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…..

(In reply to re(5): NOT enough by Gamer by Ady TZIDON)

"Jer understood that only 99 (and not 999 ) numbers have to be examined (need I explain WHY?) "

I have realized that 999 was overkill, but I don't see that 99 would suffice.  Even 3*81=243 allows that a 3-digit number could produce another 3-digit number and there might be a cycle wholly within the 100-243 range.  Certainly a proof would have to spell out what may seem obvious to you.

Posted by Charlie on 2011-03-09 15:20:18