GPT 6 Astra Ultra attempt on Math-5536

Problem 261: an exact residual search for distinct dyadic representations

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. The target is to represent every $n/2^n$ as $\sum_{a\in E}a/2^a$ with distinct positive integers $a$ and $|E|\ge2$. I read 261.tex and 261_v2.tex fully; they repeat the offset identity and two infinite families. I add an exact bounded search with a proved pruning rule, rather than treating those families as coverage of every $n$.

2) WORK

In a representation with at least two positive terms, all exponents exceed $n$: the function $a/2^a$ decreases after its initial tie at $a=1,2$, and an exponent at most $n$ already contributes at least the target. Write the selected offsets as $d=a-n>0$. The equation becomes \[ n=\sum_{d\in D}(n+d)2^{-d}. \] Define $r_0=n$ and, scanning $d=1,2,\ldots$, \[ r_d=2r_{d-1}-\epsilon_d(n+d),\qquad\epsilon_d\in\{0,1\}. \] A finite representation ends at $r_d=0$. Every viable intermediate state satisfies \[ 0\le r_d<n+d+2, \] because the total available future contribution, after multiplying by $2^d$, is $\sum_{h\ge1}(n+d+h)2^{-h}=n+d+2$. Equality would require selecting the entire infinite tail and cannot lead to a finite representation. Negative residuals cannot be repaired by positive terms.

Consequently a complete search to offset $M$ needs only the integer states $r$ in this interval, together with the number of selected terms capped at two. For states with the same residual and capped count, retaining one witness loses no possible continuation. This gives $O(M(n+M))$ transition work, apart from storing witnesses, instead of enumerating all $2^M$ subsets.

The accompanying exact-rational program searched $1\le n\le100$ through offset $200$. It found witnesses for 49 values and none within that bound for 51 values. For example, \[ \frac1{2}=\frac4{2^4}+\frac5{2^5}+\frac6{2^6},\qquad \frac3{2^3}=\frac4{2^4}+\frac6{2^6}+\frac8{2^8}. \] Every found witness was independently summed using rational arithmetic. The largest first successful offset in this run was $104$.

3) ADVERSARIAL VERIFICATION

In particular, $n=10$ had no witness through offset $200$; this is a finite exclusion, not a counterexample. The state count must distinguish zero, one, and at least two selected terms to exclude the one-term equality for $n=1$. A further necessary check is that the largest exponent in any valid representation is even: clearing the largest power of two leaves all other terms, including the target, even.

4) FINAL

UNRESOLVED. The search is exhaustive within its stated bound and yields reproducible certificates. There is no uniform termination proof and no finite certificate of impossibility for any $n$.

5) SOURCES CHECKED

Both complete supplied versions. Reproducible code and all witnesses or bounded failures are in verification/batch_1_checks.py and verification/batch_1_checks.json, executed 2026-09-23. No external theorem is needed for the search.

6) COMPLETION ESTIMATE

Subjective progress toward a representation for every positive integer. COMPLETION: 15%