Erdős problem 317 — live audit, a prime obstruction, and an exact finite threshold
Access date: 2026-07-26 (UTC).
Claim labels used throughout:
- (a) elementary-rigorous: a complete argument is supplied here.
- (b) rigorous-modulo-named-theorem: the stated published theorem is used.
- (c) plausible/structural-unverified: a heuristic, conjectural route, or
literature-search miss.
- (d) computational-only: established by the accompanying exact program
over its stated finite range, with no extrapolation beyond that range.
0. Mandatory live-page gate
I fetched the live problem page, its
LaTeX endpoint, and the complete
discussion thread through
the Bright Data browser path. This was a live browser fetch, not datacenter
curl.
The live page says OPEN, 0 claimed proofs, and **Currently working on
this problem: None**. It also says:
- Interested in collaborating:
None; - Likes this problem:
None; - difficult / tractable / formalisable / working on formalisation: all
None; - formalised statement:
Yes; - 8 comments;
- last edited: 06 January 2026.
Thus none of the mandatory stop conditions is present.
Verbatim live statement
The following is copied verbatim from the live /latex/317 endpoint:
Is there some constant $c>0$ such that for every $n\geq 1$ there exists some $\delta_k\in \{-1,0,1\}$ for $1\leq k\leq n$ with\[0< \left\lvert \sum_{1\leq k\leq n}\frac{\delta_k}{k}\right\rvert < \frac{c}{2^n}?\]Is it true that for sufficiently large $n$, for any $\delta_k\in \{-1,0,1\}$,\[\left\lvert \sum_{1\leq k\leq n}\frac{\delta_k}{k}\right\rvert > \frac{1}{[1,\ldots,n]}\]whenever the left-hand side is not zero?
Listed known result and comments
The page itself records:
- (a) the non-strict lower bound in the second question follows by putting
the sum over the common denominator \([1,\ldots,n]\);
- strictness fails at small \(n\), with
\(1/2-1/3-1/4=-1/12\);
- (b) arguments of Kovac and van Doorn give the first question's weaker
upper bound
\[ 2^{-\,n(\log\log\log n)^{1+o(1)}/\log n}; \]
- (c) van Doorn's random-spacing heuristic suggests that this weaker
scale may be the true one.
The thread says comments are not verified. I read all eight nondeleted
posts and the one rendered deleted placeholder:
1. On 2025-08-25, StijnC proposed a prime-denominator analogue, gave a
heuristic, and reported an experimental value at \(n=19\).
2. Vjeko_Kovac replied that both page questions looked affirmative and that
small-\(n\) checks supported this, while distinguishing the prime variant.
3. One post is displayed only as [Post deleted].
4. Vjeko_Kovac pointed out that
\(\operatorname{lcm}(1,\ldots,n)=\exp(n+o(n))\), not \(2^n\), invalidating
an attempted inference.
5. StijnC acknowledged that error.
6. Vjeko_Kovac observed that reciprocal sums over the primes at most \(n\)
give the weaker bound \(2^{-(1+o(1))n/\log n}\).
7. StijnC discussed the LCM-scaled minimum and reported unverified values
\(874,299,1995,460,13340,2057,4114\) at selected \(n\leq32\), together
with a random-spacing heuristic.
8. Vjeko_Kovac again separated the \(e^{n+o(n)}\) LCM scale from \(2^n\).
9. On 2026-01-04, Woett defined the distinct reciprocal subset-sum set
\(Q_n\), used the result now recorded at problem 320 and pigeonhole to
obtain the bound displayed on the main page, and gave the
\(1/|Q_n|^{2+o(1)}\) random-spacing heuristic.
None is a proof claim or a current-worker marker. I use only the result
incorporated into the main page, not the unverified numerical assertions.
1. Primary-source literature audit
1. (b) Erdős and Graham,
[*Old and New Problems and Results in Combinatorial Number
Theory*](https://mathweb.ucsd.edu/~ronspubs/80_11_number_theory.pdf),
Monographie 28 de L'Enseignement Mathématique (1980), contains the
problem on printed page 42. I inspected that scanned page directly. It
discusses the \(\{-1,0,1\}\) sums, the \(2^{-n}\) target, the common
denominator \(L_n=\operatorname{lcm}(1,\ldots,n)\), and the small equality
\(1/2-1/3-1/4=-1/12\). The live page remains authoritative for the exact
modern statement above.
2. (b) R. T. Worley,
J. Austral. Math. Soc. 21 (1976), 410--413, is directly relevant and
predates the monograph. For
\[ M(n)=\min_{\eta_k\in\{-1,1\}} \left|\sum_{k=1}^n\frac{\eta_k}{k}\right|, \]
Worley proves, for every \(\varepsilon>0\) and all sufficiently large
\(n\),
\[
M(n) His last paragraph explicitly asks whether some \(n\geq5\) has \(M(n)=1/L_n\). This is a narrower predecessor of the second page question: Worley requires every coefficient to be \(\pm1\), whereas problem 317 also permits zero. 3. (b) M. N. Bleicher and P. Erdős, [*The Number of Distinct Subsums of \(\sum_{1}^{N}1/i\)*](https://www.renyi.hu/~p_erdos/1975-45.pdf), Math. Comp. 29 (1975), 29--42, proves iterated-logarithm lower bounds for the number of distinct reciprocal subset sums. This is the classical input behind the counting approach in the live comments. 4. (b) S. Bettin, L. Grenié, G. Molteni, and C. Sanna, [*A lower bound for the number of Egyptian fractions*](https://arxiv.org/abs/2509.10030), arXiv:2509.10030 (2025), proves the explicit improvement \[
\frac{\log |Q_N|}{\log 2}\geq
\left(2-\frac{3}{\log_kN}\right)
\frac{N}{\log N}\prod_{j=3}^{k}\log_jN
\quad(\log_kN\geq3/2),
\] and computes \(|Q_N|\) through \(N=154\). This verifies that the cited paper exists and says what the page's counting discussion requires. 5. (b) S. Bettin, G. Molteni, and C. Sanna, [*Small values of signed harmonic sums*](https://arxiv.org/abs/1806.05402), arXiv:1806.05402 / C. R. Math. 356 (2018), studies only coefficients in \(\{-1,1\}\) and proves \(\mathfrak m_N(\tau)<\exp(-C(\log N)^2)\) for \(C<1/\log4\). The paper also reports exact values only through \(N=64\). This does not reach the \(2^{-N}\) scale and does not address zeros in the coefficient alphabet. 6. (b) A. Gambini, R. Tonon, and A. Zaccagnini, [*Signed harmonic sums of integers with \(k\) distinct prime factors*](https://arxiv.org/abs/1911.11969), arXiv:1911.11969 / Rend. Semin. Mat. Univ. Politec. Torino 78 (2020), and O. Klurman, M. Munsch, and Y.-C. Sun, [*Small values of signed harmonic sums and logarithmic means of multiplicative functions*](https://arxiv.org/abs/2605.04694), arXiv:2605.04694 (2026), give stronger results for all-\(\pm1\) signed sums and related restricted sets. The 2026 paper explicitly records the \(\exp(-N^{1/3-\varepsilon})\) all-integer bound. These remain much larger than \(2^{-N}\) and do not settle either exact question here. I searched exact fragments of the statement, the Erdős--Graham citation, signed-harmonic-sum papers and their references, reciprocal subset-sum papers, and LCM residues modulo \(p\) and \(p^2\). (c) I found no primary source proving either live question, treating the exact \(\{-1,0,1\}\) lattice-equality issue, or proving the uniform prime-residue lemma isolated below. This is an honest search miss, not a proof that no unindexed source exists. Put and Then, exactly, Therefore: \(F(n) sufficiently large \(n\). (a) Let \(p\) be a prime with \(n/2 Then Proof. The prime \(p\) occurs exactly once in \(L_n\), and \(p\) divides \(L_n/k\) for every \(k\ne p\). Hence If \(\delta_p=0\), any nonzero \(A_n(\delta)\) has absolute value at least \(p\). If \(\delta_p=\pm1\), its absolute value is at least the distance from \(\pm r_{n,p}\) to \(0\) modulo \(p\), namely \(D_{n,p}\). Since \(D_{n,p} In particular, which proves the strict inequality in the second question for that \(n\). Equivalently, the certificate condition is (d) The standalone verifier proves Thus, restricted to the complete range \(1\leq n\leq200000\), the second question has the sharp threshold \(n=5\). The four equality witnesses are checked as exact fractions: | \(n\) | nonzero coefficients | sum | |---:|:---|:---| | 1 | \(\delta_1=1\) | \(1=1/L_1\) | | 2 | \(\delta_2=1\) | \(1/2=1/L_2\) | | 3 | \(\delta_2=1,\delta_3=-1\) | \(1/6=1/L_3\) | | 4 | \(\delta_2=1,\delta_3=-1,\delta_4=-1\) | \(-1/12=-1/L_4\) | For every remaining \(n\), the program supplies a prime \(p\) and exact residue \(r_{n,p}\notin\{1,p-1\}\). Selected rows, including the stronger single-prime numerator lower bound \(D_{n,p}\), are: | \(n\) | \(p\) | \(r_{n,p}\) | certified \(F(n)\geq D_{n,p}\) | |---:|---:|---:|---:| | 5 | 5 | 2 | 2 | | 6 | 5 | 2 | 2 | | 7 | 7 | 4 | 3 | | 10 | 7 | 3 | 3 | | 100 | 53 | 11 | 11 | | 1,000 | 503 | 276 | 227 | | 10,000 | 5,003 | 2,989 | 2,014 | | 100,000 | 50,021 | 27,007 | 23,014 | | 200,000 | 100,003 | 43,541 | 43,541 | Of the 199,996 certified values: The 61 exceptional-to-the-first-prime cases, compressed into intervals, are The canonical serialization SHA-256 The verifier's frozen run also checks that \(L_{200000}\) has 288,578 bits. On this VM it completed in 22.37 seconds with peak RSS 58,752 KB. Run: The standalone file uses only the Python standard library and exact integer arithmetic. It: 1. builds the primes by an Eratosthenes sieve; 2. builds \(L_n\) independently from the identity \(L_n/L_{n-1}=p\) when \(n\) is a power of \(p\), and \(1\) otherwise; 3. cross-checks that recurrence against iterative gcd/lcm on a prefix; 4. computes \[
r_{n,p}=\frac{L_n\bmod p^2}{p}
\] exactly (valid because \(p\Vert L_n\)); 5. recomputes every displayed sample and every two-prime case by a second prime-factorisation product that never uses the rolling big integer; 6. checks the frozen digest, histogram, sample table, and four equality witnesses. No floating point, randomized primality test, SAT solver, or enumeration of the \(3^n\) coefficient vectors is used. The following sufficient lemma would settle the second question affirmatively: > For every sufficiently large \(n\), some prime \(p\in(n/2,n]\) satisfies > \(L_n\not\equiv\pm p\pmod{p^2}\). (a) The prime-obstruction lemma proves this implication immediately. (d) The verifier proves the proposed residue lemma for every \(5\leq n\leq200000\), in fact using one of the first two interval primes. (c) The missing step is uniformity. Bertrand's theorem supplies a prime in \((n/2,n]\), and the prime number theorem supplies many, but neither controls the varying residue \(L_n/p\bmod p\). The searches above found no theorem providing that control. A failure of this sufficient lemma would not disprove the second question; it would only defeat this one-prime certificate. Extending the same scan to \(10^6\) would still be only finite evidence. The observed quadratic big-integer scaling predicts about 0.15 core-hour, so I did not spend that larger budget here. Let coordinate modulo each \(p\in\mathcal P_n\) is \(\delta_p r_{n,p}\). For \(n\geq5\), the three choices \(-r_{n,p},0,r_{n,p}\) are distinct. The possible large-prime coefficient vectors therefore give exactly \(3^{|\mathcal P_n|}\) distinct CRT classes. Let \(R_n\) be the least absolute centered representative of a nonzero one of these classes. Then Indeed, if all \(\delta_p=0\) for \(p\in\mathcal P_n\), a nonzero numerator is a multiple of \(P_n\); otherwise its nonzero CRT class has centered absolute value at least \(R_n\). (b) The prime number theorem gives \(\log P_n=n/2+o(n)\) and \(|\mathcal P_n|=O(n/\log n)\). (c) If these \(3^{|\mathcal P_n|}\) special classes behaved like random points modulo \(P_n\), one would expect Any proved lower bound of this strength along an unbounded sequence would make \(2^nF(n)/L_n\) unbounded and therefore answer the first question negatively. What is missing is precisely deterministic anti-concentration of these structured CRT classes near zero. Counting their number alone cannot exclude the class \(1\). An exact meet-in-the-middle computation of \(R_n\) costs \(\Theta(3^{|\mathcal P_n|/2})\) states. At \(n=200000\), \(|\mathcal P_n|=8392\), so this is about \(3^{4196}\approx10^{2002}\) states, not a computation that can be run here (or realistically at all). The distinct-subset-sum machinery on the live page has the complementary wall: it currently supplies only \(|Q_n|=\exp(o(n))\) distinct values. A \(2^{-n}\) gap would require extreme clustering far below the average spacing of that subexponential set. Neither the published small-signed-sum results nor the current distinct-value counts provide the missing exponential-scale construction or the CRT anti-concentration needed for a disproof. PARTIAL: Exact modular certificates prove the second inequality for every \(5\leq n\leq200000\) (and equality defeats it exactly at \(n=1,2,3,4\) in this range); the uniform prime-residue lemma and the first question remain open.2. Exact reduction
Prime-obstruction lemma
3. Verified finite result
Exact threshold through 200,000
n:p:residue\n of all 199,996 certificates has03125f544b6d378a82d00029b8884af9eaa5cc7b1dabb279ba86fd519fbbe70a.Reproduction and independence checks
python runs/erdos317_wave5u_verify.py
4. Clean uniform reductions and the exact walls
What would close the second question by this route
A CRT packet that exposes the first question's missing lemma