[Legal counting forum game] Counting with four fours

I don't know if it's been long enough for a tip post, but oh well...
We are nearing 120. This is a very easy number to make with just two 4s - (4 + gamma (sqrt (4)))! .
Therefore, numbers above and below it can be made with the remaining two 4s.

I had/have a couple in mind but I am the latest post :3

Orrrr....

121 = (4 + gamma (sqrt (4)))! + 4 / 4

124 = (4 + gamma (sqrt (4)))! + sqrt (4) * sqrt (4)
124 = 4! * 4 + 4! + 4
124 = (4 + sqrt (4)))! / gamma (4) + 4

Another 124 that I personally liked.

(4 + gamma (sqrt (4)))! + 4 + sqrt (4) = 126

127 = ((4^4)-sqrt(4))/sqrt(4)

128 = (4^4) / (4 - sqrt(4))

(4^4) / sqrt (4) + gamma (sqrt (4)) = 129

131 = (((4!!)!! + 4!)/4!!) + (4!!)!!!

(4 ^ 4 + 4!!) / sqrt (4) = 132

Well, there's been a bit of a gap, and I can't work out 133 myself. According to google, it's 133 = (4!! - 4/4) * int (sqrt (4!!)!!). Looking ahead, a lot of them rely on rounding numbers down, so it's gonna get a lot harder and more necessary to bend the rules a bit

It is too late for me to find out 134 (I really need some sleep), but From now on Will rounding to a number be allowed, It's not preferd, but allowed.

((4!!)!!! * sqrt4) - 4! - sqrt4 = 134 (i think?)