ConwayLife.com

(Only calculated up to 4 digits)
Base 2: 11, 1 (1 step)
Base 3: 222, 22, 11, 1 (10 steps)
Base 4: 333, 123, 12, 2 (3 steps)

Definition?

11-> 1*1 -> 1 (any base)
222-> 2*2*2-> 22 (base 3) -> 2*2 -> 11 (base 3) ->1.

So in base 10:
999 -> 9*9*9 -> 729 -> 7*2*9 -> 126 -> 12 -> 2



Exponentiation and tetration and pentation, etc. Persistance exist too but are generally either very small or possibly infinitely large.
Example: 222-> 16 -> 1
333-> 19683 -> 1
Likely most numbers have an exponentiation persistence of two to four depending on how long it takes to get a 1 in front.

Every number has a finite multiplication persistence, as the product of its digits can be no more than the original number.
For exponentiation persistence, if exponents are evaluated from left to right (1234->((1^2)^3)^4) it ends when either the first digit is 1 or its expansion contains a 0 (or if it ends up with a single digit). If evaluated right to left (1234->1^(2^(3^4))), it ends if the first digit is 1, the second digit that is not a 1 is 0, or one step after the third digit is 0 (as well as some edge cases like 221x which can be enumerated by hand).
Assuming digits of large composite powers (divisible only by 2,3,5 or 7) are randomly distributed, left-to-right exponents should on average halt after reaching 10 digits and right-to-left exponents should halt after less than 3.32 iterations (less because this estimate is based off of Benford's law and doesn't consider the 0 rules).
EDIT: factoring in the second digit 0 rule for right-to-left exponents, we get an average survival time of 2.308 iterations.

toroidalet wrote:Every number has a finite multiplication persistence, as the product of its digits can be no more than the original number.
For exponentiation persistence, if exponents are evaluated from left to right (1234->((1^2)^3)^4) it ends when either the first digit is 1 or its expansion contains a 0 (or if it ends up with a single digit). If evaluated right to left (1234->1^(2^(3^4))), it ends if the first digit is 1, the second digit that is not a 1 is 0, or one step after the third digit is 0 (as well as some edge cases like 221x which can be enumerated by hand).
Assuming digits of large composite powers (divisible only by 2,3,5 or 7) are randomly distributed, left-to-right exponents should on average halt after reaching 10 digits and right-to-left exponents should halt after less than 3.32 iterations (less because this estimate is based off of Benford's law and doesn't consider the 0 rules).
EDIT: factoring in the second digit 0 rule for right-to-left exponents, we get an average survival time of 2.308 iterations.

What about tetration persistence?

All those estimates should be increased by 1, since the last iteration is the one after the first digit is a 1 or the second non-1 digit is a 0 or whatever.

Moosey wrote:What about tetration persistence?

Right to left and left to right tetration both seem to behave similarly to exponentiation in terms of when they stop, i.e. left-to-right stops after the number contains even a single 0, and so the same figures should apply. (right-to-left should last slightly longer since there are fewer edge cases of the form 221x that lead to it being a single digit).