A vector addition system with states (VASS) is a finite graph \((Q,E)\) with labels in \(\mathbb{Z}^d\) and a designated initial state \(q_0\) and final state \(q_f\). The set of configurations is \(Q \times \mathbb{N}^d\), and a step according to an edge \(e=(p,q)\) labelled with \(v\) consists of moving from the current configuration \(p,x\) to \(q, x+v\), where \(x+v\) has to be in \(\mathbb{N}^d\) again.

The standard definition of computing a function with a VASS is as follows: One of the \(d\) "counters", usually the first, is a designated input counter. Another counter, usually the second, is used for the output. The rest of the counters are called auxiliary counters. A VASS weakly-computes a function f if the following two conditions hold:

- For all \(x \in \mathbb{N}\), there exists some vector \(w' \in \mathbb{N}^{d-2}\) and \(x' \in \mathbb{N}\) such that \(q_0, (x,0,0^{d-2})\) can reach \(q_f, (x', f(x), w')\). I.e. the VASS can reach the final state with the value \(f(x)\) in the output counter, the values left in auxiliary counters are irrelevant.
- For all \(x,x',y' \in \mathbb{N}\) and vectors \(w' \in \mathbb{N}^{d-2}\), if \(q_0, (x,0,0^{d-2})\) can reach \(q_f, (x', y', w')\), then \(y' \leq f(x)\). I.e. if the VASS can reach the final state with a value \(y'\) in the output counter, again irrespective of values left in the auxiliary counters, then \(y' \leq f(x)\).

Despite this being the standard definition, surprisingly little is known about functions weakly-computable by vector addition systems. Essentially the only known properties are the following.

- They are monotone, i.e. if \(x \leq x'\), then \(f(x) \leq f(x')\).
- They are primitive-recursive, in particular computable.
- They grow at least linearly, or intuitively "have to grow steeper over time".

To resolve the current state of no progress in this direction, I suggest the following open problems:

- Is \(2^{\sqrt{x}}\) weakly-computable? It would be quite surprising, since \(\sqrt{x}\) is not.
- Which linear recursive sequences (LRS) are weakly-computable? I conjecture that exactly the monotone LRS are weakly computable (asymptotically at least).