The 'four 1s' problem

Question

There are four 1s that you get to use, and you must use ALL FOUR (which is not a problem because of mulitplying by 1) You may put math signs in between each one however you please. You may even put them together to make 11. I have figured out a lot of them but need help solving some (mostly 13, 17, and 18). Factorials and triangular numbers (represented by T) are allowed. As well as square roots, exponents, gamma, absolute value, etc. However, algebra like (if x=1+1 then 13=x•x•x+x•x+1*1)

1: 1•1•1•1
2: 1+1•1•1
3. 1+1+1•1
4. 1+1+1+1
5. (1/.1)/(1+1)
6. ?
7. ?
8. (1/.1)-1-1
9. (1/.1)-1•1
10. (1/.1)•1•1
11. (1/.1)+1•1
12. (1/.1)+1+1
13. ?
14. (T((T(T(1+1)))-1))-1
15. (T((T(T(1+1)))-1))•1
16. (T((T(T(1+1)))-1))+1
17. ?
18. ?
19. ((1+1)/.1)-1
20. ((1+1)/.1)•1

Edit: I added linebreaks

Answer 1

Why not use the successor function $s(n) = n + 1$? Then the problem is easy? You're attitude that there is a finite set of functions that make sense is troublesome ("etc"). If you gave a finite list of functions, then it becomes a well defined problem.

Maybe not an interesting problem, but at least well defined.

Answer 2

$$6 = (1+1)T(1+1)$$

$$7 = T(T(1+1))+1\cdot 1$$

$$13 = 11+1+1$$

$$17 = 11+T(T(1+1))$$

$$18 = T(1+1)T(T(1+1))$$

$$21 = 11+\frac{1}{.1}$$

$$22 = (11)(1+1)$$

$$23 = (T(1+1)+1)!-1$$

$$24 = (T(1+1)+1)!\cdot 1$$

$$25 = (T(1+1)+1)!+ 1$$

These definitely aren't all of the solutions; I encourage you to try to come up with more yourself!

Answer 3

In the same vein as YBerman's answer:

Let $R$ denote the $k$-fold iteration of $\sqrt \cdot$ applied to $2=1+1$, so that $R = 2^{1/2^k}$. Then $-\log (\log R)\big/\log 2 = k - (\log\log 2)/\log 2)\approx k + .5288,$ so finally, $\lfloor -\log (\log R)\big/\log 2 \rfloor = k$. As in $\lfloor- \log (\log \sqrt{\sqrt{\sqrt{1+1}}})\big/\log(1+1)\rfloor = 3,$ and so on.