gpt pro 5.4 attempt on Math-648

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}.