hmm

>>9120522
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

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>>9120522
lmao
42 - 4 + 2 - 1

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

>>9120522
(4+2)^2 + 4 - 1

>>9120731
You used three 2's. You're only allowed two of them.

>>9120522
(39+1) / 4 = 2*4 + 2

((24-4)*2) - 1

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

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>>9121395
Floor(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt( ((4+1)!)! ))))))))+4

I aint using your dirty 2 2's

>>9121465
This raises the question, is it possible to make every number using some digit > 1, sqrt's and factorials?

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>>9121465
You can keep your filthy fours.

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

>>9121778
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) <= F! < (X+1)^(2^n), for some positive integer n.

The "window" for F! to fall in seems to get larger as n gets larger.

((X+1)^(2^n)-X^(2^n))/(X+1)^(2^n) is the portion of numbers that fall into the window relative to the upper bound.

Now I'm not good at limits, but lemme try

Lim n->inf of ((X+1)^(2^n)-X^(2^n))/(X+1)^(2^n)

1-X^(2^n)/(X+1)^(2^n)

1-(X/(X+1))^(2^n)

1-(0)

1

That was like middle school limits but it works out. Its not very rigorous but I think this shows that the window grows to be as arbitrarily close to the entire bound as we need it to be.

Thus we can just give a fucking huge n, get a massive window that some F! will be contained within.

Any issues in logic with my beginner semi proof?

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>>9121840
Explain yourself devil

>>9120522 >>9120601
4(4*2+2)-1
= 4(8+2)-1
= 4*10-1
= 40-1
= 39

>>9120522
(4-2+1)*4^2
You didn't signify what base 39 was in :^)

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>>9121864
It should be in radians, I messed up on a few parenthesis and your calculator doesn't do factorials correctly.

>>9120522
>1, 2, 2, 4 , 4

21+24+4=39
ez
pz

>>9120522
(4+2)2 + (4-1)
(2 * 4!) - (2 * 4 + 1)

>>9121941
>(4+2)2 + (4-1)
meant (4+2)^2 + (4-1)

>>9120522
(2^4)+1+2+4

>>9122002
2^4!=32, 2^4=16=4^2 ;^)

1+2+2+4+4+26

21+4*4+2

>>9120522
4*4*2+(1^2)

>>9120522
(2*4+2)*4-1

>>9120522
44-22-1

>>9124911
Whoops, that was supposed to be 2^2 not 22