Take any whole number greater than one.
1. If the number is odd, multiply it by three and add one.
2. If it is even divide it by two.
2a. If the result is still even, continue to divide by two until the result is odd.
3. Continue steps 1 and 2 until you get the same number twice.
[For example starting with 9 -> 28 -> 14 -> 7 which is considered one iteration. The next iteration brings this to 11.]
What number(s) does this process terminate at?
What starting value less than 200 takes the most iterations to terminate?
Each iteration involves at least one division by 2. So f(x)<= (3 x + 1) /2. I will call this increase a "step".
If an iteration involves more than one division by 2 then f(x) <=(3 x +1) /4 < x for x > 1. If this happens then the sequence is determined by a lower value of x.
The key to this problem could (because I think this problem is unsolved) be to calculate primitive elements by which I mean that x is a primitive element if the number of increasing steps generated by x is greater than those generated by all y<=x.
For example we get {0,0,1,0,0,0,2,0,0,0} for x = 1..10 since x = 3 and x = 7 are the only generators of sequences
which begin with increasing sequences.
There are a number of sequences less than 10000 which begin with steps but it appears these are primitive if and only if x = 2^n-1 for n = 2, 3, ..... The number of steps in these cases is n - 1.
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Posted by goFish
on 2006-01-13 18:52:17 |
A Contribution
