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Is there some C(r)>2 such that ρ2(2(r+2)+1,r+2)≥C(r)ρ2(2r+1,r)\rho_{2}(2(r+2)+1,r+2)≥C(r)\rho_{2}(2r+1,r) holds for all r ≥2 ?

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Source
  • k-Domination invariants on Kneser graphs
  • FAR
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No person has examined this. Everything below was judged by machines. say whether it holds →

  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: In finite simple graphs, let K(n,r)K(n,r) be the Kneser graph on the rr-subsets of [n][n], with adjacency given by disjointness. Let ρ2(n,r)\rho_2(n,r) be the maximum size of a vertex set whose distinct members have graph-distance at least 33. The reconstructed question asks whether, for every integer r≥2r\ge2, there is a real number C(r)>2C(r)>2 such that

    ρ2(2r+5,r+2)≥C(r)ρ2(2r+1,r).\rho_2(2r+5,r+2)\ge C(r)\rho_2(2r+1,r).

    This matches the paper’s notation ρ2(n,r)=ρ2(K(n,r))\rho_2(n,r)=\rho_2(K(n,r)) and its stated Problem 23.

    Result: Yes. In fact,

    ρ2(2r+5,r+2)≥2ρ2(2r+1,r)+1\rho_2(2r+5,r+2)\ge 2\rho_2(2r+1,r)+1

    for every r≥2r\ge2. Hence one may take

    C(r)=2+(2r+1r)−1>2.C(r)=2+\binom{2r+1}{r}^{-1}>2.

    Proof. For s≥2s\ge2, a family F⊆([2s+1]s)\mathcal F\subseteq \binom{[2s+1]}s is a 22-packing in K(2s+1,s)K(2s+1,s) iff every two distinct A,B∈FA,B\in\mathcal F satisfy

    1≤∣A∩B∣≤s−2.1\le |A\cap B|\le s-2.

    Indeed, ∣A∩B∣=0|A\cap B|=0 means adjacency, while A,BA,B have a common neighbor iff [2s+1]∖(A∪B)[2s+1]\setminus(A\cup B) has size at least ss, i.e. iff ∣A∩B∣+1≥s|A\cap B|+1\ge s.

    Let S\mathcal S be a maximum 22-packing in K(2r+1,r)K(2r+1,r). Add four new points split as

    A={a1,a2},B={b1,b2}.A=\{a_1,a_2\},\qquad B=\{b_1,b_2\}.

    Define

    T={S∪A:S∈S}∪{S∪B:S∈S}.\mathcal T=\{S\cup A:S\in\mathcal S\}\cup \{S\cup B:S\in\mathcal S\}.

    For distinct S,S′∈SS,S'\in\mathcal S, intersections inside T\mathcal T are either ∣S∩S′∣|S\cap S'|, ∣S∩S′∣+2|S\cap S'|+2, or rr, all lying between 11 and r=(r+2)−2r=(r+2)-2. Thus T\mathcal T is a 22-packing in K(2r+5,r+2)K(2r+5,r+2) of size 2∣S∣2|\mathcal S|.

    Since r≥2r\ge2, S≠([2r+1]r)\mathcal S\ne \binom{[2r+1]}r; for example two rr-sets with intersection r−1r-1 cannot both lie in a 22-packing. Choose Y∈([2r+1]r)∖SY\in\binom{[2r+1]}r\setminus\mathcal S, and set

    Z=Y∪{a1,b1}.Z=Y\cup\{a_1,b_1\}.

    For every S∈SS\in\mathcal S,

    ∣Z∩(S∪A)∣=∣Y∩S∣+1,∣Z∩(S∪B)∣=∣Y∩S∣+1.|Z\cap(S\cup A)|=|Y\cap S|+1,\qquad |Z\cap(S\cup B)|=|Y\cap S|+1.

    Because Y≠SY\ne S, we have ∣Y∩S∣≤r−1|Y\cap S|\le r-1, so these intersections lie between 11 and rr. Hence T∪{Z}\mathcal T\cup\{Z\} is a 22-packing in K(2r+5,r+2)K(2r+5,r+2). Therefore

    ρ2(2r+5,r+2)≥2ρ2(2r+1,r)+1.\rho_2(2r+5,r+2)\ge 2\rho_2(2r+1,r)+1.

    Finally, since 1≤ρ2(2r+1,r)≤(2r+1r)1\le \rho_2(2r+1,r)\le \binom{2r+1}{r},

    2ρ2(2r+1,r)+1≥(2+(2r+1r)−1)ρ2(2r+1,r),2\rho_2(2r+1,r)+1 \ge \left(2+\binom{2r+1}{r}^{-1}\right)\rho_2(2r+1,r),

    which proves the claim.

    Audit: No extra hypotheses beyond r≥2r\ge2 were introduced. The proof uses only the standard Kneser graph definition and the exact distance characterization for odd Kneser graphs, proved above. The conclusion gives an explicit C(r)>2C(r)>2, exactly matching the reconstructed statement.

    Citation: Problem source: Brešar, Cornet, Dravec, Henning, “kk-Domination invariants on Kneser graphs,” arXiv:2312.15464, Problem 23. No prior resolution is used here.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The proof attacks the exact stated problem. The distance/intersection characterization for K(2s+1,s)K(2s+1,s) is correct: a 2-packing requires pairwise intersections between 11 and s−2s-2.

    The doubling construction gives 2ρ2(2r+1,r)2\rho_2(2r+1,r) valid vertices in K(2r+5,r+2)K(2r+5,r+2), and the added mixed vertex Z=Y∪{a1,b1}Z=Y\cup\{a_1,b_1\} is compatible with all of them because its intersections are ∣Y∩S∣+1∈[1,r]|Y\cap S|+1\in[1,r]. The existence of Y∉SY\notin\mathcal S is justified for r≥2r\ge2. Thus

    ρ2(2r+5,r+2)≥2ρ2(2r+1,r)+1,\rho_2(2r+5,r+2)\ge 2\rho_2(2r+1,r)+1,

    which yields the claimed C(r)>2C(r)>2.

    I found the source paper’s Proposition 12 gives only the weaker factor 22, and no stronger prior result surfaced in the checked arXiv/source context.

    Novelty assessment

    TYPE1

    Classification rationale: The result is a very small strengthening of Proposition 12 in the source paper: the published construction already gives the factor 22, and the accepted proof adds one extra admissible vertex to obtain 2ρ2+12\rho_2+1. This resolves the stated question because C(r)C(r) is allowed to depend on rr, but the improvement is tiny and elementary. It is not enough for a standalone combinatorics paper; at most it would be a short remark or addendum.

    Literature check: I found no source stating the 2ρ2(2r+1,r)+12\rho_2(2r+1,r)+1 strengthening or resolving Problem 23. The arXiv search for “Kneser” and “2-packing” returns essentially the source paper and the earlier Cornet–Torres paper. The source paper itself states Proposition 12 with only the weaker factor 22, then explicitly asks Problem 23. Searches for the exact expressions ρ2(2(r+2)+1,r+2)\rho_2(2(r+2)+1,r+2), ρ2(2r+5,r+2)\rho_2(2r+5,r+2), “C(r)>2C(r)>2”, and “odd graph 2-packing” did not reveal a prior resolution.

    Citation: Brešar, Cornet, Dravec, Henning, “kk-Domination invariants on Kneser graphs,” arXiv:2312.15464, Proposition 12 and Problem 23. Earlier related work: Cornet and Torres, “kk-tuple domination on Kneser graphs,” arXiv:2308.15603.

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