md5(
            pack('N', $this->keyToBytes($key1)) .
            pack('N', $this->keyToBytes($key2)) .
            $key3, true);

Lets start with key1, key2, and the 8 bytes at the end of the handshake

key1 = "18x 6]8vM;54 *(5:  {   U1]8  z [  8"
key2 = "1_ tx7X d  <  nw  334J702) 7]o}` 0"
end8 = "Tm[K T2u"
We make a number for each key by ignoring every character that is not a digit 0-9. In Python:

def numFromKey(key):
    return int(filter(lambda c: c in map(str,range(10)),key))
next we divide that number by the number of spaces in the original key string, so here is a is a function that counts the spaces in a string.

def spacesIn(key):
    return len(filter(lambda c: c==' ',key))
The two numbers resulting from the keys are:

pkey1 = numFromKey(key1)/spacesIn(key1)
pkey2 = numFromKey(key2)/spacesIn(key2)
Now we need to concatenate the bytes of pkey1, pkey2, and end8. The processed keys need to be represented as 32 bit Big-Endian numbers.

from struct import pack
catstring = pack('>L',pkey1) + pack('>L',pkey2) + end8
Then we take the md5 hash of those bytes to get the magic number that we tack on the end of the handshake

import md5
magic = md5.new(catstring).digest()
Thats how I think it works at least

byte[] clientBytes = LeseBytesVomClient();

ulong zahl1 = ExtrahiereZahl(Key1);  // ulong, da die zahlen recht gross sein können
ulong zahl2 = ExtrahiereZahl(Key2);
ulong spaces1 = AnzahlLeerschläge(Key1);
ulong spaces2 = AnzahlLeerschläge(Key2);

uint neueZahl1 = (uint)(zahl1 / spaces1);
uint neueZahl2 = (uint)(zahl2 / spaces2);

byte[] buffer = { die bytes von neueZahl1, neueZahl2, clientBytes };

buffer = MD5(buffer);

string antwort = (string)buffer;