[yorwba]

Okay, but iterating square roots like √√2 = (2^(2^-1))^(2^-1) recurses into the base, whereas the equivalent iterated log is 2^(2^-1 × 2^-1) = 2^(2^-2) = +2^(+2^(-2^(+2^0))) with the bit representation [1 1 0 0 1 0 0 ...], i.e. it recurses into the exponent.

[DannyBee]

So uh, how is that not a variation, exactly?

Dirac's solution was also arbitrarily restricted by the problem definition.

Do you believe that if you asked someone familiar with the solution to come up with a bit efficient variant they would not have trivially come up with the encoding in this post and called it a variation?

I don't, for a second, believe that.