# Is fixed key AES in counter mode a PRG?

I want to implement a PRG (Pseudorandom Generator). As AES is implemented in hardware, I am planning to use AES for efficiency. I am planning to use:

$(AES_k(seed) \| AES_k(seed+1) \| AES_k(seed+2) \|....)$

as a PRG ($k$ represents the key of AES). The receiver of this sequence does not know the random seed, but knows the AES key $k$. So, is this sequence a provably secure PRG, when the AES key is known (at least in ideal cipher model)?

Hint: given $AES_k(seed)$ and the key $k$, what can an adversary compute? Should this be possible for a secure PRG?
A better option would be standard CTR mode, i.e. $AES_{seed}(0) \| AES_{seed}(1), ...$
This is completely broken. In particular, since $k$ is known, given any block $r$ of randomness, it is possible to compute $s=AES^{-1}_k(r)$ and then generate the rest of the sequence by $AES_k(s+1),AES_k(s+2),\ldots$.