Let us assume that the magician and the assistant can perform for a number n. Then, the assistant has a way to change all possible settings of heads/tails into a setting that the magician will interpret as any given number. In other words, the duo have a way of interpreting any setting of heads/tails as a number from 1 to n, and changing any setting to a setting that reflects any given number.
How to do it: assume the audience choose k.
1. The assistant, given the setting of 2n coins, pairs each coin up with another one, to form n pairs.
[O O] [O O] [O O] ...
2. He interprets each pair as an individual coin by the following scheme: two heads or two tails = a head, a head and a tail either way = a tail. Let's call these new n coins "p-coins."
3. He calculates the remainder of k / n, replacing 0 with n. Let this be x.
4. He decides which p-coin to flip so as to make the p-coins show x. (He doesn't flip it or anything.)
5. He calculates r: if k > n, r=1, otherwise r=0.
6. He looks at the second coin of each p-coin, and computes how many heads are among this set of n coins, not p-coins.
7. He finds the result of step 6's remainder when divided by 2. Call this s.
8. He compares r (step 5) with s (step 7). If they are equal, then flip the first coin of the step 4 p-coin. If they are different, flip the second coin of the step 4 p-coin.
How the magician interprets:
1. Change the coins into p-coins by pairing them and valuing them with two alike = head and two unalike = tail like above, and calculate the p-coins' number like above.
2. Count the heads in the second coins of each p-coin. If the number is odd, add n to the step 1 number. Otherwise, do not.
This is the number.
Proof: Clearly, the p-coins can represent a number of our choice, and they are designed so that flipping any coin in a p-coin will change the p-coin, so that 2n coins can show a number from 1 to n. Now, we want to show a number from 1 to 2n, so we have the second-coin head count. Flipping any second coin will change the count's parity, while flipping any first coin will preserve it. This bit of information can decide between x and n+x. Now, from flipping the p-coin we have narrowed the flippable coins to a first coin and a second coin. The head count narrows the flippable coins to either all first coins or all second coins, and clearly now exactly one coin will work.
This statement is false.
Conclusion: "This statement is false" is not a statement.