Two (similar) concatenations
(1) The succession of the digits of S and K (above) are the same.
You get K from S by computing A + 3*B where A and B are adjacent digits of S.
S is hopefully the lexicographically earliest such sequence of distinct terms.
S = 2, 9, 1, 27,
23, 13, 11, 6, 10, 64, 19, 91, 18, 187, 28, 3, 61, ...
K = 29 12 7 23
13 11 6
10 6 4 19 9 1 18 18 7 28 36 12 4 25 11 25 ...
(2) The succession of the digits of T and U (hereunder) are the same.
You get U from T by computing A + 4*B where A and B are adjacent digits of T.
T is hopefully the lexicographically earliest such sequence of distinct terms.
T = 2, 6, 10, 1, 4, 17, 8, 29,
39, 16, 38, 21,
391, 32, 5, ...
U = 26 10 1 4 17 8 29 39 16 38
21 39 13 25 18 35 16 6 ...
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P.-S.
The herunder msg was sent to Math-Fun on April 20, 2023 without any echo – but I personaly like a lot the idea!
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Hello Math-Fun
The succession of the digits of S remains the same when 2023 is added to each term.
S = 20,4,3,202,7,2026,2,22,5,2030,40,49,2025,204,520,28,405,320,6,3207,240,48,222,72,54,3205,12,42,8,23,43,202,9,52,302,263,…
This is (hopefully) the lexicographically earliest seq of distinct terms with the «2023» property.
S builds itself easily.
Best,
É.
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(idea follow-up)
.... the smallest such «2023» term that keeps (by addition) the same digits' succession as the original sequence, is 11:
A similar task can be achieved with a multiplication (by 11):
For S I get the same,
RépondreSupprimerS = 2, 9, 1, 27, 23, 13, 11, 6, 10, 64,
19, 91, 18, 187, 28, 3, 61, 24, 25, 112,
5, 29, 132, 617, 21, 97, 14, 101, 7, 8,
4, 71, 711, 291, 210, 920, 92, 213, 52, 830,
1013, 713, 22, 31, 20, 2510, 2210, 47, 2912, 75,
12, 715, 227, 15, 85, 1018, 1126, 17, 33, 1310,
2410, 109, 81, 16, 72, 6178, 168, 51, 122, 513,
29127, 232, 287, 2310, 161, 182, 310, 162, 923, 813,
251, 147, 209, 221, 612, 610, 616, 1471, 312, 73,
311, 41, 927, 1320, 922, 3111, 1930, 238, 478, ...
Will / can a term having "40" or "50" in its digits, ever appear ?
For T(9) I don't get 39 but 3, followed by T(10)=9:
T = 2, 6, 10, 1, 4, 17, 8, 29, 3, 9,
16, 38, 21, 39, 13, 25, 18, 35, 166, 133,
91, 31, 311, 22, 93, 320, 23, 92, 5, 30,
101, 315, 391, 313, 7, 137, 59, 103, 82, 11,
51, 12, 81, 43, 917, 221, 73, 41, 413, 72,
117, 3913, 1371, 33, 111, 1331, 27, 411, 3112, 351,
66, 52, 19, 593, 412, 171, 63, 913, 291, 510,
62, 919, 198, 178, 13311, 56, 529, 193, 9131, 37,
133111, 131, 57, 55, 513, 15, 79, 302, 385, 1375,
914, 239, 2530, 26, 136, 372, 94, 121, 1989, 629, ...
If this can be useful, to compute the K or U sequence, for a given sequence S, one can copy-paste the following in PARI/gp (http://pari.math.u-bordeaux.fr/gp.html):
RépondreSupprimerS=[2,9,1]/*example*/; DIGITS=concat(apply(digits,S)); print("K = ",[DIGITS[j-1]+3*DIGITS[j] | j<-[2..#DIGITS]])/* of course you can replace "3" with "4" here, for the T/U variant */