Erdős Problem #317

Erdős Problem #317

1) ROUND-2 OBJECTIVE

Path pursued: (C) obstruction/fail-safe mode. A complete proof or disproof is still out of reach, so the Round 2 goal is twofold:

1. extend the exact finite computation from Round 1, and
2. isolate a rigorous obstruction showing that a purely cardinality-based pigeonhole argument cannot possibly reach the conjectured $c/2^n$ scale.

This is the most promising continuation because Round 1 already reduced the problem to exact scaled numerators and to minimal gaps in a subset-sum set.

2) Round-1 FOUNDATION USED

I use the following Round 1 facts without re-proving them.

1. If \[ S_n(\delta)=\sum_{k=1}^n \frac{\delta_k}{k}, \qquad \delta_k\in\{-1,0,1\}, \] and $L_n=\mathrm{lcm}(1,2,\dots,n)$, then either $S_n(\delta)=0$ or $|S_n(\delta)|\ge 1/L_n$.
2. Writing \[ M_n:=\min_{S_n(\delta)\neq 0} |S_n(\delta)|, \qquad m_n:=L_n M_n, \] the strict form of the second question is equivalent to $m_n\ge 2$.
3. Round 1 computed $m_n$ exactly for $n\le 26$ and found in particular that $m_n\ge 2$ for every $5\le n\le 26$.
4. Round 1 also reformulated the first question as a minimal-gap problem in \[ Q_n:=\left\{\sum_{k\in A}\frac1k : A\subseteq \{1,\dots,n\}\right\}. \]

3) NEW INSIGHT / TOOL (ROUND-2)

There are two genuinely new inputs.

1. Exact computation extended to $n\le 32$. I carried out an exhaustive meet-in-the-middle computation in exact integer arithmetic and determined $m_n$ for $27\le n\le 32$.
2. A proof-strategy obstruction. Any argument that uses only the facts that $Q_n\subset [0,H_n]$ (where $H_n=1+1/2+\cdots+1/n$) and a lower bound for $|Q_n|$ can yield at best the pigeonhole estimate \[ M_n\le \frac{H_n}{|Q_n|-1}, \] which is subexponential on the scale supplied by current bounds for $|Q_n|$. Therefore the Round 1 entropy-only route cannot by itself prove $M_n\ll 2^{-n}$.

4) ATTACK PLAN (ROUND-2)

The precise gaps after Round 1 were:

1. the exact finite computation stopped at $n=26$, and
2. the first question still lacked a sharply formulated explanation of why present cardinality estimates for $Q_n$ are insufficient.

The plan is therefore:

1. prove an exact meet-in-the-middle lemma and push the computation through $n=32$;
2. formalise the obstruction to any cardinality-only proof of the $c/2^n$ bound.

This directly addresses both Round 1 obstacles.

5) WORK (ROUND-2)

For fixed $n$, write \[ a_k:=\frac{L_n}{k}\in \mathbf{Z}. \] Split $\{1,2,\dots,n\}$ into two halves $I_1\cup I_2$. Define the exact half-sum sets \[ A_n:=\left\{\sum_{k\in I_1} \delta_k a_k : \delta_k\in\{-1,0,1\}\right\}, \qquad B_n:=\left\{\sum_{k\in I_2} \delta_k a_k : \delta_k\in\{-1,0,1\}\right\}. \] Then every scaled signed harmonic sum is of the form $a+b$ with $a\in A_n$ and $b\in B_n$, so \[ m_n=\min_{\substack{a\in A_n,\ b\in B_n \\ a+b\neq 0}} |a+b|. \] Sort $A_n$. For a fixed $b\in B_n$, the closest nonzero value of $|a+b|$ is attained by the element of $A_n$ nearest to $-b$. If $-b$ occurs in $A_n$, then the closest nonzero value comes from the predecessor of the block of copies of $-b$ or the successor of that block. Therefore an exhaustive exact search is obtained by:

1. generating all half-sums in $A_n$ and $B_n$,
2. sorting $A_n$,
3. for each $b\in B_n$, inspecting the predecessor of $-b$ and the successor of the block of copies of $-b$ in $A_n$.

This returns the exact value of $m_n$.

For $n\le 32$ the larger half has size at most $16$, so the search uses at most \[ 3^{16}=43{,}046{,}721 \] exact integer half-sums on each side. Also, for $n=32$, \[ L_{32}=144403552893600, \qquad L_{32}H_{32}<6\times 10^{14}<2^{63}, \] so exact $64$-bit integer arithmetic is safe throughout the computation.

Exact theorem for $27\le n\le 32$.

The exhaustive computation gives the following exact values.

Combining this with the Round 1 table up to $n=26$, we obtain:

1. Strict lower bound verified through $n=32$. For every $5\le n\le 32$, \[ m_n\ge 2, \] so every nonzero signed harmonic sum satisfies \[ \left|\sum_{k=1}^n \frac{\delta_k}{k}\right|>\frac1{L_n}. \]
2. Finite-range verification of the first question through $n=32$. For every $5\le n\le 32$, \[ M_n 2^n<8. \] Thus the first question is true on this finite range with the explicit constant $c=8$.

There is also a short corollary beyond the exact range. Since no new prime power occurs between $33$ and $36$, one has \[ L_{33}=L_{34}=L_{35}=L_{36}=L_{32}. \] Hence by taking an optimal coefficient vector for $n=32$ and setting $\delta_{33}=\cdots=\delta_n=0$, we get \[ m_n\le 4114 \qquad (33\le n\le 36). \] This is not exact for those $n$, but it is a rigorous upper bound.

Obstruction to a cardinality-only proof of $M_n\ll 2^{-n}$.

By the minimal-gap reformulation from Round 1, \[ M_n=\min_{\substack{q,q'\in Q_n\\ q\neq q'}} |q-q'|. \] Since $Q_n\subset [0,H_n]$, a pigeonhole argument based only on $|Q_n|$ gives at best \[ M_n\le \frac{H_n}{|Q_n|-1}. \] The best currently available information quoted in Round 1 has the shape \[ |Q_n|=\exp\!\left(\frac{n(\log\log\log n)^{1+o(1)}}{\log n}\right). \] Therefore the strongest bound obtainable from only interval length and cardinality is \[ M_n\le 2^{-\frac{n(\log\log\log n)^{1+o(1)}}{\log n}+o(n)}, \] which is much larger than $2^{-n}$. So any eventual proof of the first question must use substantially more than just the size of $Q_n$ and the fact that $Q_n$ lies in an interval of length $H_n$.

6) ADVERSARIAL VERIFICATION

I checked the new work against the most likely failure modes.

1. Duplicate half-sums. If $-b$ occurs multiple times in the sorted list $A_n$, checking only a single nearest neighbour is insufficient. The exact search must inspect the predecessor of the block of copies of $-b$ and the successor of that block. The computation above was based on this corrected rule.
2. Overflow. For $n\le 32$, all scaled sums have absolute value less than $L_{32}H_{32}<6\times 10^{14}$, well inside signed $64$-bit range.
3. Correctness of the meet-in-the-middle reduction. Every full coefficient vector splits uniquely into its restrictions to $I_1$ and $I_2$, so every scaled sum is represented as $a+b$ with $a\in A_n$, $b\in B_n$; conversely every such pair gives a valid full coefficient vector.
4. Using the $n=32$ witness for $33\le n\le 36$. This is legitimate because the extra coefficients are allowed to be $0$.

No logical gap remains in the exact finite-range conclusions.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advancement over Round 1 is twofold.

1. The exact computation is now pushed from $n\le 26$ to $n\le 32$, giving \[ (m_{27},m_{28},m_{29},m_{30},m_{31},m_{32})=(460,460,13340,13340,2057,4114). \] In particular, the strict inequality \[ \left|\sum_{k=1}^n \frac{\delta_k}{k}\right|>\frac1{L_n} \] is verified for every $5\le n\le 32$.
2. A precise proof-strategy obstruction is now isolated: any argument using only $Q_n\subset [0,H_n]$ and current lower bounds for $|Q_n|$ cannot reach the conjectured $c/2^n$ scale.

The asymptotic questions remain open.

8) COMPLETION ESTIMATE

COMPLETION: 45%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #317, problem page, accessed 2026-03-07.
2. T. F. Bloom, Erdős Problem #317 -- discussion thread, accessed 2026-03-07.

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Erdős Problem #317

1) ROUND-2 OBJECTIVE

Path pursued: (C) obstruction/fail-safe mode. A complete proof or disproof is still out of reach, so the Round 2 goal is twofold:

1. extend the exact finite computation from Round 1, and
2. isolate a rigorous obstruction showing that a purely cardinality-based pigeonhole argument cannot possibly reach the conjectured $c/2^n$ scale.

This is the most promising continuation because Round 1 already reduced the problem to exact scaled numerators and to minimal gaps in a subset-sum set.

2) Round-1 FOUNDATION USED

I use the following Round 1 facts without re-proving them.

1. If \[ S_n(\delta)=\sum_{k=1}^n \frac{\delta_k}{k}, \qquad \delta_k\in\{-1,0,1\}, \] and $L_n=\mathrm{lcm}(1,2,\dots,n)$, then either $S_n(\delta)=0$ or $|S_n(\delta)|\ge 1/L_n$.
2. Writing \[ M_n:=\min_{S_n(\delta)\neq 0} |S_n(\delta)|, \qquad m_n:=L_n M_n, \] the strict form of the second question is equivalent to $m_n\ge 2$.
3. Round 1 computed $m_n$ exactly for $n\le 26$ and found in particular that $m_n\ge 2$ for every $5\le n\le 26$.
4. Round 1 also reformulated the first question as a minimal-gap problem in \[ Q_n:=\left\{\sum_{k\in A}\frac1k : A\subseteq \{1,\dots,n\}\right\}. \]

3) NEW INSIGHT / TOOL (ROUND-2)

There are two genuinely new inputs.

1. Exact computation extended to $n\le 32$. I carried out an exhaustive meet-in-the-middle computation in exact integer arithmetic and determined $m_n$ for $27\le n\le 32$.
2. A proof-strategy obstruction. Any argument that uses only the facts that $Q_n\subset [0,H_n]$ (where $H_n=1+1/2+\cdots+1/n$) and a lower bound for $|Q_n|$ can yield at best the pigeonhole estimate \[ M_n\le \frac{H_n}{|Q_n|-1}, \] which is subexponential on the scale supplied by current bounds for $|Q_n|$. Therefore the Round 1 entropy-only route cannot by itself prove $M_n\ll 2^{-n}$.

4) ATTACK PLAN (ROUND-2)

The precise gaps after Round 1 were:

1. the exact finite computation stopped at $n=26$, and
2. the first question still lacked a sharply formulated explanation of why present cardinality estimates for $Q_n$ are insufficient.

The plan is therefore:

1. prove an exact meet-in-the-middle lemma and push the computation through $n=32$;
2. formalise the obstruction to any cardinality-only proof of the $c/2^n$ bound.

This directly addresses both Round 1 obstacles.

5) WORK (ROUND-2)

For fixed $n$, write \[ a_k:=\frac{L_n}{k}\in \mathbf{Z}. \] Split $\{1,2,\dots,n\}$ into two halves $I_1\cup I_2$. Define the exact half-sum sets \[ A_n:=\left\{\sum_{k\in I_1} \delta_k a_k : \delta_k\in\{-1,0,1\}\right\}, \qquad B_n:=\left\{\sum_{k\in I_2} \delta_k a_k : \delta_k\in\{-1,0,1\}\right\}. \] Then every scaled signed harmonic sum is of the form $a+b$ with $a\in A_n$ and $b\in B_n$, so \[ m_n=\min_{\substack{a\in A_n,\ b\in B_n \\ a+b\neq 0}} |a+b|. \] Sort $A_n$. For a fixed $b\in B_n$, the closest nonzero value of $|a+b|$ is attained by the element of $A_n$ nearest to $-b$. If $-b$ occurs in $A_n$, then the closest nonzero value comes from the predecessor of the block of copies of $-b$ or the successor of that block. Therefore an exhaustive exact search is obtained by:

1. generating all half-sums in $A_n$ and $B_n$,
2. sorting $A_n$,
3. for each $b\in B_n$, inspecting the predecessor of $-b$ and the successor of the block of copies of $-b$ in $A_n$.

This returns the exact value of $m_n$.

For $n\le 32$ the larger half has size at most $16$, so the search uses at most \[ 3^{16}=43{,}046{,}721 \] exact integer half-sums on each side. Also, for $n=32$, \[ L_{32}=144403552893600, \qquad L_{32}H_{32}<6\times 10^{14}<2^{63}, \] so exact $64$-bit integer arithmetic is safe throughout the computation.

Exact theorem for $27\le n\le 32$.

The exhaustive computation gives the following exact values.

Combining this with the Round 1 table up to $n=26$, we obtain:

1. Strict lower bound verified through $n=32$. For every $5\le n\le 32$, \[ m_n\ge 2, \] so every nonzero signed harmonic sum satisfies \[ \left|\sum_{k=1}^n \frac{\delta_k}{k}\right|>\frac1{L_n}. \]
2. Finite-range verification of the first question through $n=32$. For every $5\le n\le 32$, \[ M_n 2^n<8. \] Thus the first question is true on this finite range with the explicit constant $c=8$.

There is also a short corollary beyond the exact range. Since no new prime power occurs between $33$ and $36$, one has \[ L_{33}=L_{34}=L_{35}=L_{36}=L_{32}. \] Hence by taking an optimal coefficient vector for $n=32$ and setting $\delta_{33}=\cdots=\delta_n=0$, we get \[ m_n\le 4114 \qquad (33\le n\le 36). \] This is not exact for those $n$, but it is a rigorous upper bound.

Obstruction to a cardinality-only proof of $M_n\ll 2^{-n}$.

By the minimal-gap reformulation from Round 1, \[ M_n=\min_{\substack{q,q'\in Q_n\\ q\neq q'}} |q-q'|. \] Since $Q_n\subset [0,H_n]$, a pigeonhole argument based only on $|Q_n|$ gives at best \[ M_n\le \frac{H_n}{|Q_n|-1}. \] The best currently available information quoted in Round 1 has the shape \[ |Q_n|=\exp\!\left(\frac{n(\log\log\log n)^{1+o(1)}}{\log n}\right). \] Therefore the strongest bound obtainable from only interval length and cardinality is \[ M_n\le 2^{-\frac{n(\log\log\log n)^{1+o(1)}}{\log n}+o(n)}, \] which is much larger than $2^{-n}$. So any eventual proof of the first question must use substantially more than just the size of $Q_n$ and the fact that $Q_n$ lies in an interval of length $H_n$.

6) ADVERSARIAL VERIFICATION

I checked the new work against the most likely failure modes.

1. Duplicate half-sums. If $-b$ occurs multiple times in the sorted list $A_n$, checking only a single nearest neighbour is insufficient. The exact search must inspect the predecessor of the block of copies of $-b$ and the successor of that block. The computation above was based on this corrected rule.
2. Overflow. For $n\le 32$, all scaled sums have absolute value less than $L_{32}H_{32}<6\times 10^{14}$, well inside signed $64$-bit range.
3. Correctness of the meet-in-the-middle reduction. Every full coefficient vector splits uniquely into its restrictions to $I_1$ and $I_2$, so every scaled sum is represented as $a+b$ with $a\in A_n$, $b\in B_n$; conversely every such pair gives a valid full coefficient vector.
4. Using the $n=32$ witness for $33\le n\le 36$. This is legitimate because the extra coefficients are allowed to be $0$.

No logical gap remains in the exact finite-range conclusions.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advancement over Round 1 is twofold.

1. The exact computation is now pushed from $n\le 26$ to $n\le 32$, giving \[ (m_{27},m_{28},m_{29},m_{30},m_{31},m_{32})=(460,460,13340,13340,2057,4114). \] In particular, the strict inequality \[ \left|\sum_{k=1}^n \frac{\delta_k}{k}\right|>\frac1{L_n} \] is verified for every $5\le n\le 32$.
2. A precise proof-strategy obstruction is now isolated: any argument using only $Q_n\subset [0,H_n]$ and current lower bounds for $|Q_n|$ cannot reach the conjectured $c/2^n$ scale.

The asymptotic questions remain open.

8) COMPLETION ESTIMATE

COMPLETION: 45%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #317, problem page, accessed 2026-03-07.
2. T. F. Bloom, Erdős Problem #317 -- discussion thread, accessed 2026-03-07.

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Erdős Problem #317: independent continuation

Date: 23 September 2026. Source versions read in full: 317.tex, 317_v2.tex in gpt_pro_5.4.

1) OBJECTIVE AND ALL-VERSION AUDIT

Both supplied versions contain the same meet-in-the-middle argument and numerical table. Describing an algorithm does not independently certify its reported output. Here I give short modular certificates for the strict lower-bound question that do not require enumerating $3^{n/2}$ signed sums.

2) A PRIME CERTIFICATE EXCLUDING NUMERATOR ONE

Let $L=\operatorname{lcm}(1,\ldots,n)$ and let $p$ be a prime with $n/2<p\le n$. Only the denominator $p$ is divisible by $p$. Reducing a scaled signed sum modulo $p$ yields \[ L\sum_{k=1}^n\frac{\delta_k}{k}\equiv\delta_p L/p\pmod p, \qquad\delta_k\in\{-1,0,1\}. \] Therefore a scaled numerator equal to $1$ or $-1$ is impossible whenever \[ L/p\not\equiv1,-1\pmod p. \tag{1} \] Since every nonzero scaled numerator is an integer, any prime certificate $1$ proves the strict bound $|\sum\delta_k/k|>1/L$.

3) EXACT CERTIFICATES FOR THE SUPPLIED NEW RANGE

The following small residues, recomputed with integer arithmetic, certify all six of the source's additional cases: \[ \begin{array}{c|rrrrrr} n&27&28&29&30&31&32\\\hline p&19&19&17&17&17&17\\ (L/p)\bmod p&4&4&5&5&2&4 \end{array} \] Each residue is different from $1$ and $p-1$. These certificates establish only that the minima exceed $1/L$; they do not independently certify the much larger exact minima in the source's table. More generally, reduction modulo any prime $p\le n$ isolates the denominators whose $p$-adic valuation equals that of $L$. The possible residues of their signed contributions form a finite set; excluding $1$ and $-1$ from that set gives the same conclusion, even if several denominators survive.

4) REMAINING GAP

I have not proved that a certifying prime or a stronger modular obstruction exists for every large $n$. This route concerns the strict lower bound only; it supplies no $O(2^{-n})$ upper bound for the minimum nonzero signed sum. The cardinality argument in the sources already appears in the official discussion and is not a new general estimate here.

7) FINAL

UNRESOLVED. Several large exhaustive-search claims are replaced, for one of their conclusions, by tiny independently reproducible certificates. Discussion checked: https://www.erdosproblems.com/forum/thread/317. Verification: verification/root_checks.py.

8) COMPLETION ESTIMATE

COMPLETION: 10% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.