GPT 6 Astra Ultra attempt on Math-648

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%