It is very illuminating to calculate the residues for all numbers in
a large interval. Thus the residues for the entire range
2^{31}  2^{32}1 (approximately 2.1E+9
to 4.2E+9) were calculated and the highest ones determined.
The top 20 of this interval is depicted in table 1.
Interestingly the range of the top 20 is very small.
The numbers 5 to 20 do not even differ in the first eight decimal places,
and the top 4 is only marginally ahead of them.


It is equally interesting to know more about the distribution of Residues.
Table 2 shows that their distribution is certainly not homogeneous. Some
classes are empty, others are filled with millions of numbers. Obviously
all classes beyond 1.26 are empty as well.
These data very much suggest that the residue is limited. When all residues up to 2^{32} (approximately 4,290,000,000) are calculated the results show that the highest residue occurs at N = 993, its residue being equal to 1.253142144...
From the definition of the Residue it should be clear that 

The question thus remains: can any number have a higher residue? First of all we will establish that 993 can not have any predecessors with a higher residue.
So in any 3x+1 sequence the contribution of odd multiples of 3 is restricted to at most a single one. The only predecessors of 993 are therefore just 993 * 2, 993 * 4 etc., which obviously have an identical residue to 993. Can any other number, not producing 993, have a higher residue? To see that this is unlikely consider Table 1. The first six entries are all divisible by 3 and therefore have no predecessors of interest. Entry number 7 is 3,240,142,891, and it has a Residue of 1.25299194. Now assume a number X with a residue higher than Res(993) does exist. Then X must have 'lost' quite a bit of its residue already by the time it drops below 2^{32} for the first time. To get the 'best' chance assume it will produce the number above, 3,240,142,891. A simple subtraction tells us that its residue is approximately 0.00015 less than Res(993). If the residue of X is to be higher than Res(993) it must have gone through a number of odd iterates. Since we assumed that X had not been below 2^{32} before the contribution of such an odd number is limited to 1 / ( 3 * 2^{32}), or approximately 0.000000000776 (7.76 * 10^{11}).
Dividing 0.00015 by this maximal contribution we find that X must have
passed at least 1,900,000 such odd numbers. Since after every odd number
an even step occurs, the total delay for X will have to be at least
3,800,000 and that is without taking into account the further 'descent'
needed after so many incremental steps, and still assuming all odd numbers
make contributions of the maximum magnitude. Current data
however show that below 10^{15} the maximum delay is only around 1860,
which is nowhere near the number required. Thus most contributions will have to
come from numbers beyond 10^{15}, which requires X to have a delay of
several billions.
The author feels that this is enough indication to justify
Back to the general 3x+1 page.
