Erdős problem 531: exact values through \(k=3\) and a \(k=4\) construction
Audit date: 2026-07-26 UTC. Live status: OPEN; no claimed proof and no current worker. Main finite result of this run:
The first equality is elementary. The next two are exact finite computations; the \(F(3)\) upper bound has a generated LRAT refutation checked by an independent, from-scratch Python LRAT checker. The \(F(4)\) bound is an explicit 200-bit colouring checked directly against all \(2{,}535{,}064\) eligible four-sets. These finite results do not estimate the asymptotic growth of \(F(k)\), so problem 531 remains open.
Rigor key
- (a) elementary-rigorous: proved here directly from the definitions;
- (b) rigorous-modulo-named-theorem: quoted from the named and linked
source;
- (c) plausible/structural-unverified: interpretation or heuristic, not a
theorem;
- (d) computational-only: an exact finite result, with reproducible code,
but not an asymptotic theorem.
The exact \(F(2)\), exact \(F(3)\), and the \(F(4)\) lower bound below are (d). I found no primary source recording these small values, but I do not claim that they are new to the literature.
0. Mandatory live-page gate
I fetched the live problem page through the configured Bright Data Chromium path before doing any mathematics. The page displayed the following current statement (verbatim; only line wrapping has been normalized):
Let \(F(k)\) be the minimal \(N\) such that if we two-colour \(\{1,\ldots,N\}\) there is a set \(A\) of size \(k\) such that all subset sums \(\sum_{a\in S}a\) (for \(\emptyset\neq S\subseteq A\)) are monochromatic. Estimate \(F(k)\).
The live gate was clear:
- status:
OPEN; 0 comments on this problem;0 claimed proofs for this problem;Interested in collaborating: None;Currently working on this problem: None;Likes this problem: None;This problem looks difficult: None;This problem looks tractable: None;The results on this problem could be formalisable: None;I am working on formalising the results on this problem: None;Formalised statement? No;Related OEIS sequences: Possible.
The page lists these known results:
- Sanders and Folkman established existence; it also follows from Rado's
theorem and is commonly called Folkman's theorem.
- Erdős and Spencer proved
\[ F(k)\geq 2^{c k^2/\log k} \] for an absolute \(c>0\).
- Balogh, Eberhard, Narayanan, Treglown, and Wagner improved this to
\[ F(k)\geq 2^{\,2^{k-1}/k}. \]
There was therefore no proof claim or worker collision, so I proceeded. (d: live-page observation)
Convention forced by the colouring domain
The live sentence leaves “\(A\subseteq[N]\)” and “all sums lie in \([N]\)” implicit. Both cited quantitative papers make the convention explicit: \(A\subseteq[N]\) and its finite-sums set must be contained in \([N]\). Since all elements are positive, this is equivalent to \(\sum_{a\in A}a\leq N\). Every calculation below uses precisely that convention. (b)
1. Primary-source literature audit
I checked the actual papers or scans, rather than relying only on search snippets.
- Erdős (1973). In
“Problems and Results on Combinatorial Number Theory”, A Survey of Combinatorial Theory, pp. 117–138, PDF page 6 / printed page 122, Erdős states the Sanders–Folkman finite-sums result and says that, as far as he knew, there were no good upper or lower bounds for its function \(g(n)\). This is the source denoted [Er73] on the live page. (b)
- Erdős–Spencer (1989). Their two-page paper
“Monochromatic Sumsets”, J. Combin. Theory Ser. A 50 (1989), 162–163, DOI 10.1016/0097-3165(89)90011-390011-3), defines exactly the same \(F(k)\), including \(P(S)\subseteq[n]\), and proves \[ F(k)>2^{c k^2/\lg k}. \] Its closing note attributes to A. D. Taylor an upper bound which is a tower of threes of height \(4k-3\). (b)
- Taylor (1981). A. D. Taylor,
“Bounds for the Disjoint Unions Theorem”90031-5), J. Combin. Theory Ser. A 30 (1981), 339–344, gives the quantitative disjoint-unions/non-repeating-sums argument underlying that tower upper bound. (b)
- Balogh–Eberhard–Narayanan–Treglown–Wagner (2017).
“An improved lower bound for Folkman's theorem”, Bull. London Math. Soc. 49 (2017), 745–747, DOI 10.1112/blms.12058, is genuinely arXiv:1703.02473 and proves, in the nontrivial range \(k\geq2\), \[ F(k)\geq 2^{\,2^{k-1}/k}. \] Its construction colours odd integers randomly and imposes \(\chi(2x)\neq\chi(x)\). A repeated subset sum then puts both \(y\) and \(2y\) in the finite-sums set and prevents monochromaticity; if all \(2^k-1\) sums are distinct, the set meets at least \(2^{k-1}\) independent odd-part chains. (b)
- Later state check. Balogh–Lavrov–Shakan–Wagner,
“Monochromatic Hilbert cubes and arithmetic progressions”, Electronic J. Combin. 26 (2019), P2.22, explicitly records the BENTW lower bound and Taylor tower bound as the state of the art for \(F(k,2)\). The 2020 survey of Raigorodskii–Cherkashin, “Extremal problems in hypergraph colourings”, §9.3, repeats the same quantitative result. (b)
Exact-title, exact-formula, forward-citation, and “small Folkman number” searches found no later primary paper improving these quantitative bounds and no source giving \(F(3)\) or \(F(4)\). This is an honest search miss, not evidence that no such item exists.
Thus the literature gap I could verify is still
for \(k\geq2\). (b)
2. Exact finite reduction to hypergraph 2-colourability
For a finite set \(A\) of positive integers, write
Define the hypergraph
Then \(F(k)\) is exactly the first \(N\) for which \(H(N,k)\) has no proper 2-colouring. This is merely the definition rewritten as Property B for a hypergraph. (a)
Introduce a Boolean variable \(x_i\) for the colour of \(i\). Each edge \(E\) contributes the two clauses
They exclude the all-zero and all-one colourings of \(E\), respectively. Global colour exchange is a symmetry, so adding the unit clause \(\neg x_1\) loses no solution. Hence the \(H(N,k)\) colouring problem is equivalent to the CNF consisting of (3) and \(\neg x_1\). (a)
The \(k=3\) edges
Let \(A=\{a<b<c\}\). Its possible sums are
All seven are distinct except possibly \(c=a+b\); the strict inequalities between the remaining terms rule out every other collision. Thus every edge of \(H(N,3)\) has size six or seven. (a)
Direct generation from (2) gives
The verifier recomputes both rows independently. (d)
In general, different generator sets can have the same finite-sums set (collisions can make two such sets into the same interval). Repeating a constraint is harmless. For \(k=3\) and \(N=53,54\), the verifier also checks that the edges counted in (5) are pairwise distinct. The larger \(k=4\) counts below deliberately count eligible generator sets \(A\), not deduplicated hyperedges.
3. Exact first values
3.1 \(F(1)=1\)
The only colouring of a nonempty one-point interval contains a one-generator finite-sums set, so \(F(1)=1\). (a)
3.2 \(F(2)=9\)
Read each bit below as the colour of \(1,2,\ldots\). The colouring
of \([8]\) contains no monochromatic \(\operatorname{FS}(\{a,b\})=\{a,b,a+b\}\) with \(a<b\) and \(a+b\leq8\). For \(N=9\), fixing the colour of 1 leaves only \(2^8=256\) colourings. Exhausting them against all 16 eligible pairs leaves none. Therefore
(d)
This is the two-colour weak Schur threshold: \(a\) and \(b\) must be distinct because \(A\) is a set.
3.3 A lower certificate for \(F(3)\)
A SAT search produced the following 53-bit colouring; its validity is then checked without trusting the search. It is proper for \(H(53,3)\):
00101110000000000110110111111111101101111000000001110
Equivalently, its colour-1 class is
The checker regenerates all 3572 triples \(a<b<c\) with \(a+b+c\leq53\), forms the set of nonempty subset sums from scratch, and checks that each edge sees both bits. Hence
(d)
3.4 A proof certificate for the upper bound \(F(3)\leq54\)
For \(H(54,3)\), (3) produces
The one extra clause is \(\neg x_1\). CaDiCaL 2.1.2, with preprocessing disabled, generated a plain-text LRAT refutation. On the final run it had:
- 265,304 clause additions;
- 263,102 individual clause deletions;
- 34,905,742 bytes;
- SHA256
0b653ddc976b7e3505295c8128108353e89f68cc7cc851e27fc9d1f54fdf795c.
The supplied Python does not accept CaDiCaL's UNSAT answer on trust. It parses the LRAT proof independently. For every learned clause \(C\), it:
- falsifies all literals of \(C\);
- reads the positive LRAT hints in order;
- requires each hinted clause to be unit under the current partial
assignment;
- propagates that unit;
- requires the last hinted clause to be false; and
- finally requires the proof to derive the empty clause.
It also honours every clause deletion, rejects absent clause identifiers, rejects variables outside \([54]\), and rejects negative/RAT hints. The --plain proof contained only RUP additions, so this small checker covers the entire certificate. CaDiCaL's own proof checking was enabled as an additional check, but the reported result rests on the independent LRAT pass. The logical fact used here is just that adding a RUP clause preserves satisfiability; deriving the empty clause proves unsatisfiability. (d)
Combining the explicit colouring (8) and the checked refutation (9) gives
4. A concrete four-generator construction
A bounded CaDiCaL search produced this witness. Concatenate the following four 50-bit lines. The resulting bit string colours \(1,\ldots,200\):
01110111101101100001110011000000100100000100011100
10011010001011110101000110010100000101111101101010
10011110001111110100100110111010101011111110101001
10100010100100101111110000100011100011001010000011
The checker streams over every \(a<b<c<d\) with \(a+b+c+d\leq200\), without retaining the whole hypergraph, and constructs all distinct nonempty subset sums. It verifies all \(2{,}535{,}064\) eligible four-sets. Their edge-size histogram is
No row is monochromatic, so
This is an explicit construction with a from-scratch checker, but only a finite lower bound. (d)
5. Reproduction
The standalone verifier is runs/erdos531_wave5x_verify.py. It uses only the Python standard library for constraint generation, witness checking, DIMACS generation, and LRAT checking. To generate the LRAT proof it calls the CaDiCaL already installed in this VM; an alternative path can be passed with --cadical.
Run:
/usr/bin/python3 runs/erdos531_wave5x_verify.py
The salient output from a complete run was (the machine-dependent timing line is omitted):
PASS explicit k=2 lower witness on [8]: 00101110
PASS exhaustive k=2 upper check on [9]: 256 colourings
PASS explicit k=3 lower witness on [53]: 00101110000000000110110111111111101101111000000001110
PASS generated k=3 distinct-edge counts: H(53,3)=3572 {6:156,7:3416}; H(54,3)=3789 {6:169,7:3620}
PASS explicit k=4 lower witness on [200]: 01110111101101100001110011000000100100000100011100100110100010111101010001100101000001011111011010101001111000111111010010011011101010101111111010100110100010100100101111110000100011100011001010000011
PASS streamed k=4 constraints on [200]: 2535064 eligible A, sizes {10:20,11:1569,12:2952,13:148837,14:61964,15:2319722}
PASS independent DP eligible-set counts: N=250 -> 6304811, N=500 -> 104646514
PASS CaDiCaL UNSAT generation: /home/exedev/.elan/toolchains/leanprover--lean4---v4.32.1/bin/cadical (2.1.2)
PASS independent LRAT check: 265304 additions, 263102 clause deletions, 34905742 bytes
PASS LRAT SHA256: 0b653ddc976b7e3505295c8128108353e89f68cc7cc851e27fc9d1f54fdf795c
VERIFIED exact table: F(1)=1, F(2)=9, F(3)=54; explicit construction proves F(4)>=201
The 200-bit witness pass took about 11 seconds. LRAT generation and independent validation took about 21 seconds on this VM. The LRAT file is generated in a temporary directory, checked, hashed, and removed; no large opaque certificate needs to be trusted or committed.
For a quick construction-only check, omitting the UNSAT certificate, run:
/usr/bin/python3 runs/erdos531_wave5x_verify.py --elementary-only
6. What remains and why the standard approaches stall
6.1 The first-moment lower-bound ceiling
In a uniform independent two-colouring, a fixed edge \(E\) is monochromatic with probability
A dissociated \(k\)-set has the maximum \(|\operatorname{FS}(A)|=2^k-1\), so even in the most favourable case its probability is \(2^{2-2^k}\). Meanwhile,
The leading term follows by taking the volume of the positive \(k\)-simplex and dividing by \(k!\) for order; coincidences and the lattice boundary contribute only \(O_k(N^{k-1})\). (a)
Consequently, even pretending that every set is dissociated, a direct union bound only reaches
Thus the naive random-colouring/first-moment framework is intrinsically limited to a doubly exponential lower bound. The BENTW exponent \(2^{k-1}/k\) is already within a factor two of this idealized ceiling in \(\log_2N\). Its essential advance is handling the non-dissociated sets, for which uniform random colouring is much worse. (a) for (13)–(15); (b) for the BENTW theorem
A lower bound of qualitatively greater than doubly exponential size would therefore require a genuinely different ingredient: a highly correlated colouring or structural mechanism that simultaneously rules out enormous families of dissociated \(A\), not merely a sharper count of small sumsets. That is the exact missing lower-bound mechanism identified by this audit. (c)
6.2 The upper-bound gap
Taylor's quantitative finite-unions argument iterates Ramsey steps and pays one exponentiation at each stage, producing height \(4k-3\). Closing the gap in (1) from above needs a substantially stronger quantitative finite-unions/non-repeating-sums lemma which avoids that stage-by-stage tower growth. Neither the hypergraph encoding nor the exact \(k=3\) certificate supplies such a uniform lemma. (b) for Taylor's bound; (c) for the diagnosis
6.3 Why brute force will not address the asymptotic question
The number of eligible \(k=4\) generator sets already grows as \(N^4/576\). An independent dynamic program gives
The direct, non-deduplicating encoding used by the verifier would give the latter instance over 209 million clauses and as many as 3.14 billion literal occurrences: 12.6 GB even for raw 32-bit literals, before clause indices, watch lists, learned clauses, and proof logging. Deduplicating coincident finite-sums edges could only reduce this, but almost all fixed-\(k\) generator sets are dissociated, so it does not change the \(N^4\) leading order: a subset-sum collision is one of only \(3^k-1\) nonzero signed linear relations, and each relation has \(O_k(N^{k-1})\) solutions, while a positive dissociated set can be recovered greedily from its subset-sum set (take the smallest positive sum and remove translated pairs, then repeat). A realistic direct \(N=500\) run would therefore need roughly 50–100 GB of RAM and at least tens of core-hours; proof generation could cost more. This is an engineering estimate, not a benchmark. The instance was not run. The \(N=200\) construction was the last bounded probe; all heavier figures in (16) come from the \(O(kN^2)\) counting DP in the verifier, not from constructing or solving those CNFs. (d) for the exact counts; (a) for the leading-order count; (c) for the resource estimate
The exact small table is consequently real, certificate-backed progress, but it does not supply the uniformity step needed to improve either side of (1).
PARTIAL: Verified \(F(1)=1\), \(F(2)=9\), and certificate-backed \(F(3)=54\); an explicit checked colouring proves \(F(4)\geq201\), while the asymptotic tower-versus-double-exponential gap remains open.