[seqfan] Re: Collatz trajectories of 9^15-1 and 9^16-1 converge rapidly

[Jack Brennen]

Here's an explanation:

1024x+184
512x+92
256x+46
128x+23
384x+70
192x+35
576x+106
288x+53
864x+160
432x+80
216x+40

9216x+1664  == 9*(1024x+184)+8
4608x+832
2304x+416
1152x+208
576x+104
288x+52
144x+26
72x+13
216x+40

Basically, for any number N of the form 1024x+184, the numbers N and
9N+8 both converge quite quickly at the number 216x+40.

The next power of 9 which yields this form is 9^143-1, which of
course converges with 9^144-1 with the same relative trajectories.

- Jack


On 11/2/2017 2:37 PM, David Rabahy wrote:
> So far I haven't found other examples of rapid convergence like this;
> 
> 9^15-1              9^16-1
> 205,891,132,094,648 1,853,020,188,851,840
> 102,945,566,047,324   926,510,094,425,920
>   51,472,783,023,662   463,255,047,212,960
>   25,736,391,511,831   231,627,523,606,480
>   77,209,174,535,494   115,813,761,803,240
>   38,604,587,267,747    57,906,880,901,620
> 115,813,761,803,242    28,953,440,450,810
>   57,906,880,901,621    14,476,720,225,405
> 173,720,642,704,864    43,430,160,676,216
>   86,860,321,352,432           ...
>   43,430,160,676,216
> 
> 9^n-1 is always divisible by 8, e.g. 9^15-1 is a multiple of 8.
> 
> 9^(2^k)-1 plunges more rapidly than other equivalence classes, e.g.
> 9^(2^4)-1=9^16-1 is not only a multiple of 8, it has 4 more factors of 2 in
> it, i.e. 9^16-1 is multiple of 2^7=128.  9^(2^k)-1 is a multiple of 2^(k+3).
> 
> 9^31-1 and 9^32-1 do converge eventually but only after more than 100
> Collatz trajectory steps.
> 
> Am I exploring new ground or is this all been looked at before?
> 
> 
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