FANDOM


  • \(f(n)=x\), when \((10\uparrow\uparrow n)^{10\uparrow\uparrow n}=(10\uparrow)^{n+2}x\)
  • \(f(1,1,\square)=f(54,\square)\)
  • \(f(1,\square,n)=f(\frac{1}{f(1,\square,n-1)},\square)\), here \(\frac{1}{f(1,\square,n-1)}\) might not be integer, so when substituting, round it off (and so forth).
  • \(f(\blacksquare,m,1,\square)=f(\blacksquare,m-1,54,\square)\)
  • \(f(\blacksquare,m,\square,n)=f(\blacksquare,m-1,\frac{1}{f(\blacksquare,m,\square,n-1)},\square)\)

when,

  • \(\square\): vector of 1, with the length larger than or equal to 0
  • \(\blacksquare\): vector of integers larger than or equal to 1, with the length larger than or equal to 0
  • \(m>1\), and \(n>1\)

f(n) = triangle(10^^n)

f(1,1) = f(54) = triangle(10^^54)

f(1,n+1) = f(1/f(1,n)) = f(1/...f(1,1)...) (n+1 nested)

f(n+1,1) = f(n,54)

f(n+1,m+1) = f(n,1/f(n+1,m))

f(1,1,1) = f(54,1) = f(53,54) = f(52,1/f(53,53))

f(1,1,n+1) = f(1/f(1,1,n),1)

f(1,n+1,1) = f(1,n,54)

f(1,n+1,m+1) f(1,n,1/f(1,n,m))

f(n+1,1,1) = f(n,54,1)

f(n+1,1,m+1) = f(n,1/f(n+1,1,m),1)

f(n,m+1,1) = f(n,m,54)

f(x,y+1,z+1) = f(x,y,1/f(x,y+1,z))

f(1,1,1,1) = f(54,1,1)

Oe(n) = f(1,1...1,1) (54 1's)

Oe = Oe(54)

FGH

Let f_1(n) = 2*n.

f(n) ~ f_3(n) ~ f_2(f_3(n)) = f_2^{n+1}(n)

f(1,1) ~ f_2(f_3(54))

f(1,2) ~ f_2(f_3(f_2(f_3(54))^-1) ~ 0

f(1,n+1) ~ f_2(f_3(f(1,n)^-1))

f(2,1) ~ 0

f(2,n+1) ~ f_4(f(2,n)^-1)

f(n+1,1) ~ 0

f(n+1,m+1) ~ f_{n+2}(f(n+1,m)^-1)

f(1,1,1) ~ 0

f(1,1,n+1) ~ f_w(f(1,1,n)^-1)

f(1,n+1,m+1) ~ f_{w+n}(f(1,n+1,m)^-1)

f(n,1,m+1) ~ f_{w*n}(f(n+1,1,m)^-1)

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