There is a paper on this at http://uts.awardspace.info
It refers itself as insights and doesn't claim to be a proof, but it looks correct.
Will someone familiar with the problem take a look for errors, because I'm not sure.
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There is a paper on this at http://uts.awardspace.info
It refers itself as insights and doesn't claim to be a proof, but it looks correct.
Will someone familiar with the problem take a look for errors, because I'm not sure.
Appplying h to g(x) does not give odd numbers in every case. For example g(9) = 28 and
h(g(9)) = 14. The definition should therefore state just that h applied to g(x) a times produces
a number.
Then you must prove that for some a the definition produces an odd number - (for all odd
starting numbers) for the statement of (1) to hold.
(1) does not correctly express the sequences you get: it ammounts to:
3x + 1
and then division of the result by 2 for some amount of repeats, you must specify the
process to reset (using the new odd number you reached) as soon as you reach an odd
number.
The iterations must also be counted (somehow), i.e. show after how large an a the
iteration terminates (reaches 1) for every starting number.
The Y in (2) seem to assume that all the odd numbers can be ordered uniqely by the
iteration formula. This is not true. You actually get more than one order, each on a
different subset of the odd numbers (except for 1). For example, starting with 3 you do
not get all the odd numbers. Far from correct.
You must actually count every 3x+1 iteration as well.
The paper was revised and is easier to follow than the original.
It doesn't consider counting steps, because it's irrelevant. If you consider the infinite sets for each predecessor (x), they are just inflated base terms, and it is shown that all reduce to the same successor (y). As it states, this is why it is not necessary to prove each element of the set, just one term to represent the set.
The integer 1 is linked to all odd positive integers.