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Math-5536-A2 · revision r1 · on Math-5536

gpt pro 5.2 attempt on Math-5536

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem #261
  2. 1. Formal Restatement
  3. 2. Quick Literature / Context Check
  4. 3. Attack Plan
  5. 4. Work
  6. 4.1 A complete proof for (a): infinitely many $n$
  7. 4.2 Tiny-case computation (evidence toward (b))
  8. 4.3 Status of (b) and (c)
  9. 5. Verification / Sanity Checks
  10. 6. FINAL
  11. Why stuck / what remains open
  12. Strongest partial result proved here
  13. Next steps (concrete)
  14. 7. Completion Estimate

Problem #261

1. Formal Restatement

For $n\in\mathbb{N}$, consider the Diophantine equation \begin{equation} \label{wsref-0-65-71-3a-32-36-31} \frac{n}{2^n}=\sum_{k=1}^{t}\frac{a_k}{2^{a_k}}, \end{equation} where $t\ge 2$ and $a_1,\dots,a_t$ are distinct positive integers.

Questions:

  1. Are there infinitely many $n$ for which (eq:261) is solvable?
  2. Is (eq:261) solvable for all $n$?
  3. Does there exist a rational $x$ for which \[ x = \sum_{k=1}^{\infty}\frac{a_k}{2^{a_k}} \] has at least $2^{\aleph_0}$ distinct solutions (i.e. uncountably many distinct increasing sequences $(a_k)$)?

2. Quick Literature / Context Check

Borwein--Loring exhibited an explicit infinite family of $n$ admitting a solution (settling (a) positively), and proposed a greedy-type algorithm and conjectures toward (b). Tengely--Ulas--Zygadło extended computations and verified solvability for all $n\le 10^4$. The existence of a rational $x$ with continuum-many representations appears open.

3. Attack Plan

  • For (a), prove the explicit identity \[ \frac{n}{2^n}=\sum_{n<k\le n+m}\frac{k}{2^k}\quad\text{for }n=2^{m+1}-m-2, \] which yields infinitely many $n$.
  • For (b), one would need a general constructive procedure producing a finite set of indices $\{a_1<\dots<a_t\}$ for each $n$. Greedy strategies (choose the largest $a$ with $a/2^a\le n/2^n$ and iterate) work numerically but lack a proof of termination.
  • For (c), one seeks a rational $x$ with extreme non-uniqueness of representation; known constructions give irrationals with uncountably many representations, and rationals with many (finite) representations, but the "continuum many" rational case remains unclear.

4. Work

4.1 A complete proof for (a): infinitely many $n$

Theorem (Borwein--Loring/Cusick family): For every integer $m\ge 1$, let \[ n := 2^{m+1}-m-2. \] Then \begin{equation} \label{wsref-0-65-71-3a-32-36-31-2d-69-64-65-6e-74-69-74-79} \frac{n}{2^n} = \sum_{k=n+1}^{n+m}\frac{k}{2^k}, \end{equation} which gives a solution to (eq:261) with $t=m$ and $a_k=n+k$. In particular, there are infinitely many $n$ for which (eq:261) is solvable.
Proof: We use the closed form for dyadic "tails": \begin{equation} \label{wsref-0-65-71-3a-74-61-69-6c} \sum_{k=N+1}^{\infty} \frac{k}{2^k} = \frac{N+2}{2^N},\qquad N\ge 1. \end{equation} (One derivation: start from $\sum_{k\ge 1} kx^k = x/(1-x)^2$ and specialize to $x=1/2$, then subtract the first $N$ terms.)

Applying (eq:tail) twice gives \[ \sum_{k=n+1}^{n+m}\frac{k}{2^k} =\sum_{k=n+1}^{\infty}\frac{k}{2^k}-\sum_{k=n+m+1}^{\infty}\frac{k}{2^k} =\frac{n+2}{2^n}-\frac{n+m+2}{2^{n+m}}. \] Therefore (eq:261-identity) is equivalent to \[ \frac{n}{2^n}=\frac{n+2}{2^n}-\frac{n+m+2}{2^{n+m}}. \] Multiplying by $2^n$ and rearranging yields \[ \frac{n+m+2}{2^m}=2 \quad\Longleftrightarrow\quad n+m+2 = 2^{m+1} \quad\Longleftrightarrow\quad n=2^{m+1}-m-2, \] which is exactly how $n$ was defined. ∎

4.2 Tiny-case computation (evidence toward (b))

The following simple depth-first search (over finite subsets of indices) can find solutions to (eq:261) for small $n$: 
Input: n, MaxExponent M
Target := n/2^n
Search over subsets A of {1,2,...,M} with |A|>=2:
    if sum_{a in A} a/2^a == Target then output A
Prune using that the terms a/2^a decrease for a>=2.
A quick run for $1\le n\le 20$ (with $M=400$) finds at least one solution for every such $n$. For example: \[ \frac{1}{2}=\frac{3}{2^3}+\frac{6}{2^6}+\frac{8}{2^8},\qquad \frac{4}{2^4}=\frac{5}{2^5}+\frac{6}{2^6}. \] This is consistent with the verified range $n\le 10^4$ in the literature.

4.3 Status of (b) and (c)

No general proof is known (to the best of the sources consulted) that (eq:261) is solvable for every $n$. Likewise, the existence of a rational $x$ with $2^{\aleph_0}$ distinct representations of the form $\sum a_k/2^{a_k}$ appears open.

5. Verification / Sanity Checks

  • The tail formula (eq:tail) can be checked by differentiating the geometric series.
  • In Theorem Theorem (Borwein--Loring/Cusick family), all $a_k=n+k$ are distinct and $t=m\ge 2$ is achieved for $m\ge 2$; for $m=1$ the identity gives a one-term representation, but $m\ge 2$ already yields infinitely many examples with $t\ge 2$.
  • The displayed small-$n$ examples can be verified by direct arithmetic.

6. FINAL

UNRESOLVED

Why stuck / what remains open

Part (a) is completely solved (infinitely many $n$). The open parts are:
  • (b) proving solvability for every $n$ (or producing a counterexample),
  • (c) finding a rational value with continuum-many distinct representations.

Strongest partial result proved here

Theorem Theorem (Borwein--Loring/Cusick family) gives an explicit infinite family of $n$ for which (eq:261) has a solution.

Next steps (concrete)

  1. Analyze the Borwein--Loring greedy algorithm: prove termination for all $n$, or find a structural obstruction.
  2. Use structural constraints from divisibility and size bounds on $a_t$ (as in later work) to reduce (b) to finitely many configurations for each $n$.
  3. For (c), attempt to build a rational $x$ admitting infinitely many independent "local rewrite" identities to force $2^{\aleph_0}$ representations.

7. Completion Estimate

55%. Part (a) is fully proved; parts (b) and (c) remain open.