x=int(input("Starting x value: "))
for i in range(10):
if x%2 == 0:
x=x+x//2
else:
x=(x+1) - (x+1)//2
print (x)Why?It strikes me as a lot simpler than the Collatz.
You multiply by 3/2 if the number is even.Unlike the Collatz there's no multiplication
Why not?All I meant was, only processing an odd integer will result in a power of 2, never an even one, so it can never zoom off to infinity.
First, that's a different algorithm. Second, (3-1)/2=1, (1-1)/2=0Why not?
Here is a simple sequence that satisfies what you said but it will shoot off to infinity if you start e.g. with 3:
If the number is odd, subtract 1 and divide by 2. If the number is even, multiply by 4 then add 1.
Had too look up "monotonically." Still don't understand though. ##2^a## is never odd. One would have to prove it can't ascend infinitely, and that there are no looping sequences other than 2,3,2,3,2,3 for any initial value. Proving that your assertion could occur might serve as disproof of the conjecture, even if the integers involved were too large to test.Of course it won’t go to infinity monotonically. The argument I’m making is that if you start with ##2^a(odd)##, you end up with ##3^a(odd)##. For ##a\geq3##, there will be a positive ##k## for which ##3^a(odd)>2^{a+k}(odd)##, and if the sequence continues from ##3^a(odd)## to ##2^{a+k}(odd)##, then you zoom up to ##3^{a+k}(odd)##. Rinse, repeat. You have to prove that doesn’t ever happen.
Yes, I demonstrated that what you wrote is not sufficient. I changed the example but didn't update the starting number, sorry. Start with 5 -> 2 -> 9 -> 4 -> 17 -> 8 -> ...First, that's a different algorithm.
I never said that there is no algorithm that can produce an infinite series? How about if even or odd add 1?Yes, I demonstrated that what you wrote is not sufficient. I changed the example but didn't update the starting number, sorry. Start with 5 -> 2 -> 9 -> 4 -> 17 -> 8 -> ...
Your algorithm might guarantee that nothing disappears to infinite, but you didn't provide any argument against it.
The proof is trivial. ##x## is even implies that ##x=2k,k\in\mathbb{N}##. Applying the algorithm converts ##x## to ##x+\frac{x}{2}=\frac{3}{2}x=3k##. That’s your base step. Now you can use induction to prove that your algorithm converts ##2^an## to ## 3^an##.Looked at your comment some more. Hadn't noticed that 2a (where a is odd and >2) always hits 3a. Might be interesting to prove just that.
You are missing the point. You claimed that your algorithm cannot produce an infinite series because it has a specific property. I constructed a different algorithm with that property that does produce infinite series, demonstrating that your argument doesn't hold.I never said that there is no algorithm that can produce an infinite series? How about if even or odd add 1?
What property did I say the algorithm had that no other algorithm could have and ascend to infinity?You are missing the point. You claimed that your algorithm cannot produce an infinite series because it has a specific property. I constructed a different algorithm with that property that does produce infinite series, demonstrating that your argument doesn't hold.
"Only odd integers will result in powers of 2" is not a sufficient criterion to be sure it will not go to infinity.All I meant was, only processing an odd integer will result in a power of 2, never an even one, so it can never zoom off to infinity.
I meant odd integers using my function, not any function, obviously. But you're right, I haven't proven (only demonstrated) anything. I'm also not sure it isn't sufficient criterion (with my function), though, as you say, I haven't proven it is.That's not what you said, or what I said. Here is what you claimed:"Only odd integers will result in powers of 2" is not a sufficient criterion to be sure it will not go to infinity.