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    $\begingroup$ Considering the information content of each weighing, $N=13$ is an upper bound for four weighings. I haven't been able to show if this bound is sharp or not, but I suspect not. $\endgroup$ Commented Mar 19 at 4:00
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    $\begingroup$ Here is question that asks for a solution for N=7 in 4 weighings when the two counterfeits are known to be heavier. I doubt you can get N=7 when you don't know the relative weights, let alone larger N. $\endgroup$ Commented Mar 19 at 8:47
  • $\begingroup$ Do you mean maximum or minimum.? Maximum surely means you can use as many coins as you like. $\endgroup$ Commented Mar 20 at 15:26
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    $\begingroup$ @JaapScherphuis In the linked question, N is 9, not 7. $\endgroup$ Commented Mar 21 at 6:00
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    $\begingroup$ @WeijunZhou You’re right, I misread. So it’s almost a duplicate, if it weren’t for the knowledge of the relative weight $\endgroup$ Commented Mar 21 at 7:15