Conjecture:
Take any positive integer x. If x is even, divide it by 2(x/2). Else n is odd, multiply it by 3 and add 1(3*x+1). The conjecture is that no matter what number you start with, you will always eventually reach 1.
Eg:
10 is even 10/2 = 5 is odd 5*3+1 = 16 is even 16/2 = 8 is even 8/2 = 4 is even 4/2 = 2 is even 2/2 = 1 is odd 1*3+1 = 4 and the cycle continues.
I was Playing with Python and this problem.
I calculated the number up-to 10,485,760 which takes maximum steps to get to 1 :
Maximum Counts: 685(+/-1)
Even Counts: 429
Odd Counts: 256
Number with maximum Counts: 8400511
Not sure if this is correct.
Some time later I thought about trying negative numbers and I noticed 3 patterns:
pattern 0:
-2
-1
pattern 1:
-14
-7
-20
-10
-5
pattern 2:
-50
-25
-74
-37
-110
-55
-164
-82
-41
-122
-61
-182
-91
-272
-136
-68
-34
-17
Numbers from -1 to -16 either follow pattern 0 or pattern 1, but -17 on wards some of them follow pattern 2.There might be other patterns too, but I haven't come across any, yet.
Has anyone (other than me) tried it?
