Prove that if the starting string has only two digits, then the sequence will eventually repeat. Include a convincing argument.
Consider this string of digits:
It has two 0s, twelve 1s, zero 2s, and so on.
We construct another string of digits, called B, as follows: write the number of zeros in A, followed by the number of 1s, followed by the number of 2s, and so on until we write the number of 9s. Thus
String B is called the derived string of A. We now repeat this procedure on B to get its derived string C, then get the derived string of C, and so on to produce a sequence of derived strings.
Notice that the last string equals a previous string so the sequence of derived strings will now repeat.
if the starting string has 2 digits, it is either 2 times the same digit or 2 different digits.
In this case there will be either 1 2 in the sequence or 2 1
If there is 1 2, then the third string will be 0010000
Starting to the fourth one, the sequences will be 0100000000 (1 1 and 0 other digits) all the time therefore it repeats....
If there are 2 1, then the third sequence will be 0200000000
then the fourth :0010000000
Starting to the fifth, the sequences will always be 0100000000 therefore it repeats.
In conclusion, the sequences repeats after the 5th sequence for any 2 digits initial sequence.
Unfortunately your answer didn't make much sense to me, and I couldn't answer my question properly. What I need is an elaboration on that answer and more explanation, I really need to understand the answer properly! Thank you!
While you're at it could you maybe post it instead on my question here:
So that I can get an update when you post an answer (if you do). Thanks so much!
Cosinusix, sorry, I didn't quite understand your answer to this question, could you explain / elaborate on it more? Thank you so much!