gpt pro 5.2 attempt on Math-757

Erdős Problem #317

1) RESTATE

For each $n\ge 1$, consider sums \[ S_n(\delta):=\sum_{k=1}^n\frac{\delta_k}{k},\qquad \delta_k\in\{-1,0,1\}. \]

1. (Exponential smallness question) Does there exist an absolute constant $c>0$ such that for every $n\ge 1$ there is a choice of $\delta_1,\dots,\delta_n\in\{-1,0,1\}$ with \[ 0<|S_n(\delta)|<\frac{c}{2^n}\ ? \] Equivalently, is \[ M_n:=\min_{\delta\in\{-1,0,1\}^n\,\colon\,S_n(\delta)\neq 0}|S_n(\delta)| \] bounded by $O(2^{-n})$?

2. (Strict lcm lower bound question) Let $L_n:=\operatorname{lcm}(1,2,\dots,n)$. It is always true that either $S_n(\delta)=0$ or $|S_n(\delta)|\ge 1/L_n$. Is it true that for all sufficiently large $n$, \[ S_n(\delta)\neq 0\ \Longrightarrow\ |S_n(\delta)|>\frac{1}{L_n}\ ? \]

2) KNOWN FACTS

1. (Obvious non-strict lcm bound) Multiply by $L_n$: \[ L_n S_n(\delta)=\sum_{k=1}^n \delta_k\frac{L_n}{k}\in\mathbb{Z}. \] Hence $S_n(\delta)=0$ or $|S_n(\delta)|\ge 1/L_n$.

2. (Strictness can fail for small $n$) For $n=4$, $\frac12-\frac13-\frac14=-\frac{1}{12}=-\frac{1}{L_4}$.

3. (Reformulation via subset sums) Let \[ Q_n:=\left\{\sum_{k\in A}\frac{1}{k}: A\subseteq\{1,\dots,n\}\right\}. \] Every $S_n(\delta)$ can be written as $q-q'$ for some $q,q'\in Q_n$ (take $\delta_k=1$ for $k\in A\setminus B$, $\delta_k=-1$ for $k\in B\setminus A$, and $0$ otherwise). Consequently \[ M_n=\min_{\substack{q,q'\in Q_n\\ q\neq q'}}|q-q'|. \] So the first question asks whether the minimal gap in $Q_n$ is $O(2^{-n})$.

4. (Best general upper bounds are much weaker than $2^{-n}$) Using information on $|Q_n|$ together with a pigeonhole principle argument yields upper bounds of the shape \[ M_n\le 2^{-\frac{n(\log\log\log n)^{1+o(1)}}{\log n}}, \] which is far larger than $2^{-n}$.

3) PROOF STRATEGY

Two complementary routes suggest themselves.

1. (Constructive) Explicitly build, for each $n$, a signed harmonic sum with value about $2^{-n}$ or smaller.
2. (Entropy/packing) Prove that the set $Q_n$ is so large and so well-distributed in an interval of length $\asymp \log n$ that two elements must be within $O(2^{-n})$.

Current technology gives only subexponential smallness via (b).

4) ATTEMPTED PROOF

Step 1: Record the exact arithmetic obstruction.

Write $S_n(\delta)=A_n/L_n$ with $A_n\in\mathbb{Z}$. Then \[ M_n=\frac{m_n}{L_n},\qquad m_n:=\min\{|A|:A\neq 0,\ A\in\{L_n S_n(\delta)\}\}. \] The strict inequality in $1b$ is equivalent to $m_n\ge 2$.

Step 2: Small-$n$ computation (meet-in-the-middle).

For $n\le 26$, one can compute $m_n$ exactly by enumerating all $3^{\lfloor n/2\rfloor}$ partial sums on each half and searching for the closest pair. The results are: In particular, for all $5\le n\le 26$ we have $m_n\ge 2$, so the strict inequality $|S_n(\delta)|>1/L_n$ holds in this range. Also, in this range $M_n\cdot 2^n$ is at most $\approx 7.874$. This is consistent with (but far from proving) the existence of a universal $c$ in $1a$.

5) OBSTACLES

1. The pigeonhole argument depends on $|Q_n|$, but $|Q_n|$ is only known to be $\exp\big(n\,f(n)/\log n\big)$ with slowly growing $f(n)$, yielding only subexponential gaps.
2. Constructive control of signed harmonic sums at scale $2^{-n}$ seems to require highly nontrivial cancellations that are not captured by standard Egyptian-fraction identities.
3. The strictness problem $1b$ amounts to showing that a certain restricted signed subset-sum of divisors of $L_n$ cannot equal $\pm 1$ for all large $n$; no known invariant rules this out.

6) FINAL

UNRESOLVED.

1. What I tried: Rewrote the questions in terms of the integer numerator $m_n$ after scaling by $L_n$; computed $m_n$ for $n\le 26$; related $M_n$ to minimal gaps in the subset-sum set $Q_n$.
2. Where it fails / stuck: I do not see how to upgrade known size estimates for $Q_n$ (or distribution information) to force $M_n\ll 2^{-n}$, nor how to rule out/ensure the equality case $M_n=1/L_n$ for infinitely many $n$.
3. What might work next: (a) Prove stronger lower bounds on $|Q_n|$ together with equidistribution/concentration estimates; (b) develop a constructive "binary-expansion"-type algorithm that keeps the partial gap shrinking by a factor $\approx 2$ each step; (c) attack the strictness question by modular obstructions showing $\sum \delta_k(L_n/k)\neq\pm 1$ for all large $n$.
4. Confidence: High confidence in the reformulations and computations shown; low confidence on the global asymptotic behavior (open problem).

7) WRITE-UP (clean, complete)

Define $S_n(\delta)=\sum_{k\le n}\delta_k/k$ with $\delta_k\in\{-1,0,1\}$. Then $L_n S_n(\delta)\in\mathbb{Z}$, so any nonzero value has absolute value at least $1/L_n$. The strict inequality question asks whether the integer $L_n S_n(\delta)$ can equal $\pm 1$ for infinitely many $n$.

For the exponential smallness question, note that $S_n(\delta)$ equals the difference of two subset sums of $\{1,1/2,\dots,1/n\}$, so the minimal nonzero absolute value $M_n$ equals the minimal gap between distinct elements of $Q_n=\{\sum_{k\in A}1/k: A\subseteq\{1,\dots,n\}\}$. Known general bounds on $|Q_n|$ combined with pigeonhole yield only subexponential upper bounds for $M_n$, much weaker than $c/2^n$. Direct computation up to $n=26$ shows $M_n\cdot 2^n$ stays below $\approx 7.874$ in that range, but this does not settle whether $\sup_n M_n 2^n<\infty$.

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