So, I already talked about tricots previously but I think I haven’t talked about why I am using this and not only refer to maps such as “3n+1”, “3n+3”, “7n-1”…
So, let’s take tricots defined by matrixes like this for example:
( 1 a
1 b )
In other words, that’s: 2k → 1k+a / 2k+1 → 1k+b
It isn’t hard to show they will send any number to a cycle.
Also
( 1 a
1 b )
is “equivalent” to
( 1 0
1 b-a )
and
( 1 0
1 1+a-b )
in the sense that there is a bijection among the integers that preserves trajectories.
That means we can study only the tricots like:
Tc =( 1 0
1 c )
and get to the conclusion that if 2c’-1 divides 2c-1 then any trajectory under Tc’ also exists under Tc
Meanwhile, in a tricot like:
( 3 0
3 0 )
it’s easy to show only 0,1,2 and 3 have converging trajectories, all other positive numbers generate strictly increasing sequences and all negative ones generate strictly decreasing ones.
Also, it generates all the trajectories in:
( 3 0
3 1 )
plus new ones.
What I think is interesting with this is that we can work with trajectories we know are diverging branches, unlike in 5k+1 where we can only speculate.
Any thoughts about this?
Both @Failix and @sligocki want to study provably divergent trajectories. It’s great. I wonder if those trajectories are “random” (whatever that means), or possibly “programmable” to do certain tasks? Unlike 3n+1, they have plenty of run room to implement for- and while-loops.
On a related topic, here’s a promised Collatz-like rule that pushes on the boundary between divergent and convergent.
if n ends in 0 → n/10
if n ends in 1 → (n+9)/10
if n ends in 2 → (n+8)/10
if n ends in 3 → (n+7)/10
if n ends in 4 → (n+6)/10
if n ends in 5 → (n+5)/10
if n ends in 6 → (n+4)/10
if n ends in 7 → (101n+3)/10
if n ends in 8 → (3541n+2)/10
if n ends in 9 → (27961n+1)/10
By conjecture, almost all start numbers should diverge, since each step increases n by a factor of (10^{10}+1)/10^{10} = 1.0000000001 on average. But this increase is so slow that human-sized numbers get trapped in loops before they can escape.
The start number 7 runs for 129,028 steps before it crashes down into the trivial cycle. I thought it was gone for sure.
7
71
8
2833
284
29
81087
818979
2289947182
228994719
640292133796
64029213380
6402921338
2267274445786
226727444579
633952607787342
63395260778735
6339526077874
633952607788
224482618417731
22448261841774
2244826184178
794892951817430
79489295181743
7948929518175
794892951818
281471594238754
28147159423876
2814715942388
996690915199591
99669091519960
9966909151996
996690915200
99669091520
9966909152
996690916
99669092
9966910
996691
99670
9967
100667
1016737
10269044
1026905
102691
10270
1027
10373
1038
367556
… on up to …
355171320492035060885781560052708046246467433196921008377886082621056532448514764014726892243783569616488594332713314301035155763490830377482375241504881204665201838978276557754422845555859116936474732258931367672662530609921903112960561123508741137875852898181798121464776498444556631276591072304703449499990409614348332941823861427423601944248793511352512525327437250696
… and back down to …
1040105412425588082646
104010541242558808265
10401054124255880827
105050646654984396353
10505064665498439636
1050506466549843964
105050646654984397
1061011531215342410
106101153121534241
10610115312153425
1061011531215343
106101153121535
10610115312154
1061011531216
106101153122
10610115313
1061011532
106101154
10610116
1061012
106102
10611
1062
107
1081
109
304775
30478
10792260
1079226
107923
10793
1080
108
38243
3825
383
39
109048
38613897
390000360
39000036
3900004
390001
39001
3901
391
40
4
1
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