The strategy is easiest to describe when you assume n = 1 + 2 + … + k is a triangular number, in which case it is possible to succeed in 2n + k + 1 weighings (note k is approximately \sqrt{2n}). Call the scales A, B, C. There are 2n unknown trits of information. For the first 2k weighings, use A to attempt to determine the first k trits, then use B to determine the next k trits. Let SA be the set of 3^{2n-k} coins whose first k trits agree with the result of A’s weighings, similarly for SB. This means the fake coin is in SA if and only if A’s weighings were all correct, either because A was a normal scale or by chance.

For the next weighing, use C to weigh SA\SB against SB\SA (\ is set difference).

– If this weighing balances, we know first 2k weighings were correct, leaving 3^{2n-2k} possibilities for the fake. Since n – k = 1 + 2 + … + (k-1) is the previous triangular number, we can recursively apply the same strategy to these coins. By induction, 2(n-k) + (k-1) + 1 = 2n – k more weighings are needed.

– If (say) SA\SB is lighter, then either B or C must have given an incorrect weighing, so A must be a normal scale. We have already used A to measure k trits, so 2n – k more uses of A suffice to find the fake.

Either way, 2n – k more weighings are needed, which combined with the first 2k+1 weighings makes 2n+k+1, as claimed.

For n which are not triangular numbers, write n = T + j, where T is the largest triangular number less than n. Do as described in the last paragraph, except starting with A and B each measuring j trits. In the worst case this will take 2n + 1 + (k+1) weighings, where T is the kth triangular number.

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