User blog comment:Spitemaster/Golfing Large Numbers/@comment-26693597-20190425201324

I can't get mathJaX to work, sorry for the eyesore. Here's it with < for \langle, > for \rangle, e for \epsilon, and w for \omega:

Since seeing this SE question on golfing a number bigger than Loader's number, I've been pondering how one might do so. To that end, I've developed an array notation that I can evaluate in under 512 bytes that gets as large as possible as quickly as possible. To evaluate, do these steps in order (whenever 1 is available, do 1, then 2, and so on). "..." refers to the rest of the array and can be empty: 1    <...,a,0,...>[b,...] --> <...,a-1,b,...>[b,...] 2    <...,a>[b] --> <...,a-1>[(b,...,b)b] 3    <0>[...] --> [...] 4    (...,a,0,...)b --> (...a-1,b,...)b 5    [...,(...,a)b] --> [...,(...,a-1)b,...,(...,a-1)b] 6    (0)a --> a 7    [...,1] --> [...] 8    [...,1,a,...] --> [...,a,a-1,...] 9    [a,b,c,...,d] --> [[a,b-1,c,...,d],[a,b,c-1,...,d],...,d-1] 10    [a,b] --> [a+b] 11    [a] --> a

A note on rule 9: in particular, <1>[(2)3,(4)5,6] --> <1>[(2)<1>[(2)3,(4)4,6],(4)<1>[(2)3,(4)5,5],5]

We calculate the value of [(1)x]:

[x,x] = x+x = f_1(x) [x,x,2] = [[x,x-1,2],[x,x,1],1] = [x,x-1,2] + x+x > x*x+x*(x-1)/2 > x^2 = f_2(x) [x,x,3] = [[x,x-1,3],[x,x,2],2] > [[x,x-1,3],x^2,2] > x^3 ... [x,x,x] > x^x = f_3(x) This can be extended, and we find that [(1)x] is about f_{2x-1}(x), roughly f_w(x).

We next calculate the value of [(2)x]: [(1)x,x] = [(1)2x] = f_{4x-1}(2x) [(1)x,x,2] = [(1)[(1)x,x-1,2],[(1)x,x,1],1] = [(1)...[(1)x,1,2]...],f_{4x-1}(2x)] > f_{2x-1}^x(x) In fact, if g(y) = f_{2y-1}(y), then [(1)x,x,2] > g^x(x). This has power roughly f_{w2}(x)

I conjecture, though I cannot prove it, that [(1)x,x,x] > g^{f_3(x)}(x), and [(1)x,x,...,x] > g^{f_{2x-1}(x)}(x). This has power roughly f_{w3}(x) This is still not close to [(2)x], which is [(1)x,(1)x,...,(1)x]. We continue: [(1)x,(1)x] = [(1)x,x,x,...,x] = g^{f_{2x-1}(x)}(x) [(1)x,(1)x,2] = [(1)[(1)x,(1)x-1,2],g^{f_{2x-1}}(x),1] We can see the pattern here, and predict that [(2)x] is roughly equivalent to f_{w^2}(x). If so, then further conjecture puts [(x)x] at f_{w^w}(x). [(1,0)x]=[(x)x], and [(1,1)x]=[(x+1)x] is only slightly larger. But when we get to [(1,2)x], we have [(1,1)x,(1,1)x,...,(1,1)x], meaning that it takes quite a few steps before rule 4 is applied more than once. I'm fairly certain that [(1,2)x] is at least f_{e_0}(x). I really am ill-equipped to judge. Further conjecture:

[(1,x)x] ~ f_{e_w} [(2,x)x] ~ f_{e_{e_w}} [(x,x)x] ~ f_{\sigma_0}

I don't know where to go from here, except to note that <9,9,...,9>[9] (with 9e9999 9s, the largest I can currently do with 512 bytes) is much larger still. Any thoughts?