Four 4s

[James Hall]

Also, is my way of making 3 from two 4's by using infinitely many square roots (posted earlier) legal?

Normally you're only allowed a finite number of functions.

Kai, the gamma function, as far as integers are concerned, is defined thus:
Γ(n) = (n-1)!

[Neil Zussman]

Presumably Kai means we can use double factorial. Thus 4!!=8 and ((4!!)!!)/4=96.
This means 3=4!/4!!, i.e. 3 can be made using just two 4's. Thus 51, 45, 93 and 99 can now be made (see my previous post). I'm sure lots more numbers can be made as well, although interest in the puzzle seems to have waned.

[Robert Baxter]

33:

(4*4*sqrt 4)+sqrt 4
4!+(4!/4)+4
sqrt(4^4)*sqrt 4+sqrt 4

Edit: Told you I was an

[Kai Laddiman]

33:

(4*4*sqrt 4)+sqrt 4
4!+(4!/4)+4
sqrt(4^4)*sqrt 4+sqrt 4

Edit: Told you I was an

There's only 3 there, not 33. Typo perhaps?

[Robert Baxter]

33:

(4*4*sqrt 4)+sqrt 4
*4!+(4!/4)+4
sqrt(4^4)*sqrt 4+sqrt 4

Edit: Told you I was an

There's only 3 there, not 33. Typo perhaps?

Does this work?