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Inspired by croutonillion.

First, C(0)=googoltriplex.

Starting for X=C(n) and this is how to get C(n+1):


0. X^^^...^^^X (X copies of ^)

1. repeat step 0 for X![C(0),C(1),...,C(n-2),C(n-1),C(n)] times
2. repeat step 0-1 for X![C(0),C(1),...,C(n-2),C(n),C(n-1)] times
3. repeat step 0-2 for X![C(0),C(1),...,C(n-1),C(n-2),C(n)] times
...
n!. repeat step 0-(n!-1) for X![C(n),C(n-1),...,C(2),C(1),C(0)] times

1'. X+C(0)
2'. X+C(1)
...
n'. X+C(n)
n+1'. X+C(0)+C(1)
...
(do all combinations)
...
2^n'. X+C(0)+C(1)+...,C(n-1)+C(n)

2^n+1'. X*C(0)
...
(do the same thing as plus)
(do the same thing until arrow becomes C(n))
...

 1''. X{([tel:[tel:7625597484987 7625597484987] 7625597484987](([tel:[tel:7625597484987 7625597484987] 7625597484987])↑7625597484987([tel:[tel:7625597484986 7625597484986] 7625597484986])↑7625597484987([tel:[tel:7625597484985 7625597484985] 7625597484985])↑7625597484987 ([tel:[tel:7625597484984 7625597484984] 7625597484984])....↑7625597484987(3)↑7625597484987(2))![C(0)]}X
 2''. X{([tel:[tel:7625597484987 7625597484987] 7625597484987](([tel:[tel:7625597484987 7625597484987] 7625597484987])↑7625597484987([tel:[tel:7625597484986 7625597484986] 7625597484986])↑7625597484987([tel:[tel:7625597484985 7625597484985] 7625597484985])↑7625597484987 ([tel:[tel:7625597484984 7625597484984] 7625597484984])....↑7625597484987(3)↑7625597484987(2))![C(0),C(1)]}X
 ...
 n''. X{([tel:[tel:7625597484987 7625597484987] 7625597484987](([tel:[tel:7625597484987 7625597484987] 7625597484987])↑7625597484987([tel:[tel:7625597484986 7625597484986] 7625597484986])↑7625597484987([tel:[tel:7625597484985 7625597484985] 7625597484985])↑7625597484987 ([tel:[tel:7625597484984 7625597484984] 7625597484984])....↑7625597484987(3)↑7625597484987(2))![C(0),C(1),...C(n)]}X

Now, start with C(100).

  1. C(X)
  2. C(C(X))
  3. C(C(C(X)))
  4. C(C(...C(C(X))...)) /w X C's
  5. {X,X,2} solved like BEAF with {a,b}=C(C(...C(a)...)) /w b C's
  6. {X,X,X} solved like BEAF with {a,b}=C(C(...C(a)...)) /w b C's
  7. {X,X,1,2} solved like BEAF with {a,b}=C(C(...C(a)...)) /w b C's
  8. {X,X,X,2} solved like BEAF with {a,b}=C(C(...C(a)...)) /w b C's
  9. {X,X,X,X} solved like BEAF with {a,b}=C(C(...C(a)...)) /w b C's
  10. {X,X,X,X,X}
  11. {X,X(1)2}
  12. {X,X(1)3}
  13. {X,X(1)X}
  14. {X,X(1)1,2}
  15. {X,X(1)X,2}
  16. {X,X(1)X,X}
  17. {X,X(1)(1)2}
  18. {X,X(2)2}
  19. {X,X(3)2}
  20. {X,X(X)2}
  21. {X,X(0,X)2}
  22. {X,X((1)1)2}

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