On Friday, Joshua Searle posted to the Sequence Fanatics Discussion list a neat procedure: take the binary complement of an integer multiplied by 3. Iterate. For example, starting with 3 we get the binary of 9 (1001), the complement of which (0110) is 6. Continuing, from 6 we get the binary of 18 (10010), the complement of which (01101) is 13. Arriving at zero, we stop.
0 3
1 6
2 13
3 24
4 55
5 90
6 241
7 300
8 123
9 142
10 85
11 0
Eleven steps to get to zero. The largest integer reached is 300 at step 7. We can shorthand the sequence data for 3 with (11,7,300) [steps to reach zero, steps to reach a maximum, the maximum]. Here are the statistics for integer starts up to 28:
0 (0,0,0)
1 (1,0,1)
2 (2,0,2)
3 (11,7,300)
4 (12,8,300)
5 (1,0,5)
6 (10,6,300)
7 (3,1,10)
8 (4,2,10)
9 (13,9,300)
10 (2,0,10)
11 (19,15,300)
12 (80,28,328536)
13 (9,5,300)
14 (2,1,21)
15 (15,11,300)
16 (16,12,300)
17 (81,29,328536)
18 (14,10,300)
19 (11,7,300)
20 (12,8,300)
21 (1,0,21)
22 (6,2,72)
23 (83,31,328536)
24 (8,4,300)
25 (73,21,328536)
26 (22,5,661)
27 (79,27,328536)
28 (7572,2962,123130640068522377168864228132316865867184046004226894)
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