Erdos problem scan: where this laboratory's built machinery has an edge

Sub-study of hunts/rogue_frontier/. Opened 2026-08-18.

Status: exploratory. Nothing here is a result. This is a feasibility survey, not mathematics. Every "best known" line below carries the URL actually fetched while writing it; anything not fetched is marked UNVERIFIED.

Question

Of the ~1100 problems on erdosproblems.com, which OPEN ones does this tree's existing, validated machinery make unusually cheap to attack, such that a first signal is reachable in hours rather than months?

The bar is deliberately not "interesting problem". It is: this specific tooling changes what is feasible here.

Method (reproducible)

1. https://raw.githubusercontent.com/teorth/erdosproblems/main/data/problems.yaml gives structured status/prize/tags for 1217 problems (604 open, plus the finite-computation classes below). Fetched 2026-08-18.
2. All 1217 problem pages fetched from https://www.erdosproblems.com/<n> (a plain user agent is required; the default agent gets 403), rendered to text, split into statement and remarks.
3. The site carries a status field that is exactly the discriminator this scan wants, and it is not visible from the YAML alone:
4. open (604): "cannot be resolved with a finite computation"
5. falsifiable (27): "could be disproved with a finite counterexample"
6. verifiable (7): "could be proved with a finite example"
7. decidable (9): "resolved up to a finite check"
8. plus not provable / not disprovable / independent (set-theoretic)
9. Page metadata also gives a mining indicator the YAML lacks: comment count, who has flagged "currently working on this problem", and the "looks tractable" / "looks difficult" votes. Recorded per candidate below.
10. Pool restricted to number-theory / analysis / combinatorics tags where this tree's tools could bite: 437 open-ish problems. Statements read in full.

What this laboratory actually brings

Scoring

Each candidate carries: F feasibility of a first signal in hours (0-5), N chance anything new comes out (probability, honest), M how heavily mined (0 = untouched, 5 = crowded), V value if it lands.

Tier 1 candidates

C1. Problem 647: is there n > 24 with max_{m<n}(m + tau(m)) <= n + 2?

• URL: https://www.erdosproblems.com/647
• Status / prize: VERIFIABLE ("could be proved with a finite example"). Erdos offered GBP 25 in [Er92e], rendered by the site as $44. Erdos and Selfridge.
• Best known, verified against the source page (fetched 2026-08-18): the site records that n = 24 works, that n+2 is best possible since max(tau(n-1)+n-1, tau(n-2)+n-2) >= n+2, that Erdos called it "extremely doubtful" that infinitely many such n exist, and that he conjectured lim_n max_{m<n}(tau(m)+m-n) = infinity. The page records no computational search bound at all. That absence is the opening.
• Tool: large exact arithmetic-function computation. A segmented divisor-count sieve with a running maximum. Nothing more exotic is needed.
• Smallest experiment, already run here: a segmented tau sieve in C, scratchpad/e647/tau647c.c, tracking D(n) = max_{m<n}(m + tau(m)) - n, reporting the per-decade minimum. Measured, 3 min 20 s to N = 10^9 on one core:

So: no n with 24 < n <= 10^9 has D(n) <= 2, and D(n) >= 12 throughout the last decade. Extrapolating the same code: 10^10 is about 35 min, 10^11 about 6 h, 10^12 about 2.5 days on one core, less with the obvious pruning (n-1 must be 1, a prime, or a prime square, so the running max only has to be maintained near primes).

• Negative result: almost certainly what you get. It is worth recording anyway, in two forms. (i) "no example with n <= 10^11" is a bound the database currently lacks entirely, and it is exactly what a VERIFIABLE problem's page should carry. (ii) the decade table is direct evidence on Erdos's stronger conjecture that D(n) -> infinity; the growth 3, 3, 5, 8, 10, 11, 12, 12 is slow and non-monotone across decades, which is itself a fact worth publishing next to the conjecture.
• P(anything new): finding an n > 24 -- under 1%. Erdos's heuristic is overwhelming and our data agrees. Producing a search bound and a growth table the database does not have -- around 70%, conditional on nobody having posted one in the 13 comments (which are not served in the static HTML, see risk).
• Mined: 13 comments, four users flagged "currently working on this problem" (Ritvik_Nayak, pommeret, ScottHughes and others). This is the main risk: a search bound may already exist inside the comment thread. Check the live comments before spending a day of compute.

C2. Problem 311: the exponential rate of delta(N)

• URL: https://www.erdosproblems.com/311
• Status / prize: OPEN, no prize.
• Statement: delta(N) is the minimal non-zero value of |1 - sum_{n in A} 1/n| over A a subset of {1,...,N}. Is delta(N) = e^{-(c+o(1))N} for some c in (0,1)?
• Best known, from the source page (fetched 2026-08-18): trivially delta(N) >= 1/lcm(1..N) = e^{-(1+o(1))N}; Tang has shown delta(N) <= exp(-c N/(log N log log N)^3). Kovac showed in the comments that the extra condition in [ErGr80] is redundant. No table of exact values of delta(N) is recorded, and OEIS returns nothing for it.
• Tool: exact integer arithmetic and a meet-in-the-middle subset-sum engine, precisely the shape sine_gram/exact_finite_N.py already uses. Multiply by L = lcm(1..N): delta(N) = min non-zero |L - sum_{n in A} L/n| / L, an exact closest-subset-sum problem in Z.
• Smallest experiment, already run here: pure-Python meet-in-the-middle, exact ints, no floating point. Measured runtimes: N = 36 in a few seconds, N = 44 in 14.6 s. Values of -ln delta(N)/N:

(delta(44) = 4.785294e-10, computed as an exact rational, not a float.) The sequence sits near 0.5 and drifts downward, which is what a non-constant rate would look like, and is the interesting reading: it is mild evidence against the conjectured e^{-(c+o(1))N} shape and consistent with Tang's polylog correction being real.

• Cost of the next rung: naive scaling is x2.4 per +2 in N, so N = 52 is about 8 min and N = 56 about 45 min with Python ints. Replacing the sort key with numpy longdouble (80-bit, absolute resolution about 2e5 at L ~ 1e24, while the expected spacing between distinct subset sums near the target is enormous) and verifying the handful of window candidates in exact Z pushes N to 58-60 inside 15 GB. That roughly doubles the range of the table.
• Negative result: the honest likely outcome is "the exponent keeps drifting, so the data neither confirms nor refutes". That is still worth recording, because the table itself is the missing artifact. A table of exact delta(N) for N <= 56 is a contribution to the problem page and to OEIS regardless of which way it points.
• P(anything new): producing the first exact table -- 90%. Producing a statement that moves the conjecture -- 10-15%.
• Mined: 8 comments, nobody flagged as working on it, no OEIS entry. Lightly mined. Kovac and Tang have both engaged, so the analytic side is watched, but the arithmetic side appears untouched.

C3. Problem 710 (and 711): exact values of the Erdos-Pomerance function f(n)

• URL: https://www.erdosproblems.com/710 , https://www.erdosproblems.com/711
• Status / prize: OPEN. Erdos offered 2000 rupees in [Er92c] for an asymptotic formula (the site renders it as $78).
• Statement: f(n) is minimal such that the open interval (n, n+f(n)) contains distinct integers a_1,...,a_n with k | a_k for every k <= n.
• Best known, from the source page (fetched 2026-08-18): Erdos and Pomerance [ErPo80] proved (2/sqrt(e) + o(1)) n (log n / log log n)^{1/2} <= f(n) <= (1.7398... + o(1)) n (log n)^{1/2}. The gap between the two ends is a factor (log log n)^{1/2}. No values of f(n) are recorded on the page, and OEIS returns nothing for the sequence.
• Tool: exact finite-N combinatorics. f(n) is a system-of-distinct- representatives question: bipartite graph with left vertices 1..n, right vertices the integers in (n, n+L), edge k -- a iff k | a. f(n) is the least L admitting a perfect matching saturating the left side. Hopcroft-Karp, or Hall's condition directly. The lab has no matching code, but this is 60 lines.
• Smallest experiment: compute f(n) for n <= 2000 by binary search on L with a matching test. Edge count is about L log n; at n = 2000 and L ~ 3n the graph has order 10^5 edges, so each matching is milliseconds and the whole table is minutes. n <= 10^5 is an afternoon. Estimated first signal: under 1 hour, including writing the code.
• What it buys: the two proven bounds differ by (log log n)^{1/2}, which at n = 10^5 is only about 1.5, so exact values across four decades genuinely discriminate between "f(n)/n(log n)^{1/2} tends to a constant" and "f(n)/n(log n/log log n)^{1/2} tends to a constant". Erdos asked for the asymptotic formula and offered money for it; the data would say which shape to aim at, and the constants 2/sqrt(e) = 1.213 and 1.7398 are close enough together to be tested numerically.
• Negative result: if f(n)/n grows too slowly to separate the two shapes over the reachable range, that is a clean statement to record ("four decades do not separate them; here is the fitted exponent"), and the table still stands as the first computation of an Erdos prize function.
• P(anything new): first table of f(n) -- 85%. A statement that visibly picks one of the two shapes -- 35%. Proving anything -- under 5%.
• Mined: 7 comments on 710, 8 on 711; one user flagged as currently working on each (SkyYang on 710, skominers on 711) and one flagged 710 "looks tractable" (Prasannam); no OEIS sequence for f(n) was found. Lightly mined, but not untouched: correct that against the earlier draft of this file.

C4. Problem 879: exact values of G(n), the maximal sum of a pairwise-coprime set

• URL: https://www.erdosproblems.com/879
• Status / prize: OPEN, no prize.
• Statement: S in {1,...,n} is admissible if its elements are pairwise coprime; G(n) = max over admissible S of sum of S; H(n) = sum_{p<n} p + n pi(n^{1/2}). Is G(n) > H(n) - n^{1+o(1)}?
• Best known, from the source page (fetched 2026-08-18): Erdos and Van Lint proved H(n) - n^{3/2-o(1)} < G(n) < H(n) and (H(n)-G(n))/n -> infinity, and got G(n) > H(n) - n^{1+o(1)} only under "plausible (but hopeless) assumptions about the distribution of primes". No values of G(n) are recorded.
• Tool: exact finite-N combinatorial optimisation. G(n) is an exact integer program: one binary variable per integer m <= n, one packing constraint per prime p <= n ("at most one chosen m is divisible by p"), objective sum m x_m. n variables, pi(n) constraints, totally structured. scipy.optimize.milp (HiGHS) is in the venv. The LP relaxation of a set-packing this sparse is usually near-integral, so the solve is cheap.
• Smallest experiment: solve for n = 100, 200, 500, 1000, 2000, 5000 and tabulate (H(n) - G(n))/n. Estimated under 30 minutes to first numbers; n = 10^5 (10^5 variables, 9592 constraints) is plausible in an afternoon.
• What a signal looks like: Erdos and Van Lint's own theorem says (H(n)-G(n))/n -> infinity, so the data must show growth; the open question is whether H(n)-G(n) is n^{1+o(1)} or genuinely larger. Fitting log(H(n)-G(n))/log n across four decades tests exactly that exponent.
• Negative result: a fitted exponent that sits stubbornly above 1 would be evidence against the conjecture, which is more interesting than confirmation and is exactly the sort of thing this tree is built to say carefully. Either direction is publishable as measurement.
• P(anything new): first table -- 85%. A defensible read on the exponent -- 40%.
• Mined: 1 comment, nobody working, no OEIS. Essentially untouched.

C5. Problem 307: the 2-cycle reformulation, and why the search is now structured

• URL: https://www.erdosproblems.com/307
• Status / prize: VERIFIABLE, no prize. Asked by Barbeau [Ba76].
• Statement: are there finite sets of primes P, Q with (sum_{p in P} 1/p)(sum_{q in Q} 1/q) = 1?
• Best known, from the source page (fetched 2026-08-18): P and Q must be disjoint, sum over P union Q of 1/p >= 2, hence |P union Q| >= 60. Cambie has examples for the weakened version where the elements need only be coprime, e.g. 1 = (1 + 1/5)(1/2 + 1/3). No examples are known for the coprime version with 1 not in P union Q.
• Reduction (derived and checked here, elementary, quite possibly already known): for any finite set of primes S, write s(S) = sum_{p in S} 1/p = a_S/prod(S). Reducing mod p for p in S shows gcd(a_S, prod S) = 1, so that fraction is already in lowest terms. Confirmed numerically on 2000 random prime sets: the denominator of s(S) is exactly prod(S), every time. Then s(P)s(Q) = 1 with P, Q disjoint forces a_P a_Q = prod(P) prod(Q) with a_P coprime to prod(P) and a_Q coprime to prod(Q), hence a_P = prod(Q) and a_Q = prod(P) exactly. Define the partial map T(S) = the set of prime factors of numerator(s(S)), defined exactly when that numerator is squarefree. The problem is then precisely: does T have a 2-cycle? T(P) = Q and T(Q) = P.
• Why this changes the search: the site's bound |P union Q| >= 60 leaves a search over subsets of primes. The 2-cycle form makes Q a function of P, so the search is one cheap deterministic check per candidate P (compute an exact rational, factor its numerator, test squarefreeness, sum reciprocals, compare), instead of a search over pairs.
• Tool: exact rational arithmetic plus factorisation, both in the venv.
• Smallest experiment, already run here: all P of size 2 to 4 drawn from the primes below 60 (1886 with T defined), no 2-cycles, seconds. Extending to size <= 8 over primes below 200 is minutes.
• The honest feasibility read, which is negative: T maps a set of size k to a set of size omega(a_P), and a_P is a number of size about prod(P). For sum_{q in Q} 1/q to be near 1 while Q omits 2 (P and Q are disjoint and one of them lacks 2), Q needs roughly 40 or more small primes, so a_P must be an extremely smooth squarefree number of size about prod(P). For a random P that is astronomically unlikely, and a blind enumeration over subsets of the first 60-100 primes is 2^60 and infeasible. So this is not a candidate for finding a solution. It is a candidate for a small structural note plus a bounded negative ("no 2-cycle with |P| <= 8 and max P <= 500"), and it should be scored as such.
• P(anything new): finding P, Q -- essentially 0. The reduction being new -- maybe 20%, it is a two-line exercise and Cambie has clearly thought about this page.
• Mined: 6 comments, three users flagged as currently working (aditya, will0708, Ary300), two flagged "looks difficult". Moderately busy.

C6. Problem 912: the constant in h(n) ~ c (n/log n)^{1/2}

• URL: https://www.erdosproblems.com/912
• Status / prize: OPEN, no prize. Erdos and Selfridge.
• Statement: h(n) counts the distinct exponents in the prime factorisation of n!. Prove h(n) ~ c (n/log n)^{1/2} for some c > 0.
• Best known, from the source page (fetched 2026-08-18): Erdos and Selfridge proved h(n) is of order (n/log n)^{1/2} (see [Er82c]). A heuristic of Tao using the Cramer model, given in the comments on that page, predicts c = sqrt(2 pi) = 2.5066... No numerical check of that prediction is recorded.
• Tool: large exact arithmetic-function computation. h(n) is Legendre's formula plus a distinct-count; nothing else is needed.
• Smallest experiment, already run here (numpy, one core, sieve to 10^8):

Total runtime 7.2 s including the sieve. The ratio sits near 2.65, which is about 5.6% above Tao's sqrt(2 pi) = 2.5066 and is not visibly falling toward it.

• The honest caveat, which is the whole game here: a secondary term of relative size 1/log n is 5.4% at n = 10^8, which is exactly the size of the discrepancy. So the present data is fully consistent with c = sqrt(2 pi) and a 1 + O(1/log n) correction, and claiming otherwise now would be a mistake. The experiment that matters is pushing n to 10^12 to 10^15, where 1/log n falls to 3.6% and 2.9%. That is feasible without a bigger sieve: for p > sqrt(n) the exponent is floor(n/p), so the distinct values there are the v <= sqrt(n) for which the interval (n/(v+1), n/v] contains a prime, which is sqrt(n) short interval tests, and for p <= sqrt(n) there are only pi(sqrt(n)) primes to handle directly. At n = 10^12 that is about 10^6 short-interval prime tests, minutes of work.
• Negative result: the likely outcome is "the ratio is 2.6x and falling like 1 + a/log n; fitting a two-term model gives c = 2.5 +/- 0.1, consistent with sqrt(2 pi)". That is a good outcome to record: it is the first numerical support for a named heuristic constant on this problem, and the fit itself is the artifact. The other outcome, that the ratio stabilises visibly above sqrt(2 pi), would be evidence against the Cramer-model heuristic, which is a more interesting thing to say and would need care and an independent recomputation before saying it.
• P(anything new): first numerical test of the predicted constant, with a fitted two-term model -- 85%. A defensible statement that the heuristic constant is wrong -- 10%.
• Mined: 2 comments (one of them Tao's heuristic), nobody flagged as working, no OEIS sequence for h(n) found. Lightly mined, and the heuristic is fresh enough that nobody has checked it.

C7. Problem 1074 (with 1072 and 1073): EHS numbers and Pillai primes

• URLs: https://www.erdosproblems.com/1074 , /1072 , /1073
• Status / prize: all three OPEN, no prize.
• Statement (1074): S is the set of m such that some prime p with p not congruent to 1 mod m divides m!+1. Does |S cap [1,x]|/x converge, and to what? Same question for the Pillai primes P relative to pi(x).
• Best known, from the source page (fetched 2026-08-18): Erdos, Hardy and Subbarao proved S and P are both infinite. "Hardy and Subbarao computed all EHS numbers up to 2^10" and expected the density of S to be 1, and the density of P to be perhaps between 0.5 and 0.6. OEIS A064164 (EHS numbers) has a b-file of only 120 terms (Amiram Eldar); OEIS A063980 (Pillai primes) has 10000 terms (Greathouse, first 1000 from Noe). Verified at https://oeis.org/A064164 and https://oeis.org/A063980 , fetched 2026-08-18.
• The asymmetry worth exploiting: OEIS computes A064164 by fully factoring m!+1 (the PARI line on the page is factor(n!+1)), which is why it stops at 120 terms. But the conjecture is that the density of S is 1, and a density lower bound needs only one witness prime per m, not a factorisation. So the right computation is: for every prime p up to B, run k! mod p for k < p and record every k with k! congruent to -1 mod p and p not congruent to 1 mod k. That establishes membership in S from below and never needs a factorisation.
• Tool: large exact modular computation. Cost is sum over p <= B of p, which is B^2/(2 log B): B = 10^6 is 3.6e10 multiply-mods (about a minute in C), B = 10^7 is 3.1e12 (about an hour in C on one core, or 15 min on four). The same single sweep answers all three problems: f(p) from 1072 is the least k in the sweep, the composite u from 1073 come from the same table, and the Pillai primes from 1074 are the p with a qualifying k.
• Smallest experiment: sweep to B = 10^6 first (a minute), reproduce the 10000 known Pillai primes as a control, then report the observed |P cap [1,x]|/pi(x) across decades and a lower bound on the density of S.
• Negative result: if the lower bound on the density of S plateaus well below 1 the conjecture is in trouble, and that is publishable. If it climbs toward 1 that is the first quantitative support since 2002. Either way the table is new.
• P(anything new): extending the recorded EHS data past m = 120 -- 80%. A density statement that moves either question -- 30%.
• Mined: 5 comments on 1074, 1 on 1072, 0 on 1073, nobody flagged as working. The prime side (A063980) has been computed by serious people; the m side has not, because everyone reached for factorisation.

C8. Problem 168: more digits of the {n,2n,3n}-free density constant

• URL: https://www.erdosproblems.com/168
• Status / prize: OPEN, no prize.
• Best known, verified at OEIS (fetched 2026-08-18): A386439 gives 35 digits, 0.80096575500655898909042032638808241..., contributed by Sean Eberhard, 18 Sep 2025, with keyword more. His method is on the OEIS page: A004059 computed to 3335 terms via one Gurobi integer program per 3-smooth number (https://oeis.org/A004059 , script at https://oeis.org/A004059/a004059.py.txt), then a RealIntervalField(1000) partial sum with a tail bound (https://oeis.org/A386439/a386439.py.txt). Graham, Spencer and Witsenhausen [GSW77] proved the limit equals (1/3) sum over K of 1/d_k.
• Tool: exact finite-N lattice enumeration (sine_gram/exact_finite_N.py is the same shape: the constraint is exactly "no lattice corner (a,b), (a+1,b), (a,b+1) all chosen" on the staircase 2^a 3^b <= x), plus interval arithmetic for the tail. The lab has scipy.optimize.milp (HiGHS) but not Gurobi.
• Smallest experiment: reproduce Eberhard's 35 digits with HiGHS to check the pipeline, then attack the algorithmic bottleneck. Doubling the digits needs 3-smooth numbers to 10^70, which is only about 18500 lattice points; the obstacle is solving 18500 integer programs of that size, not the arithmetic.
• Honest feasibility read: this is the candidate most likely to be slower here than where it already lives. Eberhard is a strong researcher, he did this eleven months ago, and he had a commercial solver. Our edge would have to be a genuinely better algorithm (a transfer-matrix or column-generation argument on the staircase), not more compute. Score it as a maybe, not a lead.
• Negative result: "HiGHS reproduces 35 digits and does not get past 40" is a perfectly fine thing to learn in an afternoon and should be recorded as a cost measurement, not a result.
• P(anything new): 15%.
• Mined: 9 comments, nobody flagged as working, but Eberhard, Alexeev, Cambie and Chapman are all thanked on the page. Actively watched.

C9. Problem 317: exact minima of |sum delta_k / k| with delta in {-1,0,1}

• URL: https://www.erdosproblems.com/317
• Status / prize: OPEN, no prize.
• Statement: (i) is there c > 0 such that for every n some choice of delta_k in {-1,0,1} gives 0 < |sum_{k<=n} delta_k/k| < c/2^n? (ii) is it true that for large n every non-zero |sum delta_k/k| exceeds 1/lcm(1..n)?
• Best known, from the source page (fetched 2026-08-18): the second claim fails for small n, e.g. 1/2 - 1/3 - 1/4 = -1/12 with lcm(1..4) = 12. Kovac and van Doorn prove in the comments a weak version of (i) with upper bound 2^{-n (log log log n)^{1+o(1)}/log n}, and van Doorn gives a heuristic that this may be the truth.
• Why it is a good fit: multiplying by L = lcm(1..n) turns (ii) into a pure integer question: is +/-1 representable as sum delta_k (L/k)? And (i) becomes a shortest-vector / closest-subset-sum problem in Z. The lab has exact integer arithmetic, and sympy 1.14's DomainMatrix.lll is present in the venv (confirmed), so LLL on the lattice generated by the L/k is directly available.
• Smallest experiment: exhaustive meet-in-the-middle over 3^n up to n = 30 (3^15 = 14.3M per half, minutes), recording (a) the minimum non-zero |sum delta_k (L/k)| and (b) every n where that minimum equals 1. That answers (ii) empirically over a range nobody has published, and gives the first exact data for (i). LLL then extends the upper side of (i) far past n = 30 by exhibiting short vectors, which is exactly what question (i) asks for.
• Negative result: "the minimum is 1 only for n in a short list" would be the expected shape of an answer to (ii), and is worth recording. If LLL finds vectors much shorter than the Kovac-van Doorn bound at moderate n, that is a concrete data point against van Doorn's heuristic being tight.
• P(anything new): first exact table -- 80%. Something that bears on the heuristic -- 25%.
• Mined: 8 comments, nobody flagged as working, no OEIS sequence found. The analytic side is being worked by Kovac and van Doorn; the lattice side is not.

C10. Problem 1210: pairwise coprime sets and sum 1/(n-a)

• URL: https://www.erdosproblems.com/1210
• Status / prize: OPEN, no prize.
• Statement: if A is a subset of [1,n) with pairwise coprime elements, is sum_{a in A} 1/(n-a) <= sum_{p<n} 1/p + O(1)?
• Best known, from the source page (fetched 2026-08-18): the page records only Erdos's own remark that he "did not state [this] quite correctly" earlier, plus the related prime-interval form. No bounds and no computation are recorded.
• Tool: the same exact set-packing integer program as C4 (one binary per integer, one constraint per prime), with objective sum 1/(n-a) instead of sum a. Rational weights, so solve with float objective and re-verify the optimal set exactly.
• Smallest experiment: maximise sum 1/(n-a) over pairwise-coprime A for n = 100, 200, ..., 12800, and plot the excess over sum_{p<n} 1/p. Given C4 ran n = 51200 in 10.8 s with the identical constraint matrix, the first signal is under 30 minutes, most of which is writing the objective.
• Negative result: if the excess grows like log log n rather than staying bounded, that is evidence against the conjecture as stated and is a genuinely useful thing to report, since Erdos himself flagged that he had misstated it.
• P(anything new): first table -- 85%. A read on boundedness -- 40%.
• Mined: 3 comments, nobody flagged as working, but three users including the database owner (TFBloom, Basile_Beyer_de_Ryke, hunterbates) have flagged it "looks tractable", which is a mild warning that others see the same opening.

C11. Problem 302 (and 301): the density of 1/a = 1/b + 1/c free sets

• URL: https://www.erdosproblems.com/302
• Status / prize: OPEN, no prize.
• Best known, from the source page (fetched 2026-08-18): f(N) >= (5/8+o(1))N (Cambie, odd integers up to N/4 together with everything in [N/2,N]) and f(N) <= (9/10+o(1))N (van Doorn, with a linked note). Erdos asked whether f(N) = (1/2+o(1))N, which the (5/8) construction already refutes.
• Tool: exact finite-N combinatorial optimisation. f(N) is a maximum independent set in the 3-uniform hypergraph on {1,...,N} whose edges are the triples with 1/a = 1/b + 1/c. Enumerating the triples is a divisor computation; the optimisation is an integer program in HiGHS.
• Smallest experiment: exact f(N) for N up to 200-400 (triples number O(N log N), the IP is small), then f(N)/N against the 5/8 = 0.625 and 9/10 = 0.9 brackets. Under an hour.
• Honest read: finite N will not settle an asymptotic density, and the interesting quantity is whether f(N)/N at N = 400 already exceeds 5/8 by a visible margin, which would say the current record construction is not optimal and would point at a better one. That is the realistic prize here: an improved lower-bound construction read off an exact optimum, which is a standard and respectable way for a computation to beat a human construction.
• P(anything new): an improved construction -- 25%. A table -- 85%.
• Mined: 1 comment, but three users flagged as currently working (Woett, Quanyu_Tang, khanukov), and both current records are recent. Busy.

C12. Problem 985: is there always a prime primitive root below p?

• URL: https://www.erdosproblems.com/985
• Status / prize: OPEN, no prize.
• Best known, from the source page (fetched 2026-08-18): the page cites only Artin's conjecture, Hooley under GRH, and Heath-Brown's result that one of 2, 3, 5 is a primitive root for infinitely many p. No computational verification bound is recorded.
• Tool: exact prime computation. For each prime p, factor p-1 (cheap with a smallest-prime-factor sieve) and test successive primes q < p for primitive root order.
• Smallest experiment: verify for all p <= 10^8. The least primitive root is almost always tiny and usually prime, so the expected work per p is a couple of modular exponentiations. Minutes to 10^7, an hour or two to 10^9.
• Negative result: a counterexample would be a genuine surprise and would settle the problem. Failing that, "verified for all p <= 10^9" is a bound the database does not currently carry. The interesting derived statistic is the largest observed prime primitive root relative to p, which bears on how much room the conjecture has.
• P(anything new): counterexample -- under 1%. A recorded bound -- 60%, discounted because this is the kind of thing that has plausibly been done informally and never written down.
• Mined: 2 comments, nobody working.

C13. Problem 114: the Erdos-Herzog-Piranian lemniscate

• URL: https://www.erdosproblems.com/114
• Status / prize: FALSIFIABLE, $250 (reported by Borwein [Bo95]).
• Best known, verified: the site's remarks give the chain Dolzhenko f(n) <= 4 pi n, Pommerenke f(n) << n^2, Borwein f(n) << n, Eremenko-Hayman f(n) <= 9.173 n and the full conjecture for n = 2, Danchenko f(n) <= 2 pi n, Fryntov-Nazarov f(n) <= 2n + O(n^{7/8}) plus z^n - 1 is a local maximiser, and Tao [Ta25] proved z^n - 1 is the unique maximiser for all sufficiently large n. Tao's paper is arXiv:2512.12455, "The maximal length of the Erdos-Herzog-Piranian lemniscate in high degree", December 2025 (https://arxiv.org/abs/2512.12455).
• The blocker, and why it does not close: a fetch of the abstract page reports no explicit threshold n_0; the result is stated for "sufficiently large n". So the shape "prove it asymptotically, then finish the small cases by machine" is not available: without an effective n_0 there is no finite check to do. Worse, checking a single degree n means a global optimisation over monic degree-n polynomials, 2n-3 real parameters after killing translation and rotation, with a contour-length functional. That is a genuine global optimisation, not a finite verification.
• What the lab could still honestly do: multistart local optimisation of the lemniscate length at n = 5..12 with mpmath contour integration, followed by interval enclosure of the length at the reported optimum via zeta/rigor.py, producing "no configuration beating z^n - 1 was found at degree n <= 12, with enclosure-carrying lengths for the candidates examined". That is a bounded negative and is worth exactly what it says.
• P(anything new): finding a counterexample -- under 2% (Fryntov-Nazarov local maximality plus Tao's asymptotic uniqueness leaves very little room). Producing an effective n_0 -- that is analysis, not computation, and is out of scope.
• Mined: 7 comments, three users flagged as currently working (KMendoza, dahlkebj, Sam_Petkov). This page got busy immediately after Tao's paper. Crowded.

Tier 3 candidates: real but lower yield

C14. Problem 170: the sparse ruler constant

• URL: https://www.erdosproblems.com/170
• Status / prize: OPEN, no prize.
• Statement: F(N) is the least size of A in {0,...,N} with {0,...,N} contained in A - A. Find lim F(N)/N^{1/2}.
• Best known, from the source page (fetched 2026-08-18): the limit exists (Erdos and Gal [ErGa48]) and lies in [1.56, sqrt(3)], lower bound Leech [Le56] (1956), upper bound Wichmann [Wi63] (1963). Pegg's computations [Pe20] suggest sqrt(3) = 1.7320... is the truth. A general web search turned up Wolfram's 2020 sparse-ruler work and OEIS A004137 but no verified improvement on the 1.56 lower bound, so treat "1.56 is still the record" as the site's claim, not as independently confirmed.
• Tool: exact finite-N combinatorial optimisation (the same set-packing machinery as C4 and C10, but as a covering program), plus the lab's habit of structure-matched controls.
• Why the ceiling is low: improving the lower bound 1.56 means an exhaustive argument over all rulers of a given length, and the state of the art here is a heavily engineered SAT/branch-and-bound community effort (OEIS A004137 and the Wolfram work). Our HiGHS setup would be re-entering a crowded, well-optimised search from behind.
• Smallest experiment: reproduce F(N) for N <= 200 from the ILP as a calibration against A004137, an afternoon. If the solver is competitive, only then consider the frontier.
• P(anything new): 10%. Mined: 3 comments here, but the underlying sequence is a long-running community computation. Crowded outside the database.

C15. Problem 1142: n such that n - 2^k is prime for every 1 < 2^k < n

• URL: https://www.erdosproblems.com/1142
• Status / prize: OPEN, no prize.
• Best known, from the source page and OEIS (both fetched 2026-08-18): the only known such n are 4, 7, 15, 21, 45, 75, 105; Mientka and Weitzenkamp [MiWe69] proved there are no others up to 2^44 (about 1.76 x 10^13). OEIS A039669 (https://oeis.org/A039669) lists the same seven terms, cites the same 1969 paper, and has no b-file and no later extension. Vaughan [Va73] gave a density bound. The page also mentions further discussion on the Prime Puzzles site, which was not reachable in this scan.
• Tool: exact prime computation plus an elementary congruence sieve.
• The structural point that makes it cheap: if q is an odd prime for which 2 is a primitive root, then {2^k mod q} covers every non-zero residue, so once 2^k ranges far enough, n must be divisible by q. For n up to 10^25 that forces n to be a multiple of 3.5.11.13.19.29.37.53.59.61.67.83, about 4.6 x 10^16, so only about 2 x 10^8 candidates survive before the further partial constraints from primes like 7 and 17 (where <2> is a proper subgroup) cut it down again.
• Smallest experiment: reproduce the seven known values, confirm the forced divisibility, and estimate the surviving candidate count to 10^20. An hour. A full search to 10^20 or beyond then costs a day at most.
• Negative result: near-certain, and worth recording: replacing a 1969 bound of 2^44 with 10^20 or more on a database page is a clean, checkable improvement.
• P(anything new): an eighth value -- under 1%. A much larger recorded bound -- 50%, discounted hard because the Prime Puzzles community very plausibly did this already and it is simply not cited on the page. Check that first.
• Mined: 5 comments, nobody flagged as working.

C16. Problem 676: integers not of the form a p^2 + b

• URL: https://www.erdosproblems.com/676
• Status / prize: OPEN, no prize.
• Statement: is every sufficiently large integer of the form a p^2 + b with p prime, a >= 1, 0 <= b < p? Equivalently, is there always a prime p <= sqrt(n) with (n mod p^2) < p?
• Best known, from the source page (fetched 2026-08-18): the Brun-Selberg sieve gives at most x/(log x)^c exceptions in [1,x]; Erdos thought it "rather unlikely" that all large integers qualify; Selfridge and Wagstaff made a "preliminary computer search" for the variant without the primality requirement. Erdos also asked about c_n, the least c with n = a p^2 + b and 0 <= b < c_n p, suggesting limsup c_n = infinity and asking whether c_n < n^{o(1)}.
• Tool: segmented sieve. For each prime p, the condition marks p residues out of every p^2, so the exceptional set is a plain sieve computation.
• Smallest experiment: sieve to 10^10 recording the exceptional n, their count by decade, and the record values of c_n. Under an hour to 10^9 in C, given that the C6 sieve for problem 647 covered 10^9 in 3 minutes.
• Negative result: the deliverable is the exceptional-set data, so there is no negative outcome, only a more or less interesting exponent. The record values of c_n directly probe Erdos's limsup c_n = infinity suggestion, which no recorded computation addresses.
• P(anything new): first recorded exception counts and c_n records -- 75%. Something that shifts belief about the conjecture -- 20%.
• Mined: 9 comments, nobody flagged as working.

C17. Problem 1094 (with 1093): least prime factor of binomial coefficients

• URL: https://www.erdosproblems.com/1094
• Status / prize: OPEN, no prize.
• Best known, from the source page (fetched 2026-08-18): Erdos, Lacampagne and Selfridge [ELS88] conjecture that for n >= 2k the least prime factor of C(n,k) is at most max(n/k, k) with exactly 14 listed exceptions, the largest being C(284,28); in [ELS93] they add computational evidence and note it is consistent with max(n/k, 13) and 12 exceptions. Selfridge [Se77] conjectured the n >= k^2-1 form with the single exception C(62,6). The range over which ELS actually verified this is not stated on the page and I did not obtain [ELS88]: UNVERIFIED.
• Tool: large exact arithmetic. The search is efficient because of Kummer's theorem: the least prime factor of C(n,k) exceeds k exactly when, for every prime p <= k, adding k and n-k in base p produces no carry. That is a strong digit condition that can be enumerated directly per k rather than tested per n, which is what makes a modern range reachable.
• Smallest experiment: for each k <= 40, enumerate the n <= 10^7 satisfying the no-carry condition for all p <= k (cheap, the surviving density is tiny), then test the primes in (k, n/k]. A few hours to a range that is very likely well past 1988 hardware.
• Negative result: a 15th exception would be a real finding on a named conjecture; no new exception over a stated range is a citable strengthening of the evidence, and the page has no range at all.
• P(anything new): a new exception -- 10%. A stated verification range -- 60%.
• Mined: 1 comment, one user flagged as working (will0708).

Killed, with reasons

These looked good on the way in. Recording why they died is the point of the scan; several of them are the kind of candidate that would have eaten a week.

Ranking

F = feasibility of a first signal in hours (0-5). N = honest probability that something new comes out. M = how heavily mined (0 untouched, 5 crowded). V = value if it lands.

Top three, with reasoning

1. Problem 710 (and 711): exact values of the Erdos-Pomerance function f(n). This is the best-shaped item in the whole scan. Erdos put money on an asymptotic formula. Erdos and Pomerance proved bounds whose ratio is only (log log n)^{1/2}, which at n = 10^5 is about 1.5, so exact values across four decades genuinely discriminate between the two shapes. The object is a system of distinct representatives, which is a bipartite matching, which is exact finite combinatorics of precisely the kind sine_gram/ already does. Almost nothing is recorded: no values on the problem page, no OEIS sequence, 7 comments, and one self-declared worker (SkyYang). And the first signal is already in hand: a correct Kuhn matching computed f(n) for n up to 4000 in 2.1 seconds, giving

n: 100 200 500 1000 2000 4000 M_min: 160 340 877 1816 3814 7900 M/(n sqrt(log n)): 0.7456 0.7386 0.7036 0.6910 0.6917 0.6858 M/(n sqrt(log n / log log n)): 0.9214 0.9537 0.9510 0.9606 0.9851 0.9975

The two normalisations move in opposite directions, the upper-bound shape drifting down and the lower-bound shape drifting up. That is the sharpest single signal produced anywhere in this scan, and it says the experiment is already discriminating. Hopcroft-Karp instead of Kuhn takes this to n = 10^5 in an afternoon. Risk: the trend may not resolve, and the constants 1.213 and 1.7398 sit in different normalisations so care is needed comparing them.

2. Problem 912: the constant in h(n) ~ c (n/log n)^{1/2}. Chosen because it is the cheapest test of a named prediction in the database. Tao's Cramer-model heuristic in the comments predicts c = sqrt(2 pi) = 2.5066; nobody has checked it. Seven seconds of numpy gives the ratio at 2.65 across five decades, about 5.6% high, which at n = 10^8 is exactly the size a 1/log n correction would be. So this is not yet a disagreement, and saying it were would be the classic error the certainty ladder exists to prevent. What makes it a top pick is that the discriminating experiment is also cheap: for p > sqrt(n) the exponent is floor(n/p), so h(n) at n = 10^12 costs about 10^6 short-interval prime tests rather than a sieve. A two-term fit over 10^3 to 10^15 either supports sqrt(2 pi) with a measured secondary coefficient, or does not, and both are worth writing down. The tooling required is exactly what this tree has.

3. Problem 879: exact values of G(n), the largest sum of a pairwise-coprime subset of {1,...,n}. Chosen for the ratio of value to effort. One comment on the page, nobody working, no OEIS sequence, and the exact optimum is a set-packing integer program with one variable per integer and one constraint per prime, which HiGHS solves at n = 51200 in 10.8 seconds. First numbers, produced here:

n: 3200 6400 12800 25600 51200 H(n) - G(n): 3714 10285 22205 31473 69934 (H-G)/n: 1.161 1.607 1.735 1.229 1.366 log(H-G)/log n: 1.0185 1.0541 1.0582 1.0203 1.0288

Erdos and Van Lint proved (H(n)-G(n))/n tends to infinity, so the very slow and non-monotone growth visible here is a real feature to explain rather than a bug, and the measured exponent sitting at 1.02 to 1.06 is direct evidence on the open question, which asks exactly whether H(n) - G(n) is n^{1+o(1)}. The same constraint matrix immediately serves problem 1210 (rank 7), so the two should be done together.

What the next two would be

Problem 311 (rank 4) if the goal is a clean artifact with near-certain delivery: the first exact table of delta(N), already computed here to N = 44 in 15 seconds, extendable to N = 56 or so, with -ln delta(N)/N drifting from 0.62 to 0.49 in a way that is mild evidence against the conjectured pure exponential shape.

Problem 647 (rank 5) if the goal is a prize and a database gap: 3 minutes 20 seconds of C already establishes no example with 24 < n <= 10^9 and produces the per-decade growth table for Erdos's stronger conjecture. It is ranked below 311 only because 13 comments and four self-declared workers make it likely someone has already run the search and not published it. Read the live comment thread before spending compute there.

Scan accounting

• 1217 problems in the database; 604 open, plus 27 falsifiable, 9 decidable, 7 verifiable and 4 open (Lean).
• 437 open-ish problems in the number-theory / analysis / combinatorics tag pool. All 437 statements read.
• 45 problems read in full including remarks and activity metadata.
• 17 candidates survive with a stated experiment, runtime and probability.
• 20 killed rows covering roughly 35 named problems plus two whole tag families (graph theory, geometry), with the reason recorded in each case.
• 7 candidates had a first-signal experiment actually run during the scan: 647, 311, 307, 488, 710, 879, 912. Two of those (488, 307) were killed by their own experiment, which is the scan working as intended.

Reproduction

Code written during this scan lives in the session scratchpad, not in the repo: scratchpad/e647/tau647c.c (segmented tau sieve, problem 647) and the inline scripts quoted above for 311, 307, 488, 710, 879 and 912. All of them are short enough to retype from the descriptions here; none of them is a repository artifact yet and none should be committed until a candidate is actually chosen.

Git state, recorded honestly. This scan committed nothing. However, a concurrent session working the same checkout ran a broad git add and swept the 58-line header of this file into commit d775e48 ("hunts(rogue_frontier): the pairing count, proved in Lean with zero sorrys", 2026-08-18 22:20:05 +0000) while it was still being written. Everything from "Tier 1 candidates" onward is uncommitted working-tree content at the time of writing. That collision is exactly the hazard CLAUDE.md warns about under "Multiple agents / parallel sessions": two agents in one checkout rather than separate worktrees. It has not been undone here, because rewriting another session's history is worse than recording what happened.