User blog comment:Deedlit11/Is BEAF well-defined?/@comment-25418284-20121023211454/@comment-5529393-20121116164935

I agree with your equalities up to

X^X^(X^2 + 1) = {b, p ((2) 0, 1) 2}

After that I get

X^X^(2X^2) = {b, p ((2) (2) 1) 2}

X^X^(3X^2) = {b, p ((2) (2) (2) 1) 2}

X^X^X^3 = {b, p ((3) 1) 2}

X^X^(2X^3) = {b, p ((3) (3) 1) 2}

X^X^(3X^3) = {b, p ((3) (3) (3) 1) 2}

X^X^X^4 = {b, p ((4) 1) 2}

X^X^X^5 = {b, p ((5) 1) 2}

X^X^X^X = {b, p ((0, 1) 1) 2}

2X^X^X^X = {b, p ((0, 1) 1) 3}

X^(X^X^X + 1) = {b, p ((0, 1) 1) 1, 2}

X^(2X^X^X) = {b, p ((0, 1) 2) 2}

X^X^(X^X + 1) = {b, p ((0, 1) 0, 1) 2}

X^X^(2X^X) = {b, p ((0, 1) (0, 1) 1) 2}

X^X^X^(X + 1) = {b, p ((1, 1) 1) 2}

X^X^X^(2X) = {b, p ((0, 2) 1) 2}

X^X^X^X^2 = {b, p ((0, 0, 1) 1) 2}

X^X^X^X^3 = {b, p ((0, 0, 0, 1) 1) 2}

X^X^X^X^X = {b, p (((1) 1) 1) 2}

X^X^X^X^(2X) = {b, p (((1) (1) 1) 1) 2}

X^X^X^X^(3X) = {b, p (((1) (1) (1) 1) 1) 2}

X^X^X^X^X^2 = {b, p (((2) 1) 1) 2}

X^X^X^X^X^3 = {b, p (((3) 1) 1) 2}

X^X^X^X^X^X = {b, p (((0,1) 1) 1) 2}

X^X^X^X^X^X^X = {b, p ((((1) 1) 1) 1) 2}

Incidentally, it's really annoying how the base array starts at 1 and the inner arrays start at 0. Bowers should have chosen one or the other (I would have preferred everything start at 0.)

As the number of exponents goes to infinity, so does the depth of the Bowers notation. So the notation runs out at X^^X - we need a new form of notation for "pentational" arrays.