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

To who can one send a proof of this conjecture?

Professor Bryan Thwaites is the one who offered the prize, but I don't know whether he is still offering it. Any mathematical journal would be interested if you have a proof in a publishable form.

Do you think you have solved it?

It's possible but there's many a slip ....

Feeling confident.

Any email contacts for mathematical publications?

No, but Thwaites seems to be associated with the University of Southampton. It should be possible to find one by googling.

Take any number and it will be a multiple of one.Perhaps you should call this Thwaite's truism.

?

Can you explain?

I will try. Every number is a multiple of one. This means there is a route whereby you can return to the state of oneness (excuse the phraseology.)The way Thwaites equation does this is to convert the original number into a number that is (can't think of the words for this bit either) a power of two, e.g. two cubed is eight. What actually happens by halving even numbers and trebling odd numbers, then adding one to them to make them even so you can halve them, is you trawl through every number untill you join this two type scale. I hope this makes sense to you but, I have just finished work and can see it is not clearly written so please esk for clarification if necessary. As an afterthought it should be possible to write a similar equasion with the root as three rather than two.

You seem to think that it is obvious. If it is obvious, can you prove it? If you can prove it, then you're entitled to the cash (assuming Thwaites is still offering it).

It seems obvious to me Gnomon but, unfortunately that does not mean I have the mathematical ability to prove it, from what i've seen on hootoo so far that would seem more likely to be something you could do. If that is something you wish to do I would suggest looking for the link between 3n+1 and the series 2,4,8,16,32,64,128,256... etc.
My math is very rusty, not having used much more than simple addition while shopping for the last 20+years. All I can see is that this equation makes a series of numbers that is divisible by 2, then if it reaches not the 'correct conclusion,' i.e. reaches an odd number, it makes a new series of numbers using a formulae that is not in itself divisible by two.
I may have just trashed my answer there. Said I was rusty, I will go numb my brain for 8&1/2 hours selling fridges and have another look later.

Take any positive whole number. If it is even, divide it by 2. If it is odd, multiply it by 3 and add 1. Repeat this process over and over again. Eventually you will reach 1.

If n is an even number halve it until it becomes an odd number. If the odd number is one you have completed the series. If not multiply it by 3 and add one 1 to try starting from a different number.
‘n’ will continuously become a different number until, when n is an odd number, 3n+1 becomes a function of 2^x or 3n+1 over2^x =n.
When n=1 this becomes 3+1=2^2 (2 squared to clarify my use of the symbol^.)
When n=any other odd number n will not be a function of 2^x.
Q.e.d. ?
Because of the nature of this conjecture I can not present this as a mathematical proof. That doesn’t stop it being true though. Or does it?
E.g. if 3n+1(/2 if the result is an even number) (repeated whenever the result is an odd number)/2^x=n then this will fail.
((3n+1)/2 )/2 (unless the result is an odd number) ad infinitum etc.

I can't see that you've shown anything here at all.

If n is divisible by 3 do so, if not add 1.

take not add

?

One thing about doing mathematical work, is you have to be very clear in what you are saying. It is impossible for anybody to tell you whether it is correct or not if they don't understand what you are saying.

Thanks for that. I am trying to say that some things are obvious but not proveable.

It's not at all obvious to me. I can conceive that there might be a number which just keeps "going up" without ever coming back down.

I think that I now agree with that, but probability says NO.