{"schema":"vela.problem-packet.v0.1","problem":376,"statement":"Are there infinitely many $n$ such that $\\binom{2n}{n}$ is coprime to $105$?","status":"open","seam":"sealed","closureRoutes":[],"obligations":[],"attestations":[],"attempts":[{"id":"att_b42293d174e1db85","kind":"dead_end","claim":"Erdős #376 (is C(2n,n) coprime to 105=3*5*7 for infinitely many n?): honest obstruction map, Opus-verified. Cheap verifier (every base-p digit of n <= (p-1)/2; for {3,5,7}: base-3 digits in {0,1}, base-5 in {0,1,2}, base-7 in {0,1,2,3}) agrees with direct gcd(C(2n,n),105) AND Kummer carry-count, 0 mismatches n<3000; reproduces OEIS A030979 exactly. The obstruction is the SIGN of the density exponent 1+r = 1 + sum_p log((p+1)/2p)/log p: for {3,5,7} it is +0.0260 (count ~ N^0.026, infinite but sub-polynomially sparse — matches the huge b-file gaps); for {3,5,7,11} it is -0.227 (finite — only {0,1,3160}, matching Graham's conjectured last term). #376 sits at +0.026, so close to 0 that infinitude is not robust to any uniform error term — exactly why digit-distribution/exponential-sum methods (which lose factors swamping a 0.026 exponent) cannot prove it. Honest obstruction map; no proof of finiteness or infinitude.","grade":"obstruction_map","gateStatus":"verified","superseded":false}],"velaLean":[],"oeis":[{"id":"A030979","name":"Numbers k such that binomial(2k,k) is not divisible by 3, 5 or 7.","terms":"0,1,10,756,757,3160,3186,3187,3250,7560,7561,7651,20007,59548377,59548401,45773612811,45775397187,237617431723407,249919","url":"https://oeis.org/A030979"}],"generated_note":"derived view; the signed event log is the source of truth (vela check / vela reproduce)"}