#708: Round 8 — c₁₁ = 2, and the general slope falls to 2/11
The target (erdosproblems.com/708, OPEN, a $100-or-1000-rupee reward). Given n integers A with largest element M, let g(n) be the least number such that from any M consecutive integers you can pick g(n) of them whose product is divisible by the product of all of A. Erdős and Surányi (1959) proved g(n) ≥ (2−o(1))n and asked whether always g(n) ≤ 2n — which would be tight. Gallai had already got g(2)=2 and g(3)=4. The general upper bound is the open problem and we did not touch it.
Round 8 (11 August): eleven labels repair as well. The sibling sequence now begins
c₂=1, c₃=√2, and c₄=c₅=c₆=c₇=c₈=c₉=c₁₀=c₁₁=2.
The private-prime padding theorem already gives c₁₁≥2. Round 8 proves the matching upper bound c₁₁≤2.
Maximum defect, not a case-by-case matching chase. In the generic interval, the exact-difference injection makes the eleven labels the edges of a simple bipartite graph. For an edge set S write d(S)=|S|−|N(S)|, take the maximum q, and choose an inclusion-minimal core K with defect q. Every remaining edge can be matched outside N(K). Minimum degree and the bipartite edge bound force q≤4. The inherited q=2 and q=3 cores have at most ten edges. At eleven edges there are exactly eleven new q=2 classes on a 4-by-5 bipartition, three q=3 classes on 3-by-5, eight q=3 classes on 4-by-4, and the sole q=4 shape K₃,₄ with one edge removed. An exhaustive isomorphism and proper-subset audit reconstructs all 22 new classes and verifies their prime-power support blocks.
The real new obstruction is joint placement. A repair factor is supported on four or five core labels, hence has many multiples in the open interval. Treat those repair progressions and the remaining edges' outside endpoint sets as one Hall family. Any deficient set containing two repairs would trap two progressions in at most nine or ten points. The short progression lemmas force their steps to agree, while the union of their label supports forces more multiples than the trap can hold. Thus any failure has exactly one repair d and a tight set R of remaining edges.
Match R onto its tight neighborhood and let z, a d-multiple, represent one outside edge e₀. Put g=gcd(d,ae₀). A support-count argument finds a fresh g-multiple w outside all ten used points, and
zw is divisible by lcm(d,ae₀)g=d·ae₀.
So the collision loses no prime valuation. The independent hostile checker exhausted 39,663 abstract set systems satisfying the hypotheses, including every one of its 2,224 tight singleton-repair failures, and recovered this contraction in each case.
The four-defect core. For K₃,₄−e, prime-power layers define a gcd of 45 admissible four-block products. The ten pair blocks form a transversal matroid. Exact König-cover rank bounds let Rado's theorem choose one block of each of two types and three of a third type with five distinct outside representatives; discarding the appropriate third-type block leaves an admissible four-block product and four distinct repair points. A separate placement audit checked every denominator grid, all 45 products, 16,384 capped valuation patterns, and literal exact-difference kernels.
The aligned midpoint remains quarantined. Remove M, reserve b₀=x+M, and use the corrected ten-edge injection for the labels below M. Exact prime-power support can fail only on a matching of exceptional two-adic edges. New finite certificates survive every endpoint subset and every deleted submatching, retain the fifth-label slack needed by the joint placement proof, and choose all ten covering points away from b₀. Only after their product covers A∖{M} do we adjoin b₀ to supply M. The independent audit checks 907 exceptional matchings on minimal cores; the supplied total 954 also includes 47 deliberately redundant controls containing a smaller K₃,₃ core.
Sharpness and the new slope. For every fixed number r of padding primes, choose clustered primes u<v<w near a large X and then r primes qj in a fixed positive-relative-width interval inside (uv+r,vw). The prime number theorem supplies all of them, while the CRT parks the unique qj-multiples at neutral points. The four-label core still needs five points, so eleven labels need twelve, and 2u/w tends to 2. The public exact DP now exercises r=7 at (u,v,w)=(101,103,107) and returns minimum cover 12; the PNT argument, not this finite instance, proves c₁₁≥2.
Finally, subadditivity uses blocks of eleven. With c₀=0 only as a bookkeeping convention, writing n=11q+r gives
cₙ≤2⌊n/11⌋+cn mod 11≤2n/11+14/11.
The worst remainder is r=4. This is a genuine slope improvement from 1/5 to 2/11, but it is still a result about the sibling interval constant cₙ. The parent conjecture g(n)≤2n remains completely open.
Round 7 (11 August): the ten-label threshold is exact. The sibling sequence now begins
c₂=1, c₃=√2, and c₄=c₅=c₆=c₇=c₈=c₉=c₁₀=2.
The private-prime padding theorem from Round 3 already gives c₁₀≥2. Round 7 proves the matching upper bound c₁₀≤2.
Nested Hall defects. Let A contain ten distinct positive integers, M=max(A), and I=(x,x+2M)⊂[0,∞). First suppose x∉Mℤ. The two-halves injection again makes each a∈A the exact positive difference of a simple bipartite edge whose endpoints are interval multiples of a. If its ten edges have distinct endpoint representatives, they are already the required cover. Otherwise take a minimal Hall-deficient edge set. If the remaining edges also fail to match outside its neighborhood, the deficiency rises from one to two; repeat once more and it rises to three. Bipartite capacity and minimum degree then leave only two-defect kernels with 8, 9, or 10 edges and three-defect kernels with 9 or 10 edges.
The finite kernels. The eight- and nine-edge cases are the hard graphs from Rounds 5 and 6. A ten-edge/two-defect kernel has eight active vertices. Up to bipartite isomorphism there are ten types: three on a 3-by-5 bipartition and seven on 4-by-4. Each admits two disjoint four-edge blocks P,Q such that every vertex-induced prime-power level contains at least as many whole blocks as its edge/vertex excess. Charging missing copies of each prime to those blocks builds repair factors r,s with
∏ae divides (∏v)rs, r|ae for e∈P, s|ae for e∈Q.
The three-defect kernels are K₃,₃, K₃,₄ with two edges removed, and K₂,₅. The first two have three fixed four-edge repair blocks. The finite classification has 54 minimal K₃,₄ deletion patterns; another 12 same-column patterns contain a K₃,₃ core but satisfy the same certificate and were checked redundantly.
Why K₂,₅ needs a different idea. Write its labels as aij and put qj=gcd(a1j,a2j). Let R be the gcd of the ten triple products qiqjqk. Prime by prime, the missing endpoint valuation is the sum of the three smallest qj-valuations, exactly vp(R). Therefore
∏aij divides L₁L₂R₁R₂R₃R₄R₅·R.
Any three qj's cover R. A König-theorem argument finds three distinct unused interval multiples of three of them. If such a matching did not exist, a two-vertex cover would trap all remaining multiples in at most two points. Since every qj divides |L₁−L₂|, counting the intervening multiples reduces the obstruction to three small integer cases, each contradicting the number of distinct labels divisible by the same factor.
Placing the ordinary repairs. A factor supported on four distinct labels is at most M/4 and has at least seven multiples in the open interval. The only new collision possibility is that two such progressions together occupy at most nine points. Then their steps must agree: otherwise their ratio is at most 5/4 while their least common multiple is too large for the required intersection inside length 2M. Equality makes the common factor divide the union of the support blocks, producing many more interval multiples and another contradiction. Hall therefore places every repair at a distinct unused point.
The aligned endpoint needed a real repair. When x∈Mℤ, put b₀=x+M and remove M. The corrected van Doorn–Li–Tang construction draws nine edges for the remaining labels entirely inside I∖{b₀}. Exact prime-power support can fail only at one surplus two-adic level on certain symmetric edges, and those exceptional edges form a matching. Deleting such a matching from an endpoint-induced level leaves the Round-6 block certificates valid. Because all nine labels are strictly below M, the interval-multiple counts also have enough slack to choose every repair point away from b₀. The resulting nine-point product covers ∏(A∖{M}); only then do we adjoin the fresh point b₀ to supply M.
This last order matters. Our first draft incorrectly allowed b₀ to cover some smaller-label valuation and then counted the same valuation again for M. One hostile audit found the explicit counterexample M=10, A={1,…,10}, with the nine-point product on {1,2,3,4,6,7,9,10,12}. We rejected that draft, preserved the counterexample in the audit bundle, and published only after two independent audits cleared the corrected avoid-b₀ construction.
Sharpness and the new slope. The lower-bound verifier now uses six private primes at (u,v,w)=(101,103,107). Its exact valuation DP finds minimum cover size 11 in the 20,805-point bad interval, the n=10 instance of the padding family whose length ratio tends to 2. Thus c₁₀≥2 and c₁₀=2.
Subadditivity now uses blocks of ten. Writing n=10q+r and using the exact values for 0≤r<10 gives
cₙ≤2⌊n/10⌋+cn mod 10≤n/5+6/5.
The public bundle checks all 2,250 labelled new two-defect kernels, all 66 K₃,₄ deletion patterns, 279,936 K₂,₅ valuation vectors, 3,888 aligned graph/exceptional-matching pairs, every K₃,₃ exceptional matching, more than a million arithmetic-progression phases, and 48,048 literal small aligned instances. The unbounded valuation conclusion comes from the written prime-power proof, not from capped computation. This determines cₙ through n=10 and leaves the parent conjecture g(n)≤2n completely open.
Round 6 (11 August): three simultaneous defects repair too. The sibling sequence now begins
c₂=1, c₃=√2, and c₄=c₅=c₆=c₇=c₈=c₉=2.
The private-prime padding theorem from Round 3 already gives c₉≥2. The new result is the matching upper bound c₉≤2.
The extra fact at nine edges. Let A contain nine distinct positive integers, M=max(A), and I=(x,x+2M)⊂[0,∞). When x∉Mℤ, use the same two-halves graph as Round 5: the edge for a joins the last a-multiple in the left half to that point plus a in the right half. The edges are simple and distinct. More importantly, because the label is exactly the difference of its endpoints, for every prime power pk,
pk|ae iff pk divides both endpoints of e.
Thus every prime-power support is a vertex-induced subgraph. That is stronger than merely knowing that an edge label divides its endpoints, and it is what makes the third repair controllable.
Why the graph list is finite. If Hall gives nine distinct endpoints, we are done. Otherwise choose a minimal deficient edge set S. As before, |N(S)|=|S|−1 and
∏e∈Sae divides gcd(ae:e∈S)·∏v∈N(S)v.
If the remaining edges have distinct representatives outside N(S), those representatives and N(S) use eight points, and one unused gcd-multiple repairs the product. If they do not, Hall supplies a larger edge set U with |N(U)|≤|U|−2. A bipartite capacity count forces |U| to be eight or nine. Eight edges then occupy exactly six vertices and form one of Round 5's hard graphs, K₂,₄ or K₃,₃−e. Nine edges occupy six or seven vertices: six forces K₃,₃, while seven either contains an eight-edge hard core or is one of two 3-by-4 types.
Hard core plus one edge. Round 5 supplies two repair factors r and s for the core, each dividing a fixed four-label block. Choose an endpoint y of the ninth edge outside the six core vertices. Since 4r≤M and 4s≤M, each factor has at least seven interval multiples. A two-set Hall argument, strengthened by the overlap lemma below, chooses distinct r- and s-multiples outside the seven used points. Together with y and the core they cover all nine labels.
The two seven-vertex graphs. A no-core graph has bipartition 3+4. On the four-vertex side its degrees are (3,2,2,2); the three incomplete columns miss rows with multiplicities (1,1,1) or (0,1,2). In either type one can partition the nine edges into fixed blocks P and Q of sizes four and five so that every proper vertex-induced subgraph with positive edge/vertex excess contains all of P or all of Q. The full graph has excess two and contains both. Prime-power layer counting therefore constructs r,s with
∏eae divides (∏vv)rs, r|ae for e∈P, s|ae for e∈Q.
Hence 4r≤M and 5s≤M. The s-multiples leave at least two choices outside the seven graph vertices, while r leaves at least one, so distinct repair points exist.
K₃,₃ and three factors. Pair its three rows and columns by index. For i=1,2,3, let Di be the four-edge K₂,₂ left after deleting row i and column i. A vertex-induced prime-power level has positive excess only when it is K₂,₃ or K₃,₂ (excess one), or the full K₃,₃ (excess three). Charge a one-unit deficit to the block indexed by the missing row or column; at a full level charge all three blocks. Summing levels gives factors d₁,d₂,d₃ such that
∏eae divides (∏vv)d₁d₂d₃, and di divides all four labels in Di.
Each factor has at least seven interval multiples. Hall for their unused multiple sets reduces to pairs and the triple. If two factors had only one unused point between them, their full multiple sets would be the same seven-point progression, forcing the factors equal; that common divisor would divide the seven labels in the union of their blocks and would actually have at least 13 interval multiples. If all three had only two unused points, every pair of full multiple sets would have union at most eight. The following overlap lemma makes all three factors equal, after which the common divisor divides all nine labels and has at least 17 multiples. Both are contradictions.
The overlap lemma. If two interval-multiple sets each have at least seven elements and their union has at most eight, their steps d,e must be equal. Each set having at most eight points forces d,e≥2M/9. Their intersection has at least six points, so five lcm(d,e)<2M. But distinct steps have lcm(d,e)≥2min(d,e)≥4M/9, impossible.
The aligned endpoint, with every case exposed. If x∈Mℤ, put b₀=x+M. The corrected van Doorn–Li–Tang construction removes M and b₀ and gives a simple eight-edge graph for A∖{M}, entirely inside I∖{b₀}; add b₀ at the end to cover M. If the eight edges match, use their endpoints. If a minimal deficient set has size s=8, use its seven neighbors and a gcd-multiple; at s=7 use its six neighbors, an outside endpoint of the last edge, and the gcd-multiple; at s=6 use its five K₂,₃ vertices, two distinct outside endpoints of the remaining edges, and the gcd-multiple. The only collision cases are exactly the inherited six-vertex hard graphs.
All eight labels are now strictly below M. Thus for a one-factor repair, sg≤M−1, so I contains at least 2s gcd-multiples: enough to avoid the seven selected graph points and b₀. In a hard graph, 4r,4s≤M−1, giving at least eight multiples of each factor. If their unused sets collapsed to one point, r=s would divide all eight labels; 8r≤M−1 would then give at least 16 multiples. Hence the eight-point cover can always avoid b₀, and adding b₀ finishes the aligned case.
Sharpness, slope, and audit. The public padding verifier now exercises r=5 explicitly: at (u,v,w)=(101,103,107), with private primes 10427, 10429, 10433, 10453 and 10457, its exact valuation DP finds minimum cover size 10 in the 20,805-point bad interval. Thus c₉≥2 and c₉=2.
Subadditivity now uses blocks of nine. Writing n=9q+r and the exact values for 0≤r<9 gives
cₙ≤2⌊n/9⌋+cn mod 9≤2n/9+10/9.
Two independent hostile audits reconstructed the primary-source injection, the graph classification, the arbitrary prime-power layer argument, the three-factor placement, and all aligned endpoint counts. Their separate checkers covered 1,572,864 seven-vertex valuation cases, 117,649 K₃,₃ valuation vectors, both inherited hard graphs, 22,620 small arithmetic phases, 2,731 literal interval instances, and the exact n=9 lower obstruction. No gap or counterexample was found. This determines the sibling constants through n=9; it still says nothing about the parent conjecture g(n)≤2n.
Round 5 (11 August): the second Hall defect also repairs. The sibling sequence now begins
c₂=1, c₃=√2, and c₄=c₅=c₆=c₇=c₈=2.
The Round-3 padding theorem already gives c₈≥2. The new result is the sharp upper bound c₈≤2.
The two-halves graph. Let A contain eight distinct positive integers, M=max(A), and I=(x,x+2M)⊂[0,∞). First suppose x∉Mℤ. Split I at x+M. For each a∈A, let ua be its largest multiple in the left half and draw the edge (ua,ua+a) across the split. Both endpoints are positive interval multiples of a. The eight edges are distinct because their differences are the eight distinct labels a, so this is a simple bipartite graph H.
Why only two graphs are hard. Regard the edges as the left side of an incidence matching problem and their endpoints as the right side. If this graph satisfies Hall, choose one distinct endpoint for every edge; those eight points already cover ∏A. Otherwise choose a minimal Hall-deficient set S. It has |N(S)|=|S|−1 and obeys the Round-4 repair identity
∏e∈Sae divides gcd(ae:e∈S)·∏v∈N(S)v.
No simple bipartite graph with at most five edges can be deficient, since s>⌊(s−1)²/4⌋ for 1≤s≤5. Thus |S| is 6, 7 or 8. The one-gcd repair works immediately unless the whole eight-edge graph has exactly six active vertices: at size 7 the last edge otherwise supplies a new endpoint; at size 6 the obstruction is K₂,₃, and its two remaining edges have distinct outside representatives unless the full vertex set again has size six. An eight-edge simple bipartite graph on six active vertices is necessarily
K₂,₄ or K₃,₃ minus one edge.
The two-factor lemma. In either hard graph, partition the eight edges into fixed four-edge blocks P1 and P2. For K₂,₄, pair the four columns. For K₃,₃−e, put the two edges at the degree-2 left vertex in P1, the two edges at the degree-2 right vertex in P2, and split the central K₂,₂ two-and-two.
Fix a prime p. At level k, let Hk contain the edges whose labels are divisible by pk, and let V(Hk) mean its incident, non-isolated vertices. Those vertices are all divisible by pk, so the endpoint product misses at most
δk=max(0, |E(Hk)|−|V(Hk)|)
copies of p at that level. In K₂,₄, positive excess means exactly three complete columns (δ=1) or all four (δ=2). In K₃,₃−e, it means the graph with one further edge deleted, one of its two K₂,₃ subgraphs (δ=1), or the full graph (δ=2). In every δ=1 case one fixed four-edge block is wholly present; charge the missing copy of p to that block. In a δ=2 case charge one to each.
Let αp and βp count the levels charged to P1 and P2, and define r=∏pαp, s=∏pβp. A charge to a block occurs only at a level supporting all four of its edges, so αp≤mine∈P₁vp(ae) and similarly for βp; there is at most one charge to either factor at one level. Thus r divides all four P1 labels and s divides all four P2 labels. Since the six active vertex values are distinct, layer summation gives, for every p,
Σevp(ae)=Σk|E(Hk)| ≤Σk|V(Hk)|+αp+βp ≤Σvvp(v)+vp(rs),
and hence
∏e∈E(H)ae divides (∏v∈V(H)v)·r·s.
Putting both repairs into the interval. Because each repair factor divides four distinct labels at most M, 4r≤M and 4s≤M. Even with open endpoints, the number of positive d-multiples in I is ceil((x+2M)/d)−floor(x/d)−1≥ceil(2M/d)−1, so I contains at least seven multiples of each factor. Let R and S be the full interval sets of r- and s-multiples, and put Ur=R∖V(H), Us=S∖V(H). Both are nonempty. Hall for these two sets fails only if Ur=Us={z}. In that event R and S each have at least seven elements but lie in the same seven-point set V(H)∪{z}; hence both equal it. Their sorted adjacent elements differ by both r and s, forcing r=s. But r would then divide all eight labels, giving 8r≤M and at least 15 r-multiples in I, a contradiction. Thus distinct unused multiples of r and s exist, and with the six graph vertices they form the required eight-point cover.
The aligned endpoint. If x∈Mℤ, put b₀=x+M. The corrected van Doorn–Li–Tang injection removes M and b₀, then gives the other seven moduli seven distinct two-endpoint edges avoiding b₀. For a<M it uses (ua,ua+a) when a∤b₀, (b₀−2a,b₀+2a) when 2a<M and 2a∤b₀, and (b₀−a,b₀+a) otherwise. Their paper proves the three types cannot collide. Apply the one-defect repair to this seven-edge incidence graph, choosing its gcd-multiple away from b₀, then add b₀ itself to cover M. The open-interval count leaves at least eleven candidates in the smallest case, against only seven forbidden points.
Sharpness and the new slope. The private-prime padding construction from Round 3 gives c₈≥2 uniformly; a concrete n=8 instance at (u,v,w)=(101,103,107) has exact minimum cover 9. Therefore c₈=2. By subadditivity, write n=8q+r with 0≤r<8 and use the now-known exact constants for r:
cₙ≤2q+cr≤n/4+1.
Thus c₉ is now the first unresolved sibling value, with 2≤c₉≤3.
Audit. Three hostile proof passes independently reclassified the hard level subgraphs and checked the aligned construction against the primary source. The finite checker exhausts all 256 support subgraphs and all 4⁸ valuation vectors through level 3 for each hard graph. The support exhaustion checks the combinatorial classification; the written layer argument, not the bounded valuation loop, covers arbitrary valuations. A separate probe rebuilds van Doorn–Li–Tang's sharp six-neighbor K₂,₄ construction; its six endpoints already cover all eight moduli, consistent with the theorem.
Round 4 (11 August): the first two Hall defects are repairable. The sibling constants, understood as infima of admissible scale factors so that endpoint attainment does not change their values, now satisfy
c₂ = 1, c₃ = √2, and c₄ = c₅ = c₆ = c₇ = 2.
The lower bounds for the last four values come from the Round-3 theorem cₙ ≥ 2 for every fixed n ≥ 4. The new work is the other direction:
c₆ ≤ 2 and c₇ ≤ 2.
The matching input. For A with M = max(A), join a ∈ A to the positive integers b in an interval I when a divides b. Van Doorn, Li and Tang prove that every open interval of length 2M has a matching of size f(m) = min(m, ⌈2√m⌉) for every m-element A. Thus every set of at most five moduli matches completely, and every seven-element set has a matching of size at least six. For a smaller subset S ⊂ A, apply their theorem inside a subinterval of I of length 2·max(S). Passing to the interior handles closed or half-open endpoint conventions.
The one-defect repair lemma. Suppose S is inclusion-minimal among the sets that fail Hall, and write N(S) for its full neighborhood. Every S{a} matches, so |N(S)| = |S|−1 and each such matching is onto N(S). Put g = gcd(S). Then
∏a∈Sa divides g·∏b∈N(S)b.
The proof is one line per prime. Fix p, remove an ap minimizing vp(a), and match the other |S|−1 elements onto N(S). The matched points supply all the valuations from S except vp(ap), while vp(g) = vp(ap) supplies the missing valuation. The matching may depend on p; the fixed product over N(S) does not.
Six moduli. If A itself has no matching, no proper subset can be deficient, so N(A) has exactly five points. Apply the lemma. The six distinct elements of A are multiples of g = gcd(A), hence M ≥ 6g. Even an open interval of length 2M contains at least ⌈2M/g⌉−1 ≥ 11 positive multiples of g. Choose one z outside N(A). The six-point set N(A)∪{z} covers ∏A.
Seven moduli. A minimal deficient S has size six or seven. If |S|=7, repair it exactly as above. If |S|=6, let a₀ be the remaining modulus. The six-matching theorem forces a₀ to have a neighbor y outside N(S): otherwise all seven moduli together would have only the five neighbors of S. Choose y first, then choose a g-multiple z outside N(S)∪{y}. Now N(S)∪{y,z} has seven distinct points; z repairs S and y covers a₀.
Sharpness and the new general bound. Round 3 already proved cₙ ≥ 2 for every n ≥ 4, so the two upper bounds give c₆ = c₇ = 2. Also cr+s ≤ cr+cs. To avoid assuming that an infimum is attained, take any ε>0, split an interval of length (cr+cs+ε)M into two disjoint open blocks slightly longer than crM and csM, cover the two parts of A separately, and let ε decrease to zero. Writing n=7q+r, with 0≤r<7, and using
(c₀,…,c₆)=(0,1,1,√2,2,2,2), cₙ ≤ 2q+cr ≤ (2/7)n+6/7.
At the end of Round 4 this left 2≤c₈≤3. The one-defect argument alone did not close automatically: the matching theorem guarantees only six matches, so two uncovered moduli can compete for the same outside point. Round 5 above resolves exactly that two-defect collision.
Audit and provenance. Three independent high-effort proof passes converged on the same lemma; one additionally classified the elementary six-edge bracketing obstruction as K₂,₃. A standard-library checker exhausts 382,096 small translated six-set instances, repairs all seven bracketing failures, and rebuilds genuine five-neighbor c₆ and c₇ full-graph obstructions from the matching paper. Separate capped-valuation programs return minimum cover sizes seven and eight for the n=6 and n=7 lower constructions.
An unpromoted local note dated 31 July already contained essentially this repair lemma and claimed c₆=c₇=2. Round 4 independently reconstructed and red-teamed it; this is not a priority claim. As of 11 August the live #708 tracker listed no claimed proof and no current worker. Nothing here is a solution of the main g(n) problem.
Round 3 (1 August): c₅ = 2 — and the sequence has a floor. The sibling constant now reads
c₂ = 1, c₃ = √2, c₄ = 2, c₅ = 2, and cₙ ≥ 2 for every n ≥ 4.
With the trivial pairing bound cₙ ≤ ⌈n/2⌉ that boxes every later constant into [2, ⌈n/2⌉]. Erdős and Surányi reported "no good upper or lower bounds" for cₙ in general; the floor is a general lower bound, which is the part of that sentence we think is worth the most.
Which half is ours — stated up front, because it matters. The upper bound c₅ ≤ 2 is not a new theorem. It is a short corollary of van Doorn, Li and Tang, "Optimal bounds for an Erdős problem on matching integers to distinct multiples" (arXiv:2603.28636, 30 March 2026), which proves that any m-element set can be matched to distinct multiples inside any interval of length 2·max(A) in f(m) = min(m, ⌈2√m⌉) places. Since ⌈2√5⌉ = 5, f(5) = 5 is a perfect matching — five distinct integers, one multiple of each element — which immediately gives the product divisibility. Note n = 5 is the last n where this works: f(6) = 5 < 6. That paper (which solves a different Erdős problem, #650) was public four months before we touched this. Anyone holding it gets c₅ ≤ 2 in a line, and we are not going to pretend otherwise.
The five-line version, for people who don't want to cite a paper. Take any 2M consecutive integers I = {t+1,…,t+2M} and set γ = t+M+½. For each a ∈ A take the two consecutive multiples of a bracketing γ: ℓ_a = a⌊γ/a⌋ and r_a = ℓ_a + a. Both are in I. Read {ℓ_a, r_a} as an edge of a bipartite graph (left = below γ, right = above); all |A| edges cross the cut, and they are distinct because their lengths are the distinct elements a. If k of these edges met at most k−1 vertices, that simple bipartite graph would need k ≤ ⌊(k−1)²/4⌋ edges — false for every k ≤ 5, and first true at k = 6. So Hall's theorem gives a system of distinct representatives for n ≤ 5. The counting dies exactly at six (K₂,₃ is six edges on five vertices), which is the same place vDLT's f(m) leaves the diagonal. We ran 60,000 random 5-element instances against this: zero failures.
The lower bound c₅ ≥ 2 is ours, and it does not come free. It does not follow from c₄ = 2 by monotonicity — cₙ is not obviously nondecreasing, since a larger n makes the set harder but also hands you more picks. (c₃ = √2 < 2 is the sequence going down before it goes up.) The move is to take the Round-2 c₄ obstruction and pad it with a prime that is deliberately useless. Choose primes u < v < w with a prime q in (uv, vw), and set
A = {uv, v², uw, vw, q}, M = vw.
Position by CRT (t ≡ 1 mod uw, t·uw ≡ v+1 mod v², t·uvw ≡ 1 mod q) an interval of length 2uv−1 centred on x = tuvw. Inside it: x is the unique multiple of each of uv, uw, vw and has exact (u,v,w)-valuation (1,1,1); x−v is the unique v²-multiple and there is no v³-multiple at all; x−1 is the unique q-multiple and is ≡ −1 mod u, v and w, so it contributes nothing to the rest of the product. Every point other than x is divisible by at most one of u, v, w — any second colour would be another multiple of uv, uw or vw.
Now count. ∏A = u²v⁴w²·q. Ignore q first. If you skip x, single-colouring forces one point for u, one for w, and — since the only v²-multiple has valuation exactly 2 and there is no v³ — at least three for v⁴ (best pattern 2+1+1). Five points, all distinct. If you take x, it pays uvw and leaves uv³w, needing one u, one w and two v points outside x. Five again. So four points can never cover u²v⁴w². And any five-point cover of the full product must burn one slot on x−1 to get q, leaving four for the impossible core. Six picks are forced, while |I|/M = (2uv−1)/(vw) = 2u/w − 1/(vw) → 2 as u/w → 1.
We didn't trust the prose — we ran it. An exact capped-valuation dynamic program (the genuine minimum cover, not a heuristic search) returns 6 at (u,v,w,q) = (17,19,23,331), (29,31,37,907), (41,43,47,1777) and (101,103,107,10427), with every uniqueness and neutrality claim holding at each, and forcing ratios climbing 1.476 → 1.567 → 1.744 → 1.888. The published certificate reproduces exactly: A = {323, 331, 361, 391, 437}, t = 14,257,816, x = 105,921,315,064, interval [105,921,314,742, 105,921,315,386], minimum cover 6.
The uniqueness conditions are not an artifact of small primes — they hold for all u < v < w. The second v²-multiple sits outside the interval because v(v−1−u)+1 > 0 whenever v ≥ u+2, and the one below is out because u < 1+v+1/v. That is tight — the window has room for about 2u/v ≈ 2 of them and the CRT placement is what keeps the count at one — but it is unconditional. For the ratio to reach 2 we also need the primes to exist: three primes clustered tightly enough, with a prime q in (uv, vw). Checked directly — (1009,1013,1019) gives 1.98037, (4133,4139,4153) gives 1.99037, (69481,69491,69493) gives 1.99965, and (1865321,1865327,1865329) gives 1.99999142, each with a genuine prime q in range and all the uniqueness inequalities satisfied. No constant below 2 survives.
The floor, which is the better theorem. The padding prime was neutral by construction, so nothing stops you adding more. For n = 4+r, take r primes q_j ∈ (uv+r, vw), park their unique multiples at x−1, …, x−r, and choose u > r so they stay neutral. An n-point cover spends r slots on inert points and has four left for a core that needs five. Hence cₙ ≥ 2 for every fixed n ≥ 4 — permanently, for all n at once. Verified by the same exact DP: minimum cover 7 at n=6 for (41,43,47) and (101,103,107), and 8 at n=7 for (101,103,107), at ratio 1.888. At the time this did not determine cₙ for n ≥ 6; Round 4 above now determines the first two values. The padding argument rules out forever that any later constant dips below 2.
What we are not claiming, and one live risk. Our targeted search found no prior determination of c₅ and no published cₙ for n ≥ 4, and the tracker still described the constants as having no good bounds in general — but search-negative is a coverage claim, not a priority proof. At Round 3 publication Wouter van Doorn, lead author of the matching paper, was listed as currently working on #708. As of the Round 4 live check on 11 August that marker is None. Round 4 also checked the 1959 primary PDF. The new repair lemma is not Lean-formalized. And per that forum's own stated bar, "an AI verified it" is not sufficient for posting, so we have not posted this there.
Round 2: c₄ = 2, proved — and our Round-1 guess was wrong, which is the better outcome. Round 1 (below) proved √2 ≤ c₄ ≤ 2 and then guessed that √2 was the sharp value — that the upper bound would come down to meet the lower. That conjecture is false. A follow-up found a different obstruction showing the upper bound is the sharp one:
c₄ = 2.
The construction is a small but lethal modification of the √2 one. Where Round 1 used A={w, uv, uw, vw}, take instead A={uv, v², uw, vw} — the Erdős–Surányi c₃ triangle {uv, uw, vw} with the square v² added as the fourth element. For primes u<v<w, place (by CRT: t≡1 mod uw and tuw≡v+1 mod v²) an interval of length 2uv−1 around x=tuvw in which x is the unique multiple of uv, uw and vw, x−v is the unique multiple of v², and there is no multiple of v³. Then ∏A = u²v⁴w², and a clean 0/1 capacity count shows four picks can never cover it: omit x and the single-colored points force one slot for u², one for w², and at least three for v⁴ (best pattern 2+1+1) — five slots; take x and the residual u·v³·w needs four slots in the three remaining. Meanwhile the ratio (2uv−1)/(vw) = 2u/w − 1/(vw) tends to 2 as the primes cluster. We verified this decisively: an exact capped-valuation dynamic program returns a minimum cover of exactly 5 for (u,v,w)=(17,19,23), (29,31,37), (41,43,47), (101,103,107) — forcing ratios 1.476 → 1.567 → 1.744 → 1.888 — and an exhaustive search confirms no 4-cover at the concrete (17,19,23) instance. Combined with the Round-1 pairing upper bound c₄≤2, this gives c₄=2.
Why √2 was the wrong guess, honestly. The Erdős–Surányi c₃=√2 proof balances two competing thresholds — a "second pair-multiple" threshold 2uv against a "square-repair" threshold w² — and their geometric mean gives √2. The Round-1 c₄ construction inherited that balance. But adding v² to the set taxes the square-repair point: the interval has exactly one v²-multiple and no v³-multiple, so the formerly-free square repair now supplies only half the v⁴ demand and no longer closes the cover. The square branch disappears, the obstruction survives all the way out to the 2uv threshold, and the normalized ratio is 2u/w → 2. The √2 lower bound we proved is still TRUE — it is simply not sharp; c₄ sits at the upper bound, not the lower. Both Round-1 bounds were correct, and we guessed wrong about which was tight.
Round 1: the sibling constant, and the first bounds past n=3. The same paper asked a sharper quantitative version: how long must the interval be — as a multiple cₙ of M — so that n picks always suffice? They proved c₂=1 and c₃=√2, and then, in the tracker's own words, had "no good upper or lower bounds in general." We prove
√2 ≤ c₄ ≤ 2.
As far as the tracker shows, these are the first nontrivial bounds on any cₙ beyond n=3.
What c₄ ≥ √2 means, and the construction. There are 4-element sets A for which intervals of length approaching √2·M still force a fifth pick. The construction is the n=4 analog of Erdős–Surányi's own c₃=√2 trick: take primes u<v<w with 2uv<w², set A={w, uv, uw, vw} so M=vw, and position (by the Chinese remainder theorem) an interval of length 2uv−1 whose centre x is the unique multiple of uv, uw and vw inside it, with no multiple of w² inside at all. Then every pick carries at most one factor w, so three w-multiples are forced, and none of the remaining picks can supply both u and v — four picks cannot reach u²v²w³ = ∏A. As u,v,w scale with w≈√2·u the ratio (2uv−1)/(vw) tends to √2.
We didn't trust the prose — we ran it. We instantiated this with real primes at five scales and ran an exact dynamic program over capped prime-valuation states, which computes the genuine minimum cover rather than searching heuristically. Every instance forces exactly five picks, all the uniqueness claims hold, and the forcing ratio climbs 1.382 → 1.391 → 1.402 → 1.410 → 1.411 toward √2 = 1.4142… from below. Since "no 4-cover works" is inherited by every sub-interval, that certifies c₄ ≥ √2 against the page's exact definition of cₙ.
The upper bound c₄ ≤ 2 is almost embarrassingly clean once seen. Split any interval of length 2M into two disjoint M-blocks. In the first, apply g(2)=2 to two of the four integers; in the second, apply it to the other two. Four picks total, product divisible by ∏A. We verified g(2)=2 exhaustively — every pair at small sizes, every interval position: maximum cover exactly two, and some pairs genuinely need two — and ran 300 random 4-integer tests on 2M-intervals, every one of which had a 4-cover. Hence √2 ≤ c₄ ≤ 2.
Two things we are explicitly not claiming. First, the lower-bound mechanism is not ours: it is the n=4 specialization of Erdős and Surányi's 1959 c₃=√2 construction, reconstructed independently here, and we name it as theirs — what was new in Round 1 was applying it to c₄ for the lower bound; Round 2's new contribution is the v² obstruction proving c₄=2 (and showing the √2 lower, while true, is not sharp). Second, "apparently new" is not "proved first-ever": the tracker listing no general bounds is strong evidence that c₄ in particular had never been pinned down (our targeted search found no later determination of c₄ either), but a full reading of the 1959 paper and its follow-ups is still ahead of us, and we say so rather than overstate.
And g(4) itself. We confirmed g(4) ≥ 5 with the same exact machinery — a clean witness: A={13, 77, 91, 143}, interval [5935, 6077], where no four picks work and five do — but that is the paper's own n+1 lower construction, not new. The reasoner also floated a weak g(4) ≤ 24 without proof; we do not assert it. The honest conjecture is g(4)=5, and it is open.
Where it stands now. The sibling constant is determined through n=8: c₂=1, c₃=√2, and c₄=c₅=c₆=c₇=c₈=2. Pairing still gives cₙ≤⌈n/2⌉, but c₅=2 already shows it is not generally sharp. Round 5 improves the general upper bound to cₙ≤n/4+1 by subadditivity. Together with the permanent floor cₙ≥2, this leaves c₉ as the first exact value: 2≤c₉≤3. The parent problem g(n)≤2n remains open and untouched; the reasoner's clean diagnosis of why the naive fractional covering argument cheats (it drops the "use each integer at most once" capacity constraint, where the difficulty lives) still stands.