Lowest Value that Breaks the Machine

One of the beautiful properties of $2$

Since it is not told about the signatures of $a$ and $b$ , we assume that they are positive integers

Let the input be $n$ . Then we have the following possibilities :

(i) $n$ is odd $\implies$

In this case either both of $a$ and $b$ or any one of them must be odd. Let $n=2m+1$ where $m$ is a non-negative integer. Then an obvious combination of $a$ and $b$ is $a=1,b=m$ .

(ii) $n$ is an odd multiple of 2 $\implies$

In this case both $a$ and $b$ must be even. Then we continue to find $a$ and $b$ until we get $\frac{n}{2^m}$ odd, where $m$ is sum positive integer. Then we proceed as in case (i).

(iii) $n$ is of the form $2^m$ , where $m$ is some positive integer $\implies$

In this case we repeat the procedure in case (ii) until we reach an odd input. But this terminal value of the input is $1$ , which can never be expressed in the form $ab+a+b$ for any positive $a, b$ ( zero is excluded ).

In this problem, the number in the given range is $2^4=\boxed {16}$ .

Actually, the answer will be $2^m$ when the given range is $[2^{m-1}+1,2^{m+1}-1]$ .

In this problem, $16$ is the only such number between $9$ and $31$ inclusive.