So studying double encryption (double DES) and understanding why it is subject to a meet-in-the-middle attack, I tried to see if the same attack could be applied to a triple encryption with only two different keys and no decryption operation. I.e.:
$$c=E_{k2}(E_{k1}(E_{k1}(m)))$$
where $c$ is the ciphertext, $m$ the plaintext and ${k_1,k_2}$ the two keys.
It seems the normal way is to have encryption-decryption-encryption with either 2 or 3 different keys. However, with only two keys and three sequential encryptions provide any more security than double DES? I guess you could still apply the meet-in-the-middle attack, since there are intermediary values to which you can apply the same attack, but I'm a bit confused to whether it adds more security than double DES?
I guess it would require $2^{56}$ additional operations to perform a meet-in-the-middle attack since there is an extra step compared to double DES?
Additional question: Does it matter in which order you apply the keys? I.e. is there a difference between encrypting with $k_2-k_1-_1$ vs. $k_1-k_2-k_1$?
All keys in my question are of size 56-bits.