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Mathematical proof:

Mathematical proof (for the balance only):

Balance + weighing scale: TBA

For $n=5$:

1. Weigh $A$ and $B$ against $C$ and $D$.

- If the weights are equal (+), either one side is all-normal and another consists of heavy and light fakes, or both have one identical fake each. Either way, $E$ is normal, so we can use it to find out the fakes in our next 3 weighings. 4 moves.

- If the weights are different (-), let's say $A+B>C+D$.

2. Weigh $A$ against $B$. If the result is balanced (-+), $A$ and $B$ are both normal or heavy. Then...

3. Weigh $A$ against $E$. A balanced result (-++) means the fakes are $C$ and $D$. 3 moves. If $A<E$ (-+-), weigh $C$ against $D$, and the fakes are $E$ and the lighter one from this measurement. 4 moves. If $A>E$ (-+-), weigh $C$ against $D$. If it's balanced (-+-+), the fakes are $A$ and $B$, and otherwise (-+--) $E$ and the lighter one from this measurement. 4 moves.

2b. The second measurement isn't balanced either (--). Weigh $C$ against $D$. If the results are balanced (--+), they're both real. Then compare the lighter one out of $A$ and $B$ to $C$ or $D$. If the result is balanced (--++), the fakes are the heavier other and $E$, otherwise (--+-) both $A$ and $B$. If the results from $C$ vs. $D$ aren't balanced (---), 2 more weighings are needed to differentiate between $hn-nl-n$, $hn-h^-n-n$ and $nl-nl^--n$. 5 moves, fitting for the formula.

($h$: heavy, $n$: normal, $l$: light, minus superscript: same state but lighter than the other coin with the same letter).

For $c=5$:

1. Weigh $A$ and $B$ against $C$ and $D$.

- If the weights are equal (+), either one side is all-normal and another consists of heavy and light fakes, or both have one identical fake each. Either way, $E$ is normal, so we can use it to find out the fakes in our next 3 weighings. 4 moves.

- If the weights are different (-), let's say $A+B>C+D$.

2. Weigh $A$ against $B$. If the result is balanced (-+), $A$ and $B$ are both normal or heavy. Then...

3. Weigh $A$ against $E$. A balanced result (-++) means the fakes are $C$ and $D$. 3 moves. If $A<E$ (-+-), weigh $C$ against $D$, and the fakes are $E$ and the lighter one from this measurement. 4 moves. If $A>E$ (-+-), weigh $C$ against $D$. If it's balanced (-+-+), the fakes are $A$ and $B$, and otherwise (-+--) $E$ and the lighter one from this measurement. 4 moves.

2b. The second measurement isn't balanced either (--). Weigh $C$ against $D$. If the results are balanced (--+), they're both real. Then compare the lighter one out of $A$ and $B$ to $C$ or $D$. If the result is balanced (--++), the fakes are the heavier other and $E$, otherwise (--+-) both $A$ and $B$. If the results from $C$ vs. $D$ aren't balanced (---), 2 more weighings are needed to differentiate between $hn-nl-n$, $hn-h^-n-n$ and $nl-nl^--n$. 5 moves, fitting for the formula. If 5 is really the solution, that means $6c(c-1)$ is right.

($h$: heavy, $n$: normal, $l$: light, minus superscript: same state but lighter than the other coin with the same letter).

For $n=5$:

1. Weigh $A$ and $B$ against $C$ and $D$.

- If the weights are equal (+), either one side is all-normal and another consists of heavy and light fakes, or both have one identical fake each. Either way, $E$ is normal, so we can use it to find out the fakes in our next 3 weighings. 4 moves.

- If the weights are different (-), let's say $A+B>C+D$.

2. Weigh $A$ against $B$. If the result is balanced (-+), $A$ and $B$ are both normal or heavy. Then...

3. Weigh $A$ against $E$. A balanced result (-++) means the fakes are $C$ and $D$. 3 moves. If $A<E$ (-+-), weigh $C$ against $D$, and the fakes are $E$ and the lighter one from this measurement. 4 moves. If $A>E$ (-+-), weigh $C$ against $D$. If it's balanced (-+-+), the fakes are $A$ and $B$, and otherwise (-+--) $E$ and the lighter one from this measurement. 4 moves.

2b. The second measurement isn't balanced either (--). Weigh $C$ against $D$. If the results are balanced (--+), they're both real. Then compare the lighter one out of $A$ and $B$ to $C$ or $D$. If the result is balanced (--++), the fakes are the heavier other and $E$, otherwise (--+-) both $A$ and $B$. If the results from $C$ vs. $D$ aren't balanced (---), 2 more weighings are needed to differentiate between $hn-nl$, $hn-h^-n$ and $nl-nl^-$. 5 moves, fitting for the formula.

($h$: heavy, $n$: normal, $l$: light, minus superscript: same state but lighter than the other coin with the same letter).

For $n=5$:

1. Weigh $A$ and $B$ against $C$ and $D$.

- If the weights are equal (+), either one side is all-normal and another consists of heavy and light fakes, or both have one identical fake each. Either way, $E$ is normal, so we can use it to find out the fakes in our next 3 weighings. 4 moves.

- If the weights are different (-), let's say $A+B>C+D$.

2. Weigh $A$ against $B$. If the result is balanced (-+), $A$ and $B$ are both normal or heavy. Then...

3. Weigh $A$ against $E$. A balanced result (-++) means the fakes are $C$ and $D$. 3 moves. If $A<E$ (-+-), weigh $C$ against $D$, and the fakes are $E$ and the lighter one from this measurement. 4 moves. If $A>E$ (-+-), weigh $C$ against $D$. If it's balanced (-+-+), the fakes are $A$ and $B$, and otherwise (-+--) $E$ and the lighter one from this measurement. 4 moves.

2b. The second measurement isn't balanced either (--). Weigh $C$ against $D$. If the results are balanced (--+), they're both real. Then compare the lighter one out of $A$ and $B$ to $C$ or $D$. If the result is balanced (--++), the fakes are the heavier other and $E$, otherwise (--+-) both $A$ and $B$. If the results from $C$ vs. $D$ aren't balanced (---), 2 more weighings are needed to differentiate between $hn-nl-n$, $hn-h^-n-n$ and $nl-nl^--n$. 5 moves, fitting for the formula.

($h$: heavy, $n$: normal, $l$: light, minus superscript: same state but lighter than the other coin with the same letter).