So, several days ago, I was introduced to the "Four game" again. The object of the game is to use four 4's to produce as many integers as you can. You are allowed to use addition, subtraction, multiplication, division, square root, and factorial functions in addition to concatenation (44), use of a decimal point (.4), and possibly the floor and ceiling functions. With these abilities, one can get all integers between 1 and 156 and my friends and I are not even done yet. What is particularly helpful is being able to define a large range of integers with just one four and the single-input functions. For instance...
$1 = \lceil .4 \rceil$
$2 = \sqrt{4}$
$3 = \left\lceil \sqrt{\sqrt{4!}} \;\;\right\rceil$
$4 = 4$
$5 = \left\lceil \sqrt{4!} \right\rceil$
$6 = \left\lceil \sqrt{\sqrt{4!}} \;\;\right\rceil!$
$7 = \left\lceil \displaystyle\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{ \left\lfloor \sqrt{ \;\;\left\lfloor \sqrt{\sqrt{ \left\lceil \sqrt{4!} \right\rceil ! }} \right\rfloor !!\;\; } \right\rfloor ! }}}}} \;\;\right\rceil$
More legibly, though less impressively, set $$T=\sqrt{ \left\lceil \sqrt{4!} \right\rceil ! }\qquad S=\sqrt{ \;\left\lfloor \sqrt{T\; } \right\rfloor !!}\qquad R=\sqrt{\sqrt{\sqrt{ \left\lfloor S \right\rfloor ! }}}\qquad;$$ then $7=\left\lceil \displaystyle\sqrt{\sqrt{R}} \;\;\right\rceil$.
$8 = \left\lfloor \sqrt{\sqrt{ 7! }} \right\rfloor$
$9 = \left\lceil \sqrt{\sqrt{ 7! }} \;\;\right\rceil$
$10 = \left\lfloor \sqrt{ \left\lceil \sqrt{ 4!} \right\rceil ! } \right\rfloor$
$11 = \left\lceil \sqrt{ \left\lceil \sqrt{ 4!} \right\rceil ! } \right\rceil$
...
I haven't looked for one for 12 yet, but I doubt it would be hard to find it. Thus, my question is:
Is it possible to express every integer using only a single four and any number of floor, ceiling, square root, and factorial functions?