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2 1 3 + n is roughly equal to
1 + in the numerator has little
2 3 n , i.e. the
numbers
effect on the value for large numbers. Thus in the long run we are likely to be
2 3 roughly half the time, i.e. after an odd number, and by 2 1 the
multiplying by
rest of the time, i.e. for even numbers. The net result of this is to be multiplying
4 3 =× many times over. As 4
2 3
2 1
3 is less than 1, this means that over time
by
the sequence will get smaller and smaller, eventually resulting in 1.
Now, whilst a plausible argument, the previous paragraph is nowhere near the rigor needed in Mathematics. There are too many assumptions, not least the premise that there will be roughly the same number of odd and even numbers in a sequence – this isn’t true for the sequence that starts with 64, for example.
A more rigorous approach (taken in ‘The 3 x + 1 conjecture and its generalizations’ by Jeffrey C. Lagarais [1]) starts by labeling the next number in the sequences after
n as
)( nT . The sequence is then n ,
)( nT ,
)) (( nTT ,
))) ((( nTTT , ... where
2 n
)( nT =
if n is even
2 1 3 + n
=
if n is odd
)( 2 nT for
)( 3 nT for
We save time by writing
)) (( nTT and
))) ((( nTTT etc.
It is then noticed that any sequence produced by the conjecture can be represented by the starting value plus a string of zeros and ones, e.g. (1, 0, 0, 1, 1,
= nx ,
0, 1, 1,...), where each zero or one is labeled
, i.e. in this case
)( nx
1 )(
0
k
)( nT k is odd and
1 )( = nx etc., with
= nx ,
= nx ,
being 1 if
)( nx
0 )(
0 )(
3
k
1
2
)( nT k is even.
being 0 if
)( nx
k
)( nT k , i.e. the result of applying T k times
We then notice that a formula for
to n , is:
k
k λ
ρ
)( nT =
nn +
)(
)(
n
k
k ... )( 1 − ++
n x )(
nx
3
0
)( =
k λ
where
n
k
2
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