On Collatz Conjecture

Let us play a game with only 2 simple rules. Choose any number (n) , if the number is an even number then divide it by 2 (i.e. n/2), if not then multiply with 3 and add one (i.e. 3n + 1) and then continue this with numbers obtained by operating n.

The numbers obtained from this rules can be represented in a sequence.

Let us take number 5 , an odd number, so the sequence would be

5,16,8,4,2,1,4,2,1,4,2,1,…..

Let’s take another number, say 6, an even number, so the sequence would be

6,3,10,5,16,8,4,2,1,4,2,1,4,2,1,….

Now taking number 52, an even number we would get sequence

52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,…..

If you observe, all the sequences get trapped and ends in a loop of 4, 2 and 1! Like the numbers we took as example, all the numbers which were put to the 2 rules men- tioned above ended up been in a loop of 4, 2 and 1! This is like a magic and this magic is called Collatz Conjecture or Hailstone problem or simply 3x+1 problem.

Can you find any number which would not follow Collatz Conjecture ? Indeed this is one of the most difficult and unsolved questions in Mathematics and Paul Erdo ̋s said about the Collatz conjecture: ”Mathematics may not be ready for such problems.”. If we see from Visualization point of view then you can refer Fig. 2. and the question becomes,”Which thread does not join with the main thread at bottom?”

Can you by yourself put up such easy looking but difficult to prove conjecture?

Let me now modify the Collatz Conjecture. For this we first need to see one function which would be helpful for stating the Conjecture. I hope that this introduction might inspire you to read more about 3x+1 problem.

2. Modified Collatz Conjecture

2.1 Priliminary Function

Highest Factor Function : Let the function be H(x) which has three rules over x: H(x)= (the highest divisor of x except x when x is a composite number), H(x)=x (if x is a prime number), H(x)=1 ( if x=1).

2.2 New Collatz Conjecture

Let CJ(x) be the function over x defined by 2 rules : if x is an even number then divide it by 2 otherwise multiply by 3 and add 2n+1, (where n is any whole number) i.e. 3x+(2n+1) and x is a odd integral multiple of H(2n+1). Repeat the same function over the numbers further obtained. Then the sequence would get trapped in loop of 4(2n+1),2(2n+1) and (2n+1).

2.3 Practical Observation

• Let n = 7, then 2n + 1 = 15 and H(15) = 5. As x can be any integral multiple of 5 we can choose 5 itself. So the sequence would be : 5,30,15,60,30,15,60,30,15…… If we take x = 10 then sequence would be: 10,5,30,15,60,30,15,60,30,15…
• Let n = 5, then 2n + 1 = 11 so H(11) = 11 and so x can be any integral multiple of 11. If we take x = 33 then Sequence would be:33,110,55,176,88,44,22,11,44,22,11,…..

What if we take n = 0? Then we would get back to Collatz Problem! and as 1 is itself’s biggest positive factor and x has to be positive integral multiple of 1, then x can be any positive integer value, which is what Collatz Problem’s states.

Now you can try this Modified Conjecture by your own with any values n.

What’s Next?

I hope you had fun reading this article and would even try to gain more information about the conjucture, and might even try proving it. Good Luck!