Thank you Alex from The Grammar School at Leeds for your solution to this problem.

All the sequences starting from 1 to 91 inclusive are 4 cycles (the first, fifth, ninth terms etc. are all equal) for example the sequence starting from 53 is: 53, 25, 48, 76, 53, 25, 48, 76, .... and so on. Alternate numbers in the sequences add up to 101 or 0.

However this is not a general rule. For example, sequences starting with numbers between 92 and 99 also go into 4-cycles but these 4-cycles start from the second term of the sequence. For example the sequence starting with 92 is: 92, 11, 9, 90, - 9, 11, 9, 90, - 9, 11, ...

The number 0 is a fixed point of this system. The problem can be generalised to apply to all integers (expressed as 10$a + b$ as above) and sequences starting from 101, 202, 303 etc. end up at 0.