h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

I can prove a special case of the conjecture:

The conjecture is true for the case n=2^m where m is any positive whole number.

Proof

Since 2^m is even (a multiple of 2) ,the iteration requires you to divide it by 2 giving 2^(m-1), which again should be divided by 2 since it is even. Repeating this process leads to 2^0 which is 1. This holds for all values of m, since I have stated that m is any positive whole number.

This case may seem trivial, but notice that all n of the Collatz conjecture, its coresponding sequence ends with a sequence of the form 2^m, 2^(m-1),2^(m-2),....,2^0=1.

Also when proving the Collatz conjecture one only needs to prove it for when n is odd, since when n is even, by the iterative process, it will be divided by two successively until it produces an odd number. So proving Collatz for n=2m+1, where m is 0 or a positive whole number, (i.e for all odd numbers) implies it is true for all n=2m.

Lastly, intuitively I think the conjecture is true because I think its equivalent to saying that:

If n is a positive whole number, then let the iteration be defined as

if n is even n/2

if n is odd n+1

will always eventually come to zero.

However I cannot formally prove this.

Let me leave by posting my own conjecture, shall we call it De Mello's conjecture:

For the 3n +1

replace the 3n by n, 3n, 5n, 7n,.....,(2m+1)n

and an equivalent iteration will always end in zero.

Now prove or disprove that!

Ps. I will also offer £1000 to anyone who proves or disproves De Mello's conjecture.

I have found a contradiction to De Mello's Conjecture, it doesn't work
for 5n+1.

If you take n=5 you get (5,26,13,66,33,166,83,416,208,104,52,26,13,...etc) but it never goes back to 1. Well that's a relief I can pay myself the £1000 I offered as a prize. Next time I thouroughly check something before I conjecture it.