{"schema":"vela.problem-packet.v0.1","problem":684,"statement":"For $0\\leq k\\leq n$ write\\[\\binom{n}{k} = uv\\]where the only primes dividing $u$ are in $[2,k]$ and the only primes dividing $v$ are in $(k,n]$. Let $f(n)$ be the smallest $k$ such that $u&#62;n^2$. Give bounds for $f(n)$.","status":"open","seam":"sealed","closureRoutes":[{"type":"witness","verifierKind":"kummer_no_carry","note":"extended no-carry witness table re-checked by the frozen verifier"},{"type":"formal_proof","verifierKind":"lean","note":"Lean patch building clean under the math CI profile (no sorry, no new axioms)"},{"type":"obstruction_report","verifierKind":"review","note":"precise, artifact-backed reason a route cannot work"}],"obligations":[{"findingId":"vf_ba50ac1f1f60aa27","banked":"f(M_K-1) > K verified for K=3..12 via Kummer no-carry (effective, subsequential lower bound)","open":"promote the subsequential bound to the full claim for all n.","dependents":1,"lease":null}],"attestations":[],"attempts":[{"id":"att_7eba29f25c2bd57c","kind":"partial_proof","claim":"Erdos #684: verified effective subsequential lower bound f(n) >= (1/2+o(1)) log n. Explicit construction M_K = prod_{p<=K} p^{floor(log_p K)+1}; computationally verified (K=3..12) that all v_p(C(M_K-1,k))=0 for p<=k<=K (no Kummer carries), so the smooth part u=1 < n^2 for every k<=K, hence f(M_K-1) > K; with log M_K = psi(K)+theta(K) = (2+o(1))K this gives the stated bound along n=M_K-1. The pointwise lower bound f(n)->infinity is known only INEFFECTIVELY (Mahler); this effective statement is subsequential, not pointwise. UPPER side (GPT-Pro, not re-verified here): an elementary Kummer+Dusart argument gives f(n) = O((log n)^2). UNVERIFIED claims (cited preprints not confirmable from here, treat with caution): an 'APSSV' optimized constant 24/(pi^2-6) and a 'Bae 2026' limsup f(n)/log n = infinity.","grade":"partial_proof","gateStatus":"verified","superseded":false}],"velaLean":[],"oeis":[{"id":"A392019","name":"If binomial(n,k) = u*v where the only primes dividing u are <= k and the only primes dividing v are > k, then a(n) is the least k such that u > n^2, or 0 if such a k doesn't exist.","terms":"0,0,0,0,0,0,0,0,0,7,0,0,0,0,0,11,0,13,0,0,8,13,0,17,12,13,13,5,17,7,19,9,12,11,11,7,11,17,15,14,17,17,19,13,13,19,23,17,","url":"https://oeis.org/A392019"}],"generated_note":"derived view; the signed event log is the source of truth (vela check / vela reproduce)"}