User blog comment:Googleaarex/Tetrational Arrays/@comment-5150073-20130311193232/@comment-5150073-20130312193939

Maybe (0 (^^) 1) instead of (1 / 2)? So I believe that we can extend it further:

(1 (^^) 1) = (X^^X)*X

(2 (^^) 1) = (X^^X)*(X^2)

(3 (^^) 1) = (X^^X)*(X^3)

(0,1 (^^) 1) = (X^^X)*(X^X)

((1) 1 (^^) 1) = (X^^X)*(X^X^X)

(0 (^^) 2) = (X^^X)*(X^^X) = (X^^X)^2

(0 (^^) 3) = (X^^X)^3

(0 (^^) 4) = (X^^X)^4

(0 (^^) 0,1) = (X^^X)^X

(0 (^^) 1,1) = (X^^X)^(X+1)

(0 (^^) 2,1) = (X^^X)^(X+2)

(0 (^^) 3,1) = (X^^X)^(X+3)

(0 (^^) 0,2) = (X^^X)^(2X)

(0 (^^) 0,3) = (X^^X)^(3X)

(0 (^^) 0,0,1) = (X^^X)^(X^2)

(0 (^^) 0,0,0,1) = (X^^X)^(X^3)

(0 (^^)(1) 1) = (X^^X)^(X^X)

(0 (^^)(2) 1) = (X^^X)^(X^X^2)

(0 (^^)(3) 1) = (X^^X)^(X^X^3)

(0 (^^)(0,1) 1) = (X^^X)^(X^X^X)

(0 (^^)((1) 1) 1) = (X^^X)^(X^X^X^X) = (X^^X)^(X^^4)

(0 (^^)(^^) 1) = (X^^X)^(X^^X) = (X^^X)^^2 ~ X^^(X+1)

(0 (^^) 1 (^^) 1) = (X^^X)^((X^^X)+1)

(0 (^^) 0,1 (^^) 1) = (X^^X)^((X^^X)+X)

(0 (^^) 0,0,1 (^^) 1) = (X^^X)^((X^^X)+X^2)

(0 (^^) 0,0,0,1 (^^) 1) = (X^^X)^((X^^X)+X^3)

(0 (^^)(1) 1 (^^) 1) = (X^^X)^((X^^X)+X^X)

(0 (^^)(2) 1 (^^) 1) = (X^^X)^((X^^X)+X^X^2)

(0 (^^)(0,1) 1 (^^) 1) = (X^^X)^((X^^X)+X^X^X)

(0 (^^)(^^) 2) = (X^^X)^((X^^X)*2)

(0 (^^)(^^) 3) = (X^^X)^((X^^X)*3)

(0 (^^)(^^) 4) = (X^^X)^((X^^X)*4)

(0 (^^)(^^) 0,1) = (X^^X)^((X^^X)*X)

(0 (^^)(^^) 0,0,1) = (X^^X)^((X^^X)*X^2)

(0 (^^)(^^) 0,0,0,1) = (X^^X)^((X^^X)*X^3)

(0 (^^)(^^)(1) 1) = (X^^X)^((X^^X)*X^X)

(0 (^^)(^^)(2) 1) = (X^^X)^((X^^X)*X^X^2)

(0 (^^)(^^)(0,1) 1) = (X^^X)^((X^^X)*X^X^X)

(0 (^^)(^^)(^^) 1) = (X^^X)^(X^^X)^2

(0 (^^)(^^)(^^)(^^) 1) = (X^^X)^(X^^X)^3

(0 (^^)(^^)(^^)...(^^)(^^)(^^) 1) (with n (^^)'s) = (X^^X)^(X^^X)^(n-1)