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 $\sqrt{9}! - \frac{9}{9} = 5$
 A: For $n=9$ and $k=9$ here is a solution:
$1=\left(9+\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+\left(9-99\right)\right)\right)}\right)}\right)$
$2=\frac{9}{\left(9+\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{99}\right)}\right)}\right)}$
$3=\left(9-\left(9+\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)\right)\right)$
$4=\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+\left(9+\left(9+9\right)\right)\right)\right)}\right)}\right)$
$5=\left(9+\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+\left(9+99\right)\right)}\right)}\right)\right)$
$6=\left(9+\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)}\right)$
$7=\left(9+\left(9+\frac{9}{\left(9 \times \frac{9}{\left(9-\left(9+99\right)\right)}\right)}\right)\right)$
$8=\left(9+\left(9 \times \frac{9}{\left(9+\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)\right)$
$9=\left(9 \times \left(9 \times \frac{9}{\left(9-\left(9+\left(9+\left(9-99\right)\right)\right)\right)}\right)\right)$
$10=\left(9-\left(9 \times \frac{9}{\left(9+\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)\right)$
$11=\frac{9}{\left(9 \times \frac{9}{\left(9-\left(9+\left(9-\left(9+99\right)\right)\right)\right)}\right)}$
$12=\left(9-\frac{9}{\left(9-\frac{9}{\left(9 \times \frac{9}{\left(9+99\right)}\right)}\right)}\right)$
A: For $n=4$ and $k=5$ here is a solution:
$\frac{4}{\left(4+\left(4 \times \left(4-4\right)\right)\right)}=1$
$\left(4-\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=2$
$\left(4+\frac{4}{\left(4-\left(4+4\right)\right)}\right)=3$
$\left(4+\left(4+\left(4-\left(4+4\right)\right)\right)\right)=4$
$\left(4-\frac{4}{\left(4-\left(4+4\right)\right)}\right)=5$
$\left(4+\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=6$
$\frac{4}{\left(4 \times \frac{4}{\left(4+4!\right)}\right)}=7$
$\left(4 \times \left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=8$
$\left(4-\left(\frac{4}{4}-\frac{4!}{4}\right)\right)=9$
$\left(4+\frac{4}{\left(4 \times \frac{4}{4!}\right)}\right)=10$
$\frac{4}{\left(4 \times \frac{4}{44}\right)}=11$
$\left(4-\left(4-\left(4+\left(4+4\right)\right)\right)\right)=12$
A: For $n=3$ and $k = 3$.
$1 = 3^{3-3}$
$2 = 3-\frac{3}{3}$
$3 = 3+3-3$
$4 = 3+\frac{3}{3}$
$5 = 3!-\frac{3}{3}$
$6 = 3*3-3$
$7 = 3!+\frac{3}{3}$
$8 = \pi(3)*\pi(3)*\pi(3)$
$9 = 3+3+3$
$10 = 3!+\pi(3)+\pi(3)$
$11 = 3!+3+\pi(3)$
$12 = 3*3+3$
A: Now with $n = 5$ and $k = 5$.
With $n = 5$ and $k = 5$ (missing a $9$ for now but I'll come back to it later).
$\dfrac{55}{5}-5-5=1$
$\dfrac{5+5}{5}-5+5=2$
$\dfrac{5+5}{5}+\frac{5}{5}=3$
$\dfrac{5+5+5+5}{5}=4$
$5 - 5 + 5 - 5 + 5 = 5$
$5 + \dfrac{5}{5} - 5 + 5 = 6$
$5 + \dfrac{5}{5}+\dfrac{5}{5} = 7$
$5 + 5 - \dfrac{5+5}{5} = 8$
$5 + \dfrac{5(5) - 5}{5}=9$
$\dfrac{55}{5} - \dfrac{5}{5} = 10$
$\dfrac{55}{5} - 5 + 5 = 11$
$\dfrac{5+5}{5} + 5 + 5 = 12$
Thanks to tzador for $9$.
A: For $n=2$ and $k=12$ here is a solution:
$1=\left(2 \times \left(2 \times \left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$
$2=\left(2+\left(2 \times \left(2+\left(2+\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$
$3=\left(2 \times \left(2+\left(2 \times \frac{2}{\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$
$4=\frac{2}{\left(2 \times \left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)}$
$5=\frac{2}{\left(2 \times \frac{2}{\left(2-\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$
$6=\left(2-\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$
$7=\frac{2}{\left(2 \times \frac{2}{\left(2-\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$
$8=\left(2-\left(2+\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$
$9=\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$
$10=\left(2+\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$
$11=\left(2+\left(2+\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)}\right)}\right)\right)$
$12=\left(2-\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$
A: For $n=-1$ and $k=8$ here is a solution:
$1=\left(-1-\left(-1 \times \left(-1+\left(-1-\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$
$2=\left(-1+\left(-1+\left(-1-\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$
$3=\left(-1-\left(-1 \times \left(-1-\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$
$4=\left(-1+\left(-1-\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$
$5=\left(-1+\left(-1 \times \left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$
$6=\left(-1-\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$
$7=\left(-1 \times \left(-1+\left(-1+\left(-1+\left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)\right)$
$8=\left(-1 \times \left(-1+\left(-1-\left(\left(-1+-1\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)$
$9=\left(-1 \times \left(-1-\left(\left(-1+-1\right) \times \left(-1+\left(-1+\left(-1+-1\right)\right)\right)\right)\right)\right)$
$10=\left(-1 \times \left(-1-\left(\left(-1+\left(-1+-1\right)\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)$
$11=\left(-1-\left(\left(-1+-1\right) \times \left(\left(-1+-1\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)$
$12=\left(-1 \times \left(\left(-1+-1\right) \times \left(\left(-1+-1\right) \times \left(-1+\left(-1+-1\right)\right)\right)\right)\right)$
A: Making numbers out of 4 fours is a common problem:
$$1=\frac {44}{44}$$
$$2=\frac {4\cdot 4}{4+4}$$
$$3=\frac{4+4+4}{4}$$
$$4=\frac{4-4}{4}+4$$
$$5=\sqrt{4!+\frac{\sqrt 4+\sqrt 4} 4}$$
$$6=\sqrt{\frac{4!\cdot 4-4!}{\sqrt 4}}$$
$$7=\sqrt{4!\sqrt 4+\frac 4 4}$$
$$8=\sqrt{\frac{4^4}{\sqrt4+\sqrt 4}}$$
$$9=(4-\frac 4 4)^{\sqrt 4}$$
$$10=\frac{4!} 4 - (4-\sqrt 4)$$
$$11=\frac{4!}{\sqrt 4}-\frac 4 4$$
$$12=\sqrt{\frac{4!4!}{\sqrt 4+\sqrt 4}}$$
You should clarify what operations you want. If you allow for any kind of rounding function, factorials and logs you can almost certainly do it with one of any number (though the resulting expressions may not fit on a clock).
A: solution for n = 1, k = 12:
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1 = 2
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1 = 3
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1 = 4
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1 = 5
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1 = 6
$$
$$
1 \times 1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1 = 7
$$
$$
1 \times 1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1 = 8
$$
$$
1 \times 1 \times 1 \times 1+1+1+1+1+1+1+1+1 = 9
$$
$$
1 \times 1 \times 1+1+1+1+1+1+1+1+1+1 = 10
$$
$$
1 \times 1+1+1+1+1+1+1+1+1+1+1 = 11
$$
$$
1+1+1+1+1+1+1+1+1+1+1+1 = 12
$$
A: Seems like $2$ would do it:
$$
1: 2^2 - 2 - 2/2
$$
$$
2:2^2 - 2^2 + 2
$$
$$
3: 2 + 22/22
$$
$$
4: 2^{2^2}/2^2
$$
$$
5: 2^2 - 2/2 + 2
$$
$$
6: 2^2 + 2 - 2 + 2
$$
$$
7:2^2 + 2 + 2/2
$$
$$
8:2^{2}(2) + 2 - 2
$$
$$
9:2^2(2) + 2/2
$$
$$
10:22/2 - 2/2
$$
$$
11 : (2^2)!/2 - 2/2
$$
$$
12: 2^{2^2} - 2^2
$$
That should do it. Thanks to Phira for $10$ and $11$ and Peter for $3$.
A: Six ones.  For ease of reading, I write $n$ for the sum of $n$ 1s, so for example I mean
$1 = ((1+1+1)-(1+1)) \times 1$.
Here are the expressions:
$1 = (3-2) \times 1$
$2 = 4-2$
$3 = (4-1) \times 1$
$4 = 5-1$
$5 = 5 \times 1$
$6 = 6$
$7 = (3 \times 2) + 1$
$8 = 4 \times 2$
$9 = 3 \times 3$
$10 = (3! - 1) \times 2$
$11 = (3! \times 2) - 1$
$12 = 3! \times 2 \times 1$
Any larger number of 1s is possible (just multiply these expressions by 1 as many times as necessary); I don't think five ones is possible.
