45
++ − k
... )(
)( n x
n x
∑ − = 1 0 k i
3 )(
+
1
1
i
= )(
k n ρ
and
nx
i
−
ik
2
This looks pretty horrendous, but it is actually just an observation on the form of the terms of the sequence. For example, suppose the sequence is a string of odd numbers, then, starting with n we get:
0
n nT = )(
+
1 3
n
1
)( = nT
2
+
1 3
n
×
+
3
1
2
×
3
3
2 1
n
2 2
2
)( = nT
=
+ +
2
2
2
2
+
2
×
3
4 3
2 1
n
×
++
3
1
3
2
4
×
3
2 3
3
2 1
n
2
)( = nT
=
+ + +
3
3
2
2
2
2
The formula with λ ’s and ρ ’s is just this with some terms taken out for the even
terms. For example, we can see that the coefficient of n in the last line is just
k λ
the
as stated.
)( n
Many more results follow from having started out in the route above, all of which can be found in the article by Lagarais (copies available in the Mathematics Office). The article concludes by saying that no final proof or disproof of the conjecture is known yet though.
Feinstein’s argument
The article shown by Mr Kulatunge to Sir Bryan, called ‘The Collatz 3 n + 1 conjecture is unprovable’ by Craig Alan Feinstein [2], picks up from the above.
The essence of the argument is:
(i) First of all to notice that for any sequence of zeros and ones as mentioned, e.g. (1, 0, 0, 1, 1, 0, 1, 1,...), there will be a value of n which produces this sequence when applying T k times to n as described. (This is proved in the Lagarais article).
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