|
| |
|
|
A358708
|
|
Starting from 1, successively take the smallest "Choix de Bruxelles" (A323286) which is not already in the sequence.
|
|
3
|
|
|
|
1, 2, 4, 8, 16, 13, 23, 26, 46, 43, 83, 86, 166, 133, 136, 68, 34, 17, 27, 47, 87, 167, 137, 174, 172, 171, 271, 272, 236, 118, 19, 29, 49, 89, 169, 139, 178, 278, 239, 269, 469, 439, 478, 474, 237, 267, 467, 437, 837, 867, 1667, 1337, 1367, 687, 347, 177, 277, 477, 877, 1677, 1377, 1747, 1727, 1717, 1734, 1732, 866, 433, 233, 263, 163, 323, 313, 316, 38, 76, 73, 143, 123, 63, 33, 36, 18, 9
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
The Choix de Bruxelles doubles or halves some decimal digit substring and rows of A323286 are all ways this can be done.
So a(n) is the smallest term of the row a(n-1) of A323286 which is not among {a(0..n-1)}.
The sequence is finite since having reached 18 -> 9 the sole Choix for 9 would be back to 18, which is already in the sequence.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
Below, square brackets [] represent multiplication by 2 (e.g., [6] = 12); curly brackets {} represent division by 2 (e.g., {6} = 3); digits outside the brackets are not affected by the multiplication or division (e.g., 1[6] = 112 and 1{14} = 17).
We begin with 1 and, at each step, we go to the smallest number possible that hasn't yet appeared in the sequence:
1 --> [1] = 2
2 --> [2] = 4
4 --> [4] = 8
8 --> [8] = 16
16 --> 1{6} = 13
13 --> [1]3 = 23
23 --> 2[3] = 26
26 --> [2]6 = 46
... and so on.
|
|
|
PROG
|
(C#) //(see in links)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy,base,fini,full
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
