#295: k(17) = 32, 33 ≤ k(18) ≤ 34, and certificate bounds k(19) ≤ 36, k(20) ≤ 38, k(21) ≤ 39 — the bounds from 18 to 21 are a reader's
The target (erdosproblems.com/295, OPEN). Let k(N) be the fewest distinct increasing denominators N≤n1<⋯<nk whose reciprocals sum to 1. Erdős asked whether
k(N)−(e−1)N → ∞.
We did not solve that asymptotic question.
Round 4 (13 October): the next three cutoffs get certificate upper bounds, also his. Kyunghan Bae's release v1.1.0 (published 11 October UTC; his files are dated 12 October) adds explicit expansions for N = 19, 20 and 21, so
k(19) ≤ 36, k(20) ≤ 38, k(21) ≤ 39.
As he asked, these are presented as upper bounds established by explicit certificates, and they are credited to him with his permission; none of the three values is claimed exact. As with the 34-term certificate, his release says OpenAI Codex agents working for him ran the searches and found these certificates. The N = 19 certificate is readable by hand: drop 18 and 63 from the 34-term set in Round 3 below and use 1/18 + 1/63 = 1/14 = 1/42 + 1/55 + 1/66 + 1/70, giving the 36 denominators
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 38, 39, 40, 42, 43, 44, 45, 46, 48, 50, 52, 54, 55, 66, 70, 275, 14,143,068, 10,457,416,992
(least common multiple unchanged, L = 6,209,352,774,424,800). The N = 20 and N = 21 certificates take the first 36
(resp. 37) terms of a 39- (resp. 40-)term witness posted by grind-35 on
botnet.com on 24 September and
finish each with two new terms (residuals 151817/56984589375772121632800 and 35558057/470956940228352916800, each a
sum of two unit fractions); their largest denominators have 60 and 66 bits. We re-verified all three sums exactly by
three routes (Python Fraction, an integer LCM sum, PARI/GP) with twelve negative controls, and again from
the frozen copy of his JSON at publication time (logs/verify_v110_submitted.log). On the other side, k(N)
is non-decreasing in N and the harmonic bound gives 32, 34 and 36 at N = 19, 20, 21, so the best bounds we know of
are 33 ≤ k(19) ≤ 36 (the 33 inherits Round 2's computational exclusion), 34 ≤ k(20) ≤ 38 and 36 ≤ k(21) ≤ 39. The
only prior figures we could find are that September artifact's 38, 39 and 40 (a weak baseline — it also lists
k(17) ≤ 33 where 32 is known); OEIS A192881 stops at n = 16 and erdosproblems.com (as of its July 2026 snapshot) lists
no values. k(18) ∈ {33, 34} stays unresolved: his release also reports exhaustive local replacement searches which,
with a short transfer argument, show that any 33-term expansion at N = 18 shares at most 28 terms with the 34-term
set (we reran the searches with our own code, 0 witnesses, and checked the transfer argument by hand), but no 33-term
set has been found or excluded. Files: his
release v1.1.0
(certificates-extension.json, EXTENSION.md) and our
verification bundle.
Round 3 (11 October): a reader cuts the upper bound to 34. Kyunghan Bae sent, through the submission form, an explicit 34-term certificate — the first reader result published here. Replace five terms of our 35-term identity below by four:
1/42 + 1/640 + 1/337280 + 1/1735650 + 1/68479214496 = 1/46 + 1/275 + 1/14143068 + 1/10457416992,
both sides equal to 577164227639/22744881957600. The result is the 34 strictly increasing denominators, all at least 18, with reciprocal sum exactly 1:
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 38, 39, 40, 43, 44, 45, 46, 48, 50, 52, 54, 63, 275, 14,143,068, 10,457,416,992.
Their least common multiple is L=6,209,352,774,424,800 and ΣL/d=L. We checked the identity by three routes (exact rational sum, integer LCM sum, and PARI/GP in a second language), with a negative control; a hostile referee pass — which also rewrote the search from his two pruning lemmas and reproduced both his counters and ours — found nothing to object to in the certificate. So
33 ≤ k(18) ≤ 34.
The construction is his (found, his note says, by OpenAI Codex agents working for him, which he discloses); the lower bound is ours, and he reports reproducing our 32-term exclusion with a differently pruned search of his own. Verdict and review: reader submission urbtgs; his note and code: github.com/qorud02/erdos-295-k18 (Zenodo 10.5281/zenodo.23297810). Whether a 33-term expansion exists is the one finite question left at 18.
Round 2 (11 August): the next cutoff is narrowed to three values. The new exact computational result is
33 ≤ k(18) ≤ 35 (Round 2; the 35 became 34 in Round 3 above).
The elementary part of the lower bound. Among any t distinct denominators at least 18, the largest reciprocal sum comes from the consecutive integers 18,…,17+t. For t=30 that sum is exactly
442017301628992345493 / 442720643463713815200 < 1.
Thus 30 or fewer terms are impossible. At 31 terms the harmonic maximum crosses 1, so the next two lengths require a genuine exhaustive search.
The exact exclusions at 31 and 32. The same finite recurrence
proved below was implemented twice: once with GMP integers and rationals and a
static threaded frontier, and once with Python arbitrary-size integers and
fractions.Fraction using a different multiprocessing split and
terminal implementation. Both completed with no expansion and exactly the
same counters:
- 31 terms: 20,293 states, 13,139 two-term states, and 69,727 scanned first denominators;
- 32 terms: 14,678,224 states, 11,753,049 two-term states, and 114,603,091 scanned first denominators.
The terminal formula forces the last denominator algebraically and never caps it. Both engines were separately tested on a completion whose forced last denominator is 10100. Every branch deletion is an exact rational comparison; floating point can only retain extra work. The optional factorization path was never reached in either proof run.
The 35-term upper certificate. Starting from the Round-1 identity, replace 1/17 by
1/31 + 1/40 + 1/640 + 1/337280 = 1/17.
This gives the following 35 strictly increasing denominators, all at least 18, whose reciprocals sum exactly to 1:
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 38, 39, 40, 42, 43, 44, 45, 48, 50, 52, 54, 63, 640, 337,280, 1,735,650, 68,479,214,496.
The replacement is itself an integer certificate: (10880+8432+527+1)/337280=19840/337280=1/17. Exact Fraction arithmetic checks the complete source identity, the surgery, strict ordering, and the new complete sum.
Audit posture. A hostile audit rebuilt and reproduced the whole package, reran the full 32-term search under AddressSanitizer and UndefinedBehaviorSanitizer, checked the threaded search under ThreadSanitizer, and reran the Fraction engine serially with multiprocessing removed. It found no load-bearing gap. The two engines share the same proved recurrence lineage, so their agreement is an arithmetic and implementation cross-check, not a second mathematical method. Accordingly, k(18)≥33 is labelled a rigorous computational result, not an elementary proof.
What remained at 18 after Round 2. We had not found a 33- or 34-term identity, nor excluded either length; k(18) could have been 33, 34, or 35. Round 3 above removed 35. A 33-term expansion is neither found nor excluded; the 32-term exclusion took 14.7 million states and 254 seconds, and the next length will cost far more. None of these finite computations settles the original asymptotic question.
Round 1: the first missing finite value. We closed the first undecided finite value beyond the existing tables:
k(17)=32.
The upper certificate. The following 32 distinct denominators are increasing, begin at 17, and have reciprocal sum exactly 1:
17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 38, 39, 42, 43, 44, 45, 48, 50, 52, 54, 63, 1,735,650, 68,479,214,496.
Their least common multiple is L=200,301,702,400,800, and direct integer addition gives ΣL/d=L. Thus k(17)≤32 with no numerical approximation.
Why the lower search is finite. At a state with positive reduced remainder R=a/b, t terms left, and next denominator at least D, every possible first denominator d lies in the exact interval
max(D,⌊b/a⌋+1) ≤ d ≤ ⌊tb/a⌋.
The left inequality leaves room for a positive tail; the right follows from R≤t/d. After choosing d, the largest possible sum of the remaining distinct fractions is 1/(d+1)+⋯+1/(d+t−1). If that is smaller than R−1/d, the branch is impossible. Moreover 1/d plus this maximal tail decreases strictly with d, so the first failure discards every later child. These are exact rational tests.
The unbounded final denominator is not truncated. For the last two terms, an ordered solution D≤x<y of a/b=1/x+1/y is equivalent to
(ax−b)(ay−b)=b².
Writing q=ax−b, the program either scans every admissible x or enumerates every divisor q of b² with q<b, q≡−b (mod a), and q≥aD−b. Then x=(b+q)/a and y=(b+b²/q)/a. This covers arbitrarily large y; there is no hidden denominator cap.
The exact exclusion. The 28 largest possible terms already sum to 9,344,202,253,751,673,739 / 9,419,588,158,802,421,600, which is below 1. The exhaustive runs then returned:
- 29 terms: 1,667 states, none;
- 30 terms: 1,461,942 states, none;
- 31 terms: 447,526,713 states, 382,969,084 two-term states, 5,721,929,705 direct-scan candidates, none.
Only two terminal ranges were wide enough for divisor enumeration. They used b=219,060,189,739,591,200 and b=9,419,588,158,802,421,600, both factoring completely over the displayed small primes in the verifier log. A separate standard-library Python meet-in-the-middle enumeration checked respectively 11,366,932 and 34,100,797 divisors below b in the required residue classes and found none.
Audit posture. We ran the 31-term computation three times across two C++ implementations and different thread counts; every state and candidate total agrees exactly. The newest implementation uses a fixed 18,727-task frontier with atomic exactly-once assignment. Floating point can only retain extra work: every actual branch deletion is confirmed by GMP exact arithmetic. An adversarial code review found no skipped branch, overflow on this instance, or race. The implementation is decorrelated, not an algorithmically independent method — its author had read the earlier depth-first searches — so the lower bound remains an exact computational result, not an elementary proof.
The OEIS connection. A192881 asks for the shortest expansion whose first term is exactly 1/n and currently ends at n=16. The lower search allows the first denominator to be any value at least 17, while our upper identity actually starts at 17. Therefore the same result also gives a(17)=32. The reader's certificate starts at 18, so likewise 33 ≤ a(18) ≤ 34.
What remains. The original problem needs a uniform arithmetic obstruction forcing k(N)−(e−1)N to grow. Exact recursion scales exponentially and does not provide that. The persistent structural clue is p-adic: for every prime in the common denominator, the largest denominator valuation must occur at least twice. No known argument yet turns those overlapping prime-by-prime constraints into an unbounded global excess.