Everybody is looking for:
h = f(x), where h < TARGET
I think you're asking if there is some weakness in SHA256 (or hashing in general) that can make solving that problem trivial.
It seems unlikely to me, but I'm not a professional cryptographer. If TARGET were '4', then the problem is equivalent to:
h = f(x) where h = 0 OR h=1 OR h=2 OR h=3
Maybe there's some tricky method that reuses work and makes solving the TARGET=4 case more than 4 times easier than solving for TARGET=1 (which is "find this specific hash")... but I just don't see that doing anything more than what has already been pointed out in previous posts: it is just a quicker way of hashing, so difficulty would go up.