How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.

I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem: 
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok. 
If you're answering with an example then use one pair per answer.
I just want to see that clock with another pair of numbers :)
Notes for the current clock:
1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.
5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$

solution for n = 1, k = 12:
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1 = 2
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1 = 3
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1 = 4
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1 = 5
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1 = 6
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1 = 7
$$
$$
1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1 = 8
$$
$$
1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1+1 = 9
$$
$$
1 \times 1 \times 1+1+1+1+1+1+1+1+1+1 = 10
$$
$$
1 \times 1+1+1+1+1+1+1+1+1+1+1 = 11
$$
$$
1+1+1+1+1+1+1+1+1+1+1+1 = 12
$$
