Express numbers 1 - 100 using only 1, 2, 3, and 4

Question

The least number that cannot be written using the numbers 1, 2, and 3, each exactly once, and any combination of standard arithmetic operations (including factorials) is 41. What is the least such number if the numbers 1, 2, 3, and 4 are allowed?

Allowed operations are addition, subtraction, multiplication, division, factorials, exponents, square roots, intermediate non-integer results, and any amount of parentheses and brackets. No other digits besides one of each of 1, 2, 3, and 4.

(This is basically what user Bernardo Recamán Santos asked 4 years ago but just without using 0. I have all the numbers except 86 and 93...)

Answer 1

91 is also possible without concatenation

$((3!)!)/(2\times4) + 1$

Answer 2

Here comes $76$:

$76 = 4 \sqrt{\frac {(3!)!}{2}+1}$

If we are allowed double factorials we can do $85$:

$85 = \frac{2^{4!!}-1}{3}$

Answer 3

Following Albert Lang's second solution, if we extend to double factorials, we can get $86$ and $93$

$86 = ((3!)!! - 4 - 1) \times 2 $
$93 = ((3!)!! \times 2) - 4 + 1$

Answer 4

An answer for $99$:

$(3!+4)^2-1$

Regarding $85$, $86$, $93$:

I think these are impossible without concatenation, double factorial (deemed acceptable by OP's comment), or other operations of unknown status. I could not find answers even with computer aid.

With concatenation:

$85 = 43 \times 2 - 1$ (based on hexomino's comment)

$86 = 43 \times 2 \times 1$ (based on hexomino's comment)

$93 = 12 + 3^4$

If arbitrary multifactorials are allowed, we can obtain any positive integer in a rather silly way:

Suppose we want to reach $N$. Let $I(n, m)$ be the itererated factorial function, applying the ordinary factorial $m$ times starting with $n$. Choose $m$ such that $I(5, m) > 2N$. Then $$N = I(4+1, m)!^{(I(5, m) - N)} / I(2+3, m)$$ because $$I(5, m)!^{(I(5, m) - N)} = I(5, m) \times (I(5, m) - (I(5, m) - N)) = I(5, m) \times N$$ Note, the next term in the multifactorial expansion would be $$N - (I(5, m) - N) = 2N - I(5, m) < 0$$