Count in balanced ternary

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You know, the numeral system some computers, somewhere, once used. There are three digits, standing for the values -1, 0 and 1, and they?re weighted by powers of 3. We could make ↓ the digit for -1, for example, and ↑ the digit for 1. This way, 46 = 1×3⁴-1×3³-1×3²+0×3+1 is represented as ↑↓↓0↑, and -8 = -1×3²+0×3+1, as ↓0↑. Positive numbers start with ↑, negative numbers, with ↓, and zero stays 0.

Just as in the more usual positional systems, you count by increasing the value of the rightmost digit as long as possible, and then you jump to the lowest-valued digit in that place, and increase the one on its left: ↑, ↑↓, ↑0, ↑↑, ↑↓↓, ↑↓0, ↑↓↑, ↑0↓, ↑00, ↑0↑, ↑↑↓, ↑↑0, ↑↑↑, ↑↓↓↓, ↑↓↓0, ?

So ?

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Lord Almighty! My math books and my own notes files are buried somewhere. Do I have to dig them up? -- Now that you showed the way, I can of course copy-paste.

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*backs out of thread slowly*

Yeah, very well!

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This is easier than I thought.

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*still backing away*

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(9-3-1 = 5)