Pick any whole number greater than 1.
1. Write down all of its proper divisors (including 1 and the number itself)
2. Add the digits of these divisors.
3. Use this sum to repeat steps 1 and 2 until your number does not change.
Must the process terminate?
At what number(s) can the process terminate at?
What numbers <1000 take the most steps to terminate?
(In reply to
re: Proof? by Bob Smith)
The root of the function that I have used to set an upper bound on f(n) i.e. 9 (1+Log(n) ) 2 Sqrt(n) = n at about 7444.
So instead of choosing n>10000 for my proof, I could have chosen any value greater than 7444.
Bob writes
"In fact, 48 is the largest value for which f(n) > n."
That this is a "fact" only follows because we can exclude all but a finite number of cases which can be computed. A smaller bounding function is possible and would reduce the number of cases to be evaluated but since that number is already computable, there seems little point.
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Posted by goFish
on 2006-01-11 13:30:39 |
re(2): Proof?
