A closed solution of F. G. Frobenius Coin Problem for $n=3$??

Question

Whether?: I chose to try and find a closed equation for $n=3$.

Whither?: I hope that my attempt to find a formula is well-accepted.

Whence?: I read several article on the internet about the Coin Problem.

When?: The problem became a curiosity in 1888.

Who?: F.G.Frobenius tried to find an expression that J.J. Sylvester later proved.

What?:The ultimate goal of Frobenius Coin Problem for $2$ variables is to find the largest amount that cannot be represented using only two denominations using a formula. e.g. $5$, $7$; $g_2(a=5,b=7)=4\cdot6−1=23$, and $24$ through $28$ can be produced and beyond.

For $n=3$, I'm trying to achieve a closed formula, $g_3(a,b,c)$ e.g. $a\geq3!,b>a,c>a$ and $b$, and $gcd(a,b,c)=1$.

$g_3(a=6,b=9,c=20)=$

$+1[(a−2)(b−2)(c−2)]+8$

$−2[(a−1)(b−1)]-1$

$-2[(a−1)(c−1)]-1$

$−2[(b−1)(c−1)]-1$

$+3[(a+b+c)+1]=$

$504+8−81−191−305+108=43$.

Can someone elaborate on my observation? Thanks, Bill Bouris

Answer 1

The case $(a,b,c)=(6,9,20)$ is a well-known example, related to Chicken McNuggets from McDonald's, which were sold in these quantities, see here from $2003$. Indeed, the answer is $43$.

Answer 2

According to your formula, $g_3(2,3,4) = 13$. But the correct answer is obviously $1$.

Also, your formula says $g_3(2,3,9) = -4$. This does not even make sense. It is impossible for the answer to be negative.