We have 152x3=456 bits, 76 are nulls after xoring, that leaves as with 380 bits, then, we have a set of 17 bits doubled, perhaps a 2nd key, making yet another 17 bits nullified, that's 380-17=363. What is 363? Well, Code-128-B, that can encode Base58 and Minikeys uses START code and END code, which are 11 bits each, also uses CRC code, also 11 bits, that's 363-11-11-11, so 330 bits left, 330 bits is exact bit count needed to encode 30-char miniprivate key...

Here: 1F4crLa8TuSNKfDiUhjkCsDqKx6VTSifG2  thx!