More on 3x+1 Residues
It is very illuminating to calculate the residues for all numbers in a large interval. Thus the residues for the entire range 231 - 232-1 (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.
Table 1: The 20 highest residues found in the range 231 - 232-1.
#NumberDelayResidue #NumberDelayResidue
14,164,943,8721151.25314214 112,880,127,0152981.25299194
23,702,521,8561201.25302385 122,430,107,1692901.25299194
32,311,520,2561401.25299452 134,100,805,8482831.25299194
42,739,585,0241481.25299203 143,645,160,7542881.25299194
52,808,977,2802231.25299194 152,733,870,5662851.25299194
62,733,870,5642851.25299194 163,645,160,7552881.25299194
73,240,142,8912931.25299194 174,100,805,8502831.25299194
83,645,160,7532881.25299194 183,075,604,3882801.25299194
93,840,169,3533011.25299194 194,156,754,2322521.25299194
102,733,870,5652851.25299194 203,075,604,3892801.25299194

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 232 (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 Res(2N) is equal to Res(N), and there are therefore infinitely many numbers with this residue. In the first table the first number is in fact 993 * 222.

Table 2: Distribution of the residues
of all numbers in the range 231 - 232-1.
RangeOccurrences RangeOccurrences
1.00-1.01130,830,809 1.13-1.1464,822,789
1.01-1.022,564,106 1.14-1.1579,366,825
1.02-1.030 1.15-1.16126,652,603
1.03-1.040 1.16-1.1797,543,128
1.04-1.050 1.17-1.18164,282,715
1.05-1.060 1.18-1.19201,251,512
1.06-1.0713,853,197 1.19-1.20192,895,689
1.07-1.0811,996,113 1.20-1.21232,868,907
1.08-1.0936,423,834 1.21-1.22123,482,008
1.09-1.1035,954,326 1.22-1.2356,574,525
1.10-1.1163,492,053 1.23-1.24142,674,336
1.11-1.12169,732,539 1.24-1.25115,263,600
1.12-1.1353,788,735 1.25-1.2631,169,299
 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. Lemma: Every 3x+1 sequence contains at most 1 odd multiple of 3. Proof: Any number Si formed by taking 3 * Si-1 + 1 is obviously no multiple of 3, nor are any of its immediate descendants formed by divisions. 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 232 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 232 before the contribution of such an odd number is limited to 1 / ( 3 * 232), 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 1017 the maximum delay is only around 2100, which is nowhere near the number required. Thus most contributions will have to come from numbers beyond 1017, which requires X to have a delay of several billions. The author feels that this is enough indication to justify Conjecture 2b : For all N: Res ( N ) <= Res ( 993 ) . Back to the general 3x+1 page.