GENERAL IDEA:

Interpret each metal as a digit in base 6; the metals are however many quicksilver you need to promote lead into that metal (lead is 0, gold is 5). Use two of the bits to split the problem of encoding into four quadrants, turning each of the base-6 digits into base-3 digits--the quadrants decide if the metals start as lead (0) or copper (3). Then convert the remaining three-bit number into a 2-digit base 3 number. You can just sorta add stuff together with projection.

In base 3, the values "11", "02", and "01" are the binary counting we recognize as 4, 2, and 1 in base 10. The only hard operation here is the carry--I build a stick of up to 4 quicksilver and conditionally disassemble it to project the correct digits. This is WAY more feasible in base 3 than it is in base 6.

Converting 3 bits to 2 trits means that the ternary value "22" goes unused. Long story short, this machine skips the following pairs in its encoding: iron-iron, iron-gold, gold-iron, and gold-gold.

The reason I don't simply do the binary counting in base 6, where I get to use more convenient presence/absence conditionals on gold, is that this way requires less quicksilver. The binary counting in base 6 can take as little as 13 quicksilver with good strategy; this method takes 11 quicksilver. I built both. This one's better.

RATE NOTES:

Up to 11 quicksilver may be used in this design, and the conditionals can operate at pseudo-period 3 at best, limiting this design's rate to 33. My actual rate is 34; the fire reader chokes slightly on the fastest bit.