Hmm

You have the numbers 1, 2, 2, 4 , 4

Make the number 39. You need to use all of them once.

2(4^2 + 4) -1

[eqn]4\,\star\,1\,-\,2[/eqn]
Where [math]a\,\star\,b\,=\,10\,a\,+\,b[/math].

g(1,2,2,4,4) = 39

4(4*2+2)-1

lmao
42 - 4 + 2 - 1

4!+4^2-1, I aint using your dirty 2nd 2.

(4+2)^2 + 4 - 1

You used three 2's. You're only allowed two of them.

(39+1) / 4 = 2*4 + 2

((24-4)*2) - 1

(4!+sqrt(4))*(1+2)/2

Floor(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt( ((4+1)!)! ))))))))+4

I aint using your dirty 2 2's

This raises the question, is it possible to make every number using some digit > 1, sqrt's and factorials?

You can keep your filthy fours.

floor(√(√(√(√(√(√(sinh(√(sinh(sinh(sinh(cosh(sinh(1))!)!)))))!)))))))

I think yes - given a target integer X, you want to find a factorial start F so that X = floor(sqrt(......sqrt(F!)))

Another way of saying this is that we need an F such that
X^(2^n)

Explain yourself devil

4(4*2+2)-1
= 4(8+2)-1
= 4*10-1
= 40-1
= 39

(4-2+1)*4^2
You didn't signify what base 39 was in :^)

It should be in radians, I messed up on a few parenthesis and your calculator doesn't do factorials correctly.

>1, 2, 2, 4 , 4

21+24+4=39
ez
pz

(4+2)2 + (4-1)
(2 * 4!) - (2 * 4 + 1)

>(4+2)2 + (4-1)
meant (4+2)^2 + (4-1)

(2^4)+1+2+4

2^4!=32, 2^4=16=4^2 ;^)

1+2+2+4+4+26

21+4*4+2

4*4*2+(1^2)

(2*4+2)*4-1

44-22-1

Whoops, that was supposed to be 2^2 not 22