Consider a series of numbers, defined as follows:
Starting with any natural number, each member is a sum of the squares of the previous member`s digits.

Prove : The series always reaches either a stuck-on-one sequence: 1,1,1… or a closed loop of the following 8 numbers: 145,42,20,4,16,37,58,89, ...

Ex1: 12345,55,50,25,29,85,89,145….. etc
Ex2: 66,72,53,34,25,29,85,89,145…
Ex3: 91,10,1,1,1…..

The in-my-head D1 part at least.

No 3 (or more) digit number can ever be followed by a larger term.  This means if the starting number has more than 3 digits, some successive term will have fewer than 3 digits.

If you then continue finding terms there must eventually be a repeat and you have a cycle.

I suppose this doesn't prove you get one of the two cycles given in the problem.  I suppose if I had the patience to check where all the numbers 1 to 99 go that would count as D2.

Posted by Jer on 2011-03-07 14:32:20