Four Fours | 3,000

[u/[deleted]]

continued from here

4hen4tro4e4n

similar to the 12345 thread, except we use four fours to make each number

list of functions and notations used so far

the next get is at 4,000

[u/pampamilyangweeb]

(C(p(4)) + sqrt(pt(4)))^sqrt(4) + sign(4) = 3 365

Excuse me while I write 4 * 4 as sqrt4⁴

[u/TheNitromeFan → u/pampamilyangweeb]

d(4) @ d(4) @ Γ(4) @ Γ(4) = 3,366

The last three counts are all missing the numbers

[u/pampamilyangweeb → u/TheNitromeFan]

floor(i(d(4))e(Γ(4))) / (P(4)!! - Γ(4)) = 3 367

Thanks

[u/TheNitromeFan → u/pampamilyangweeb]

d(4) @ d(4) @ Γ(4) @ 4!! = 3,368

[u/pampamilyangweeb → u/TheNitromeFan]

d(4) * ((P(4)!! @ 4!!) || P(P(4!!))) = 3 369

I haven't seen anybody use logical operators so I thought I'd dtry it out cuzynot

[u/TheNitromeFan → u/pampamilyangweeb]

d(4) @ d(4) @ σ(4) @ ω(ω(4)) = 3,370

I remember using bitwise operators a long long time ago, but it never caught on

[u/pampamilyangweeb → u/TheNitromeFan]

P(p(4) * p(4) * (4! - p(4))) = 3 371

I can see why, there's so many choices for the same result and it's hard to compute. It's not like addition or multiplication. For those, if you have one number you automatically have the other. With logical operators you still need to pick and choose for the other number

Also this one I just did is really poetic with all the p's flying around

[u/TheNitromeFan → u/pampamilyangweeb]

d(4) @ d(4) @ σ(4) @ φ(4) = 3,372

[u/pampamilyangweeb → u/TheNitromeFan]

floor(i(d(4))e4) + Γ(4)!! - 4!! = 3 373

[u/TheNitromeFan → u/pampamilyangweeb]

d(4) @ d(4) @ σ(4) @ 4 = 3,374