I recently discovered the Suanpan, which quickly became my favorite abacus design. Like the more common Soroban (Japanese abacus), the Suanpan has an upper rack and a lower one. The Soroban has 4 beads on the lower, and one bead on the upper. The Suanpan has 5 beads on the lower, and two beads on the upper. This is to allow for calculations involving base 16, like the classic Chinese weight system.
I was intrigued by this versatility and I decided to see if bases other than 10 and 16 could be represented on a Suanpan. I discovered that bases 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, and 18 could be represented with relative ease. But 14 and 17 seemingly had no solution.
Eventually I found an okay solution for 14, but I'm at a loss for 17. So I task you to help solve this mystery.
Rules
These rules are stated in reference to one column.
The solution must be for a Suanpan, meaning it works for an abacus with 5 lower beads, and 2 upper beads.
The lower beads are of value one (2 lower beads=2), but the top beads can be of any value.
You may choose to not use any number of the beads. For example, Suanpans mimic Sorobans if you forget 1 lower bead and 1 upper bead. Suanpans can also perform binary math if you use 1 of the lower beads and nothing else.
The value of an upper bead must be within 1 of the amount of lower beads used (this includes the same value as beads used). For example, in Soroban mode, the upper bead is worth 5, 1 more than the amount of beads used in the lower rack. Additionally, in hexadecimal mode, the upper beads are worth 5, which is the same as the amount of beads used in the lower rack.
The layout/counting rules must not change when counting to the maximum value for the base (16). In another word: overflow must be consistent. The rule 3 lower beads not overflowing, but if 1 upper bead is counted it overflows would not be allowed.
These rules are intended to have the solutions mimic the traditional (and intuitive) use of the Suanpan.
I'm happy to provide any information on how my solutions work, to help you find the solution for base 17.
For base $14$, each lower bead is $1$ and each higher bead is $4$. That allows you to represent each number from $0$ to $13$ in a column, which is all you need. I don't see a solution for base $17$ without relaxing your rules a bit. The closest I could come is $1$ for each lower bead and $6$ for each higher bead as you do for base $18$, then don't use the combination with all beads up. This doesn't seem too bad to me. You normally have one rule for carrying from the bottom to the top and a different one for carrying from one column to the next. You still need those, but the number of lower beads that need to be up for a carry to the next column is not all of them.