset and isolated sums).

Write \[ A+A := \left\{a+b:\ a,b\in A\right\}\subset \mathbb{Z}. \] An element $s\in A+A$ is called $\pm 1$--isolated if \[ s-1\notin A+A \quad\text{and}\quad s+1\notin A+A. \] Let \[ I(A):=\left\lvert \left\{s\in A+A:\ s\pm 1\notin A+A\right\}\right\rvert \] denote the number of isolated sums.

Problem statement (one clean formulation).

Is it true that \[ \forall M\in \mathbb{N}\ \exists n_0(M)\ \forall\text{ finite Sidon }A\subset \mathbb{N}\ (\,\left\lvert A\right\rvert\ge n_0(M)\ \Rightarrow\ I(A)\ge M\,)? \] Equivalently, does $I(A)\to \infty$ along every sequence of finite Sidon sets with $\left\lvert A\right\rvert\to\infty$?

Edge cases.

• If $\left\lvert A\right\rvert=1$, then $A+A$ is a singleton and that unique sum is isolated.
• If $\gcd(A)\ge 2$ (e.g. $A\subset 2\mathbb{Z}$), then $A+A\subset 2\mathbb{Z}$ and every element of $A+A$ is isolated (since neighbors have opposite parity). This makes the problem nontrivial mainly in the "$\gcd(A)=1$" regime.
• Translating $A$ by an integer $t$ replaces $A+A$ by $(A+A)+2t$, which preserves adjacency; hence $I(A)$ is translation invariant.

2) Quick literature/context check (high-level, only what's needed)

This question originates in work of Erdős--Sárközy--Sós on the structure of sumsets of Sidon sets, where they study the decomposition of $A+A$ into intervals of consecutive integers and related gap statistics [ESS94,ESS95]. A convenient summary of results and open questions appears in O'Bryant's Sidon set bibliography [OBryant04]; among other things it records that [ESS94] proves a quadratic lower bound on the number of "interval starts" in $A+A$ (defined below), but the number of intervals of length one (i.e. isolated points) is left as an open problem [OBryant04].

3) Attack plan

Let $S:=A+A$.

1. Relate isolated points in $S$ to the interval decomposition of $S$ into maximal runs of consecutive integers.
2. Use known lower bounds for the number of runs (intervals) in $S$ from [ESS94] to show that $S$ is "highly fragmented" (average run length bounded).
3. The key missing step is to upgrade "many runs" to "many singleton runs". One way this could happen is if one can show that run lengths cannot be almost always equal to $2$, or that the number of runs is $>(1/2+\varepsilon)\left\lvert S\right\rvert$ for large $\left\lvert A\right\rvert$.
4. Alternatively, attempt a counterexample construction in which $A+A$ is (asymptotically) a disjoint union of runs of length $\ge 2$ only.

4) Work (partial results and computations)

Throughout let $n:=\left\lvert A\right\rvert$.

Lemma 4.1 (size of the sumset).

If $A$ is Sidon then \[ \left\lvert A+A\right\rvert=\frac{n(n+1)}{2}. \]

Runs, endpoints, and isolated points.

Write the elements of $S$ in increasing order $s_1<s_2<\cdots<s_t$ where $t=\left\lvert S\right\rvert$. A run (maximal interval) in $S$ is a set of the form $\{m,m+1,\dots,m+\ell-1\}\subset S$ such that $m-1\notin S$ and $m+\ell\notin S$. Let $B(S)$ be the number of runs, and let $I(S)$ be the number of runs of length $1$. Then $I(S)=I(A)$ by definition.

Define the left-endpoints set \[ L(S):=\left\{s\in S:\ s-1\notin S\right\}. \] Then $\left\lvert L(S)\right\rvert=B(S)$, since each run has exactly one left endpoint.

Similarly define right endpoints \[ R(S):=\left\{s\in S:\ s+1\notin S\right\}, \] and $\left\lvert R(S)\right\rvert=B(S)$ as well.

Lemma 4.2 (a deterministic inequality: isolated points from many runs).

For any finite $S\subset \mathbb{Z}$, \[ I(S)\ \ge\ 2B(S)-\left\lvert S\right\rvert. \]

A quadratic lower bound for the number of runs.

Erdős--Sárközy--Sós define, for $d\ge 1$, \[ B(S,d):=\left\{s\in S:\ s-d\notin S\right\}. \] In particular $B(S,1)=L(S)$ and $\left\lvert B(S,1)\right\rvert=B(S)$. According to [OBryant04, summary of [ESS94]], there is an absolute constant $c_1>0$ such that for every finite Sidon set $A$ and every $d\ge 1$, \[ \left\lvert B(A+A,d)\right\rvert\ \ge\ c_1\, n^2. \] For $d=1$ this yields \[ B(S)=\left\lvert L(S)\right\rvert=\left\lvert B(S,1)\right\rvert\ \ge\ c_1 n^2. \] Combining with Lemma 4.1 gives that the average run length is bounded: \[ \frac{\left\lvert S\right\rvert}{B(S)} \le \frac{\tfrac12 n(n+1)}{c_1 n^2} \le \frac{1}{2c_1}\left(1+\frac1n\right). \] So $S=A+A$ is forced to have many runs (components)---indeed $\Omega(n^2)$ of them---but this alone does not force $I(S)\to\infty$, since it is still consistent with (say) all runs having length $2$.

Trivial "easy" family where the conclusion holds strongly.

If $\gcd(A)\ge 2$ then every $s\in A+A$ is isolated, so $I(A)=\left\lvert A+A\right\rvert=\tfrac12 n(n+1)$.

Small computational sanity check.

A quick backtracking search (not exhaustive over all diameters, but over sets with maximum element $\le 50$ after translation) found no Sidon set of size $4\le n\le 9$ with $I(A)=0$. This is evidence (not a proof) that "zero isolated sums" may be impossible.

Pseudocode sketch:

backtrack(A, Sums):
  if |A|=n: compute sumset S and I(A); record
  else:
    for x from last(A)+1 to Max:
      if all x+a not in Sums and 2x not in Sums:
        recurse with A \cup {x} and updated Sums

5) Verification / consistency checks

• Lemma 4.1 is standard and follows immediately from injectivity of pair-sum map.
• Lemma 4.2 is a one-line inequality from the run decomposition.
• The implication "many run starts $\Rightarrow$ bounded average run length" is immediate from $\left\lvert S\right\rvert=\sum \ell_i$.
• The "$\gcd(A)\ge 2$" family is checked directly.

6) Final

UNRESOLVED.

1. Strongest proved partial result included here. Let $A$ be a finite Sidon set of size $n$ and $S=A+A$. Then $\left\lvert S\right\rvert=\tfrac12n(n+1)$ (Lemma 4.1), and known results of Erdős--Sárközy--Sós imply that the number of runs $B(S)$ satisfies $B(S)\ge c_1 n^2$ for an absolute $c_1>0$ [ESS94,OBryant04]. Hence the average run length in $S$ is bounded by an absolute constant.
2. First precise obstacle. A lower bound on the number of runs does not control the distribution of run lengths; it is consistent with $S$ being a disjoint union of runs of length exactly $2$, which would force $I(A)=0$.
3. Most plausible next lemma. Show a strict inequality of the form $B(S) > (1/2+\varepsilon)\left\lvert S\right\rvert$ for all sufficiently large Sidon sets (or along all large Sidon sets), which by Lemma 4.2 would give $I(A)\ge 2B(S)-\left\lvert S\right\rvert\ge 2\varepsilon\left\lvert S\right\rvert\gg n^2$. Any mechanism forcing an asymptotic average run length $<2$ would already yield $I(A)\to\infty$.

7) Completion estimate

\[ \textbf{Completion: }35\%. \] (We have rigorous structural partial results and a clear obstruction, but no route currently to force singleton runs.)