Erdős Problem #28

Round 4: Exact exceptional-set exponent for the Chen basis

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction. Round 3 proved that the explicit basis \[ A=E\cup O,\qquad E:=\left\{\sum_{i\ge 0}\varepsilon_i4^i:\ \varepsilon_i\in\{0,1\}\right\},\qquad O:=2E, \] satisfies $r_A(n)=2$ for density-one many $n$, while still having unbounded spikes. The remaining conceptual gap was quantitative: how sparse can the exceptional set \[ \mathcal X:=\{n\ge 1:r_A(n)\ne 2\} \] actually be for this construction?

In this round I sharpen the Round 3 obstruction from a qualitative density-one statement to an exact polynomial-order count. In particular, I prove that for this same explicit basis the exceptional set has exact exponent \[ \gamma=\log_4 3, \] and along powers of $4$ one has a closed formula \[ \#(\mathcal X\cap[1,4^m))=5\cdot 3^{m-1}-F_{2m+1}-2. \] This is strict progress: it rules out a much larger class of frequency-based proof strategies than Round 3 did.

2) Round-3 FOUNDATION USED

I use the following Round 3 results without reproving them.

1. The explicit basis $A=E\cup O$ above, with $E\cap O=\{0\}$.
2. Every $n\ge 0$ has a unique decomposition \[ n=e(n)+o(n),\qquad e(n)\in E,\ o(n)\in O, \] obtained by separating the binary digits of $n$ into even and odd positions.
3. Consequently, for every $n\ge 1$, the ordered pairs \[ (e(n),o(n))\quad\text{and}\quad(o(n),e(n)) \] are always representations of $n$, so $r_A(n)\ge 2$.
4. Round 3 proved the qualitative obstruction $r_A(n)=2$ for all but $O(N^\theta)$ integers $n\le N$, where $\theta=\log 7/(3\log 2)$. The present round strengthens this to the exact exponent $\log_4 3$ and, along powers of $4$, an exact counting formula.

3) NEW INSIGHT / TOOL (ROUND-4)

The new tool is an exact base-$4$ representation count for the same-parity pieces $E+E$ and $O+O$.

For $n\ge 0$, let $d_0(n),d_1(n),\dots\in\{0,1,2,3\}$ be the base-$4$ digits of $n$, and define \[ \omega(n):=\#\{i:\ d_i(n)=1\}. \] Define also \[ \rho(n):= \begin{cases} 2^{\omega(n)},&\text{if every base-4 digit of }n\text{ lies in }\{0,1,2\},\\ 0,&\text{otherwise.} \end{cases} \] Then:

1. the number of ordered representations of $n$ as $e_1+e_2$ with $e_1,e_2\in E$ is exactly $\rho(n)$;
2. the number of ordered representations of $n$ as $o_1+o_2$ with $o_1,o_2\in O$ is exactly \[ \mathbf 1_{2\mid n}\,\rho(n/2). \]

From this I obtain:

1. an exact characterization \[ \mathcal X=((E+E)\cup(O+O))\setminus\{1\}; \]
2. exact counts \[ |(E+E)\cap[0,4^m)|=3^m,\qquad |(O+O)\cap[0,4^m)|=2\cdot 3^{m-1}; \]
3. the exact intersection formula \[ |(E+E)\cap(O+O)\cap[0,4^m)|=F_{2m+1}, \] where $F_0=0,F_1=1$ are the Fibonacci numbers.

Combining these gives the exact growth law for the exceptional set.

4) ATTACK PLAN (ROUND-4)

The precise gap left after Round 3 was this:

Round 3 showed only that the exceptional set has density $0$. It did not determine its true scale.

The claims to prove in this round are:

1. compute the exact same-parity representation counts $r_{E,E}(n)$ and $r_{O,O}(n)$;
2. deduce an exact characterization of the exceptional set $\mathcal X$;
3. count $(E+E)$, $(O+O)$, and their intersection below $4^m$;
4. infer an exact formula for $\#(\mathcal X\cap[1,4^m))$;
5. deduce the sharp obstruction exponent $\gamma=\log_4 3$, improving the Round 3 exponent $\theta=\log 7/(3\log 2)$.

This overcomes the earlier obstacle because it replaces the qualitative binary-block argument from Round 3 by an exact base-$4$ digit analysis.

5) WORK (ROUND-4)

5.1 Exact same-parity representation counts

For $n\ge 0$, let \[ r_{E,E}(n):=\#\{(e_1,e_2)\in E^2:\ e_1+e_2=n\}, \qquad r_{O,O}(n):=\#\{(o_1,o_2)\in O^2:\ o_1+o_2=n\}. \]

Lemma 5.1. For every $n\ge 0$, \[ r_{E,E}(n)=\rho(n). \]

Proof. Write $n=\sum_{i\ge 0} d_i4^i$ in base $4$. Every element of $E$ has base-$4$ digits in $\{0,1\}$. Hence a representation \[ n=e_1+e_2,\qquad e_1,e_2\in E, \] is obtained by adding, independently at each base-$4$ position, two digits from $\{0,1\}$. Since $1+1=2<4$, there are no carries at all. Therefore:

• if some base-$4$ digit $d_i$ equals $3$, then no such representation exists;
• if $d_i=0$, the split is uniquely $0+0$;
• if $d_i=1$, the split can be either $0+1$ or $1+0$, giving exactly $2$ choices;
• if $d_i=2$, the split is uniquely $1+1$.

Thus the total number of ordered representations is exactly $2^{\omega(n)}$ when all digits lie in $\{0,1,2\}$, and $0$ otherwise. This is precisely $\rho(n)$. $\square$

Lemma 5.2. For every $n\ge 0$, \[ r_{O,O}(n)=\mathbf 1_{2\mid n}\,\rho(n/2). \]

Proof. Since $O=2E$, a representation $n=o_1+o_2$ with $o_1,o_2\in O$ is equivalent to \[ n=2e_1+2e_2=2(e_1+e_2),\qquad e_1,e_2\in E. \] Hence $r_{O,O}(n)=0$ when $n$ is odd, and for even $n$, \[ r_{O,O}(n)=r_{E,E}(n/2)=\rho(n/2) \] by Lemma 5.1. $\square$

5.2 Exact representation formula for $r_A$

Let \[ P_{EO}(n):=\{(e,o)\in E\times O:\ e+o=n\}, \qquad P_{OE}(n):=\{(o,e)\in O\times E:\ o+e=n\}. \] By the Round 3 unique mixed decomposition, for every $n\ge 1$, \[ |P_{EO}(n)|=|P_{OE}(n)|=1. \] Moreover, for $n\ge 1$, these two sets are disjoint because $E\cap O=\{0\}$ and the two summands cannot both be $0$.

The actual ordered representations counted by $r_A(n)$ are the union of the four sets \[ P_{EO}(n),\ P_{OE}(n),\ P_{EE}(n),\ P_{OO}(n), \] where $P_{EE}(n)$ and $P_{OO}(n)$ are the sets counted by $r_{E,E}(n)$ and $r_{O,O}(n)$.

The only overlaps are:

• $P_{EE}(n)\cap P_{EO}(n)$ and $P_{EE}(n)\cap P_{OE}(n)$, each of size $1$, exactly when $n\in E$ (equivalently, $o(n)=0$);
• $P_{OO}(n)\cap P_{EO}(n)$ and $P_{OO}(n)\cap P_{OE}(n)$, each of size $1$, exactly when $n\in O$ (equivalently, $e(n)=0$).

There is no overlap between $P_{EE}(n)$ and $P_{OO}(n)$ for $n\ge 1$, since $E\cap O=\{0\}$.

Therefore, for every $n\ge 1$, \begin{equation}\label{wsref-0-65-71-3a-65-78-61-63-74-2d-72-41} r_A(n)=2+r_{E,E}(n)+r_{O,O}(n)-2\mathbf 1_{E}(n)-2\mathbf 1_{O}(n). \end{equation} Combining (eq:exact-rA) with Lemmas 5.1 and 5.2 gives the exact formula \begin{equation}\label{wsref-0-65-71-3a-65-78-61-63-74-2d-72-41-2d-72-68-6f} r_A(n)=2+\rho(n)+\mathbf 1_{2\mid n}\,\rho(n/2)-2\mathbf 1_E(n)-2\mathbf 1_O(n) \qquad(n\ge 1). \end{equation}

5.3 Exact characterization of the exceptional set

Let \[ \mathcal X:=\{n\ge 1:\ r_A(n)\ne 2\}. \] Since $r_A(n)\ge 2$ for all $n\ge 1$, this is the same as \[ \mathcal X=\{n\ge 1:\ r_A(n)>2\}. \]

Proposition 5.3. One has \[ \mathcal X=((E+E)\cup(O+O))\setminus\{1\}. \]

Proof. First suppose $n\ge 1$ does not lie in $(E+E)\cup(O+O)$. Then $r_{E,E}(n)=r_{O,O}(n)=0$, so (eq:exact-rA) gives $r_A(n)=2$. Hence $n\notin\mathcal X$.

Conversely, suppose $n\in O+O$, $n\ge 1$. If $n\notin O$, then the $O+O$ representation(s) are distinct from the two mixed representations, so $r_A(n)>2$. If $n\in O$, write $n=2e$ with $e\in E$. Then \[ n=e+e \] is an $E+E$ representation distinct from the two mixed representations $(n,0)$ and $(0,n)$. So again $r_A(n)>2$.

Now suppose $n\in E+E$, $n\ge 1$, and $n\neq 1$. If $n\notin E$, then any $E+E$ representation is distinct from the two mixed representations, so $r_A(n)>2$. If $n\in E$, write $n$ in base $4$. Since $n\neq 1$, there are two possibilities:

1. $n$ has at least two base-$4$ digits equal to $1$. Then Lemma 5.1 gives $r_{E,E}(n)=2^{\omega(n)}\ge 4$, so $r_A(n)>2$.
2. $n$ has exactly one base-$4$ digit equal to $1$, so $n=4^j$ for some $j\ge 1$. Then \[ 4^j=2\cdot(2\cdot 4^{j-1}), \] and $2\cdot 4^{j-1}\in E+E$ because \[ 2\cdot 4^{j-1}=4^{j-1}+4^{j-1}. \] Hence $4^j\in O+O$, and the previous paragraph shows $r_A(4^j)>2$.

Therefore every $n\in((E+E)\cup(O+O))\setminus\{1\}$ belongs to $\mathcal X$.

Combining both directions yields \[ \mathcal X=((E+E)\cup(O+O))\setminus\{1\}. \] $\square$

5.4 Counting $E+E$, $O+O$, and their intersection below $4^m$

Fix $m\ge 1$, and define \[ B_m:=(E+E)\cap[0,4^m),\qquad C_m:=(O+O)\cap[0,4^m). \]

Lemma 5.4. \[ |B_m|=3^m. \]

Proof. By Lemma 5.1, $n\in E+E$ if and only if every base-$4$ digit of $n$ lies in $\{0,1,2\}$. Below $4^m$, there are exactly $m$ base-$4$ positions, and each position has $3$ choices. Hence $|B_m|=3^m$. $\square$

Lemma 5.5. \[ |C_m|=2\cdot 3^{m-1}. \]

Proof. Since $O+O=2(E+E)$, an integer $n\in[0,4^m)$ lies in $O+O$ if and only if $n=2t$ for some $t\in E+E$ with \[ t<2\cdot 4^{m-1}. \] Now $t\in E+E$ means that the base-$4$ digits of $t$ all lie in $\{0,1,2\}$. The additional bound $t<2\cdot 4^{m-1}$ says that the leading base-$4$ digit has only $2$ possibilities, namely $0$ or $1$, while each of the remaining $m-1$ digits has $3$ possibilities. Therefore \[ |C_m|=2\cdot 3^{m-1}. \] $\square$

Lemma 5.6. Let $I_m:=|B_m\cap C_m|$. Then \[ I_m=F_{2m+1}, \] where $F_0=0,\ F_1=1,\ F_{k+1}=F_k+F_{k-1}$.

Proof. An element of $B_m\cap C_m$ is an integer \[ n=2t<4^m \] such that both $t$ and $n$ have all base-$4$ digits in $\{0,1,2\}$.

Write \[ t=\sum_{i=0}^{m-1} x_i4^i,\qquad x_i\in\{0,1,2\}. \] We double $t$ in base $4$, processing digits from $i=0$ upward. Let $c_i\in\{0,1\}$ be the carry entering position $i$. The transition rules are: \[ \begin{array}{c|c} \text{incoming carry }0 & x_i=0,1 \text{ keep carry }0;\ x_i=2 \text{ creates carry }1,\\ \hline \text{incoming carry }1 & x_i=0 \text{ gives carry }0;\ x_i=2 \text{ keeps carry }1;\ x_i=1 \text{ is forbidden} \end{array} \] because $2\cdot 1+1=3$, which would create a forbidden base-$4$ digit $3$.

Let $u_m$ be the number of admissible strings $(x_0,\dots,x_{m-1})$ that finish with carry $0$, and let $v_m$ be the number that finish with carry $1$. We start with no incoming carry, so \[ u_0=1,\qquad v_0=0. \] The transition rules give \[ \binom{u_{m+1}}{v_{m+1}} = \begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix} \binom{u_m}{v_m}. \] Now \[ \begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}^{\!2}, \] so by induction \[ \binom{u_m}{v_m} = \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}^{\!2m} \binom{1}{0} = \binom{F_{2m+1}}{F_{2m}}. \] Finally, the condition $n<4^m$ is exactly the requirement that the final carry be $0$. Hence \[ I_m=u_m=F_{2m+1}. \] $\square$

5.5 Exact exceptional-set count and sharp exponent

Define \[ X(N):=\#\{1\le n\le N:\ r_A(n)\ne 2\}. \]

Theorem 5.7. For every $m\ge 1$, \begin{equation}\label{wsref-0-65-71-3a-65-78-61-63-74-2d-65-78-63-65-70-74-69-6f-6e-61-6c} X(4^m-1)=5\cdot 3^{m-1}-F_{2m+1}-2. \end{equation} Consequently, with \[ \gamma:=\log_4 3, \] one has \[ X(4^m-1)=\Theta(4^{m\gamma})=\Theta((4^m)^\gamma), \] and for all $N\ge 2$, \[ X(N)\ll N^\gamma. \]

Proof. By Proposition 5.3, \[ \mathcal X\cap[1,4^m)=\left((B_m\cup C_m)\cap[1,4^m)\right)\setminus\{1\}. \] Since $0\in B_m\cap C_m$ and $1\in B_m\setminus C_m$, this gives \[ X(4^m-1)=|B_m\cup C_m|-2. \] By inclusion--exclusion and Lemmas 5.4--5.6, \[ |B_m\cup C_m| = |B_m|+|C_m|-|B_m\cap C_m| = 3^m+2\cdot 3^{m-1}-F_{2m+1} = 5\cdot 3^{m-1}-F_{2m+1}. \] Therefore \[ X(4^m-1)=5\cdot 3^{m-1}-F_{2m+1}-2, \] proving (eq:exact-exceptional).

Since $F_{2m+1}=O(\varphi^{2m})$, where $\varphi=(1+\sqrt5)/2$, and $\varphi^2<3$, the main term is $5\cdot 3^{m-1}$. Thus \[ X(4^m-1)=\Theta(3^m)=\Theta(4^{m\log_4 3}). \] This proves the sharp exponent along powers of $4$.

For general $N$, choose $m$ with $4^{m-1}\le N<4^m$. By monotonicity, \[ X(N)\le X(4^m-1)\ll 3^m\ll N^{\log_4 3}. \] $\square$

5.6 Consequence: a sharper obstruction than in Round 3

Round 3 showed that the set $\{n:r_A(n)=2\}$ has density $1$. The present round proves the stronger quantitative statement \[ \#\{1\le n\le N:\ r_A(n)\ne 2\}=O(N^{\log_4 3}), \] and along infinitely many $N$ (namely $N=4^m-1$) this bound is best possible up to constant factors.

Equivalently: for this explicit basis, all deviations from the baseline value $2$ live on a set of precise polynomial order $N^{\log_4 3}$, where \[ \log_4 3\approx 0.79248. \] Thus any strengthening of Erdős--Turán that attempts to force more than $N^{\log_4 3+o(1)}$ nontrivial spikes for every basis is false.

6) ADVERSARIAL VERIFICATION

1. Overlap issue in (eq:exact-rA). The only possible overlaps between the mixed representations and same-parity representations occur when one summand is $0$, i.e. precisely for $n\in E$ or $n\in O$. There is no $P_{EE}$--$P_{OO}$ overlap for $n\ge 1$ because $E\cap O=\{0\}$.
2. Are all numbers in $O+O$ really exceptional? Yes. If $n\notin O$, any $O+O$ representation is extra beyond the two mixed ones. If $n\in O$, writing $n=2e$ with $e\in E$ produces the extra representation $n=e+e$.
3. Are all numbers in $E+E\setminus\{1\}$ really exceptional? Yes. If $n\notin E$, any $E+E$ representation is extra. If $n\in E$ and has at least two base-$4$ digits equal to $1$, then $r_{E,E}(n)\ge 4$. If $n=4^j$ with $j\ge 1$, then $n\in O+O$, already handled. The only positive element of $E+E$ not covered is $n=1$, and indeed $r_A(1)=2$.
4. Carry automaton for the intersection. The forbidden case $2\cdot 1+1=3$ is exactly the only way a base-$4$ digit $3$ can arise when doubling a digit string over $\{0,1,2\}$. Final carry $0$ is equivalent to the condition $2t<4^m$, so the automaton count matches $|B_m\cap C_m|$ exactly.
5. Consistency with earlier rounds. Round 3 gave the upper bound $O(N^\theta)$ with $\theta=\log 7/(3\log 2)\approx 0.936$. The present round improves this to the exact exponent $\log_4 3\approx 0.792$ and an exact formula on powers of $4$, so this is strict progress.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 4 strengthens the Round 3 obstruction substantially. For the explicit Chen basis $A=E\cup O$, the exceptional set is now understood sharply: \[ \{n\ge 1:\ r_A(n)\ne 2\}=((E+E)\cup(O+O))\setminus\{1\}, \] and \[ \#\{1\le n<4^m:\ r_A(n)\ne 2\}=5\cdot 3^{m-1}-F_{2m+1}-2. \] Hence the true exceptional-set exponent for this construction is exactly \[ \gamma=\log_4 3. \]

This rules out an entire class of proof strategies more strongly than Round 3 did: any universal strengthening of Erdős--Turán that would force "many" spikes---for example more than $N^{\log_4 3+o(1)}$ deviations from the baseline value $2$---is false.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 63%

9) REFERENCES

1. P. Erdős and P. Turán, On a problem of Sidon in additive number theory, and on some related problems, J. London Math. Soc. 16 $1941$, 212--215.
2. Y.-G. Chen, On the Erdős--Turán conjecture, C. R. Acad. Sci. Paris, Ser. I 350 $2012$, 933--935.

10) OUTPUT FORMAT

This file is the requested drop-in LaTeX continuation, with no preamble and no \begin{document}.

Problem 28: moments expose the spikes in the digit basis

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. The Erdős--Turán target is unboundedness of the two-term representation function of every asymptotic additive basis. I read the supplied gpt_pro_5.4/28.tex completely. It computes the exceptional-set exponent for one basis $A=E\cup2E$, where $E$ consists of nonnegative integers with base-4 digits 0 or 1. Here $r_A$ counts ordered pairs. I test how large the rare spikes are by estimating all fixed moments greater than one.

2) WORK

For $n<4^m$, let $\rho(n)$ be zero if a base-4 digit is 3, and otherwise $2^{j(n)}$, where $j(n)$ is the number of digits equal to 1. There are exactly $\rho(n)$ ordered representations using two members of $E$: each digit 1 has two choices and digits 0 or 2 have one, without carries. Hence $r_A(n)\ge\rho(n)$.

For fixed real $q>1$, independence of the $m$ digit choices gives \[ \sum_{0\le n<4^m}\rho(n)^q=(2+2^q)^m. \] On the other hand the union of four representation types gives, for $n\ge1$, \[ r_A(n)\le2+\rho(n)+\mathbf1_{2\mid n}\rho(n/2). \] The constant 2 counts the two unique mixed representations; overcounting their overlap with same-type representations only strengthens the inequality. Convexity gives $(x+y+z)^q\le3^{q-1}(x^q+y^q+z^q)$ for nonnegative arguments. Summing, and observing that $n/2<4^m$, proves \[ \sum_{1\le n<4^m}r_A(n)^q =\Theta_q\left((2+2^q)^m\right). \] The lower bound subtracts only the term $n=0$ from the digit sum; the upper bound absorbs $4^m$ because $2+2^q>4$.

Thus, with $X=4^m$, the normalized $q$th moment grows as $X^{\log_4(2+2^q)-1}$. In particular the normalized second moment has order $X^{\log_4 6-1}$, despite density-one baseline representation count 2. At $n=1+4+\cdots+4^{m-1}$, $r_A(n)\ge2^m$, so these are explicit spikes of square-root size along a sequence.

3) ADVERSARIAL VERIFICATION

All moment estimates concern this particular basis, not every basis. Ordered counts include diagonal pairs once, consistent with the digit rule. The supplied set identity must be corrected at zero: since its exceptional set is defined only for $n\ge1$, it is $((E+E)\cup(2E+2E))\setminus\{0,1\}$, not removal of 1 alone. The supplied finite counting formula already subtracts both 0 and 1 and is compatible with this correction.

4) FINAL

UNRESOLVED. This verifies quantitatively that the supplied construction has very large spikes. It gives no argument forcing spikes in an arbitrary asymptotic basis, which remains the exact missing universal step.

5) SOURCES CHECKED

The complete supplied version on 2026-09-23. The digit calculation and moment inequality are proved directly; no claim of novelty or unverified external theorem is used.

6) COMPLETION ESTIMATE

Subjective progress toward the universal unboundedness assertion. COMPLETION: 12%

Round 2: Continuation and Gap-Closure Mode (Erdős--Turán, order 2)

1) ROUND-2 OBJECTIVE

Path (A). Continue the Round 1 investigation in the difficult "thin" regime by adding a generating-function/Fourier inequality that constrains a hypothetical bounded representation function. The aim is strict progress (not a full resolution): obtain a quantitative necessary condition for bounded $r_A$, and isolate a rigorously proved obstruction to several common proof strategies.

2) ROUND-1 FOUNDATION USED

I use the following Round 1 results as given.

1. Lemma 28.1. If $A+A$ contains every integer $\ge N_0$, then with $S_N:=A\cap[1,N]$ and $m_N=|S_N|$ one has \[ \frac{m_N(m_N+1)}{2}\ge N-N_0+1,\quad\text{hence }m_N\gtrsim\sqrt N. \]
2. Lemma 28.2. For any finite $S\subseteq\{1,\dots,N\}$, \[ \max_{2\le n\le 2N} r_S(n)\ge \frac{|S|^2}{2N-1}. \]
3. Corollary 28.3. If $|A\cap[1,N]|/\sqrt N$ is unbounded along a sequence, then $\limsup_{n\to\infty} r_A(n)=\infty$.
4. Truncation: for $n\le N$, every representation $n=a+b$ uses $a,b\le N$, hence $r_A(n)=r_{A\cap[1,N]}(n)$ for $n\le N$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a generating-function inequality (a Sándor-type estimate) that links $\limsup r_A$ to $\liminf r_A$.

In addition, two rigorously checked contextual facts (used as consequences/obstructions, not as black boxes inside the proof above) are:

1. For an asymptotic basis $A$, one has $r_A(n)\ge 8$ for infinitely many $n$ (hence $\limsup r_A(n)\ge 8$).
2. There exists an asymptotic basis $A\subset\mathbb N$ such that $\{n: r_A(n)=2\}$ has natural density $1$. Thus large values of $r_A(n)$ cannot be forced on a positive-density set in general.

4) ATTACK PLAN (ROUND-2)

Round 1 gap. Round 1 proves Erdős--Turán under the extra hypothesis that $|A\cap[1,N]|/\sqrt N$ is unbounded; it leaves open the "thin" regime $|A\cap[1,N]|\asymp\sqrt N$, where bounded representation could conceivably occur.

Round 2 plan. Assume $\limsup r_A(n)=K<\infty$ and try to contradict eventual lower bounds on $r_A(n)$ by analytic control of the generating function $F(z)=\sum_{a\in A} z^a$, noting that \[ F(z)^2=\sum_{n\ge 0} r_A(n)z^n. \] Compare $F(z)^2$ to a tuned geometric series $\sum C z^n = C/(1-z)$, integrate on $|z|=r$ with $r=1-1/N$, and combine a $\log N$ upper bound for $\int |1-re^{2\pi i\alpha}|^{-1}\,d\alpha$ with a $\sqrt N$ lower bound for $F(r^2)$ forced by the growth constraints coming from the basis property.

5) WORK (ROUND-2)

5.1 Setup and notation

Let $A=\{a_1<a_2<\cdots\}\subseteq\mathbb N_0$. Define the ordered representation function \[ R(n):=r_A(n)=\#\{(i,j): a_i+a_j=n\}. \] Define the generating function (for $|z|<1$) \[ F(z):=\sum_{k\ge 1} z^{a_k}. \] Then \[ F(z)^2=\sum_{n\ge 0} R(n)z^n. \] Assume $\limsup_{n\to\infty} R(n)=K<\infty$.

5.2 A growth dichotomy lemma

Hence, in proving the main inequality, we may restrict to the complementary case: \begin{equation}\label{wsref-0-65-71-3a-67-72-6f-77-74-68-2d-63-6f-6e-64} \exists n_1\ \forall n\ge n_1:\quad a_n\le \frac{n(n+1)}{2}. \end{equation}

5.3 Sándor-type inequality: bounded $\limsup$ forces small $\liminf$

5.4 Consequences for the Erdős--Turán setting

If $A+A$ is cofinite (i.e. $r_A(n)\ge 1$ for all $n\ge N_0$), then \[ \liminf_{n\to\infty} r_A(n)\ge 1. \] If (contrary to Erdős--Turán) $\limsup r_A(n)=K<\infty$, Theorem Theorem forces \[ 1\le (\sqrt K-1)^2, \] hence $K\ge 4$.

Independently (as a known global constraint for asymptotic bases of order $2$), one has $r_A(n)\ge 8$ for infinitely many $n$, so $\limsup r_A(n)\ge 8$. Moreover, there exists an asymptotic basis with $r_A(n)=2$ on a set of density $1$, showing that any approach attempting to force "many" large values of $r_A$ (e.g. positive density) cannot work in full generality.

6) ADVERSARIAL VERIFICATION

1. Ordered vs unordered: throughout, $R(n)$ counts ordered pairs $(i,j)$, matching Round 1.
2. Use of compactness in (eq:delta-ineq): since $f$ is continuous and $f(x)<1$ pointwise on compact $I$, one has $\max_I f=1-\eta$ for some $\eta>0$; setting $(1-\delta)^2:=1-\eta$ is valid.
3. Integral estimate: the bound $\int_0^1 |1-re^{2\pi i\alpha}|^{-1}d\alpha\ll\log N$ is obtained directly via $|1-z|\ge \max\{1/N,\alpha\}$ on $[0,1/2]$, so no hidden external analytic input is used.
4. Lower bound on $F(r^2)$: for $n\le \sqrt N$, $2a_n\le n(n+1)\le N$, hence $r^{2a_n}\ge r^N$. The constant lower bound on $(1-1/N)^N$ for large $N$ is standard.
5. Interaction with Round 1: the new theorem targets the Round 1 gap (bounded $\limsup$ scenario) and is consistent with the existence of bases where $r_A(n)=2$ for density $1$.

7) FINAL (exactly one)

UNRESOLVED (BUT STRICTLY ADVANCED). Round 2 adds a fully detailed generating-function argument yielding Theorem Theorem, a quantitative constraint on bounded $\limsup r_A$, and isolates a rigorously proved obstruction to "positive-density large values" strategies (via existence of bases with $r_A(n)=2$ on density $1$).

8) COMPLETION ESTIMATE (mandatory)

COMPLETION: 40%

9) REFERENCES

1. Cs. Sándor, A note on a conjecture of Erdős--Turán, Integers 8 $2008$.
2. Y.-G. Chen, On the Erdős--Turán conjecture, C. R. Acad. Sci. Paris 350 $2012$.
3. Y. Ding and L. Zhao, A new upper bound on Ruzsa's number on the Erdős--Turán conjecture, arXiv:2307.12311.