A last boomerang?
The "boomerang" idea here is to replace every digit "d" of the sequence S by the smallest integer unused so far in which "d" is present.
We want of course S to be, as usual, the lexicographically earliest sequence of distinct nonnegative terms with this property – and Giorgos Kalogeropoulos was quick to check, correct (my first attempt) and extend S – many thanks to him!-)
S = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 12, 30, 11, 21,
13, 40, 14, 15, 22, 16, 17, 23, 24, 50, 18, 34, 19, 25, 26, 27, 31, 36,
41, 37, 28, 32, 29, 42, 35, 60, 51, 38, 33, 43, 61, 39, 52, 45, 62, 46, 72, 47, 53, 71, 63, 56, 44, 81, 73, 57, 82, 48, 83, 92, 102, 49, 54, 112, 93, 55, 64, 70, 58, 91, 103, 68, 113, 123, 74, 130, 65, 100, 131, 59, 75, 120, 84, 85, 66, 121, 94, 67, 76, 122, 104, 77, 95, 132, 78, 101, 69, 133, 105, 86, 114, 124, 80, 106, 79, 134, 115, 87, 88, 125, 140, 89, 98, 135, 90, 126, 107, 108, 127, 141, 96, 145, 142, 109, 110, 128, 97, 136, 150, 151, 116, 143, 117, 160, 152, 118, 99, 111, 119, 170, 137, 146, 138, 129, 139, 153, 144, 162, 163, 147, 148, 149, 173, 180, 156, 154, 155, 190, 200, 157, 183, 158, 159, 169, 167, 165, 161, 172, 201, 168, 164, 178, 175, 166, 176, 171, 182, 174, 179, 184, 186, 177, 187, 196, 181, 192, 202, 185, 203, 194, 197, 207, 189, 195, 188, 193, 204,...
Examples
For a(11) = 10 in S, we replace "1" by 10 and "0" by 20 (the smallest integers not used so far for S' where respectively "1" and "0" are present):
S = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
S' = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20,
For a(12) = 20 in S, we replace "2" by 12 and "0" by 30; we get:
S = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20,
S' = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 12, 30,
For a(13) = 12 in S, we replace "1" by 11 and "2" by 21; we get:
S = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 12,
S' = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 12, 30, 11, 21,
For a(14) = 30 in S, we replace "2" by 12 and "0" by 30; we get:
S = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 12, 30,
S' = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 12, 30, 11, 21, 13, 40, etc.
In the end S' is equal to S.
The hereunder graph (almost looking like a boomerang?) was computed by GK.
(Dall-e creation)
The "boomerang" idea has been already explored in the hereunder links.
BOOMERANG #1
BOOMERANG #2
BOOMERANG #3
BOOMERANG #4
BOOMERANG #5 (update)
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