Are you kidding me?

I threw this into the wiki on the Frobenius number .

The easiest way to compute this is to note that the largest such number is $g(13,31) = 13\cdot 31 -13-31 = 359,$ and then to use the fact (proved in the wiki) that any two nonnegative integers that sum to $359$ have the property that exactly one of them is possible to buy. To maximize the impossible one, minimize the possible one. $0$ is obviously the smallest (pairing with $359$ ) and $13$ is obviously the second smallest, which pairs with $346$ . In general, the answer will be $ab-a-b-\text{min}(a,b),$ for coprime integers $a,b$ .

The problem is same as finding the second largest non-negative integer $a$ such that the equation $31x + 13y = a$ , has no solutions when $x$ and $y$ are non-negative integers.

After solving the equation when $x$ and $y$ are integers we get:

$x = -5a + 13k$ , $y = 12a - 31k$ , where $k$ is an integer

For fixed $a$ , the equation above has a solution for non-negative integers $x, y$ iff there exist an integer $k$ for which $\dfrac{5a}{13} \leq k \leq \dfrac{12a}{31}$ .

$k$ doesn't exist iff $\dfrac{5a}{13}$ and $\dfrac{12a}{31}$ have the same integer part, but aren't integers.
This is the same as saying that $k$ doesn't exist iff there exists positive integers $n < 13,m < 31$ such that $\dfrac{5a-n}{13} = \dfrac{12a-m}{31} \implies a = 13m - 31n$ .

Now we can generate ALL values of $a$ which cannot be expressed in form $31x + 13y$ where $x,y$ are positive integers.
We maximize the value of $a$ by keeping $m$ as large as possible and $n$ as small as possible, and thus get $a = 13 \cdot 30 - 31 \cdot 1= 359$ .

Now we get the second smallest $a$ by lowering the value of the largest $a$ by the smallest possible amount.

Since lowering the value of $m$ by $1$ is better than adding $1$ to the value of $n$ and since we cannot add more to the value of $m$ to compensate with larger $n$ we get $a = 13 \cdot 29 - 31 \cdot 1= \boxed{346}$