If you take a 32-bit sequence, and perform a CRC32 on it, you get another 32-bit sequence as the result; if you do CRC32 of this, you get another, and so on. It is easy to show that if you keep doing this, you end up with a single loop of 2^32 bit sequences, before starting over.
Simple question: does anyone know if the same holds true (or not) for SHA256, starting with a 256-bit sequence? Would a similar process cycle through a loop of all 2^256 possible 256-bit sequences before starting over? Or are there known (or likely) shorter loops within this hash?
Brian