Same (boomerang) player shoot again

mars 02, 2024

To produce the sequence T, replace each digit d in the sequence S by the smallest number N which contains exactly two copies of d (N missing from T).

At the end of the replacements we have T = S.

S = 11, 101, 110, 100, 112, 113, 114, 200, 115, 300, 400, 116, 117, 22, 118, 119, 33, 121, 131, 44, 122, 500, 600, 141, 151, 55, 133, 700, 800, 144, 900, 1001, 161, 171, 66, 181, 191, 77, 202, 212, 211, 311, 88, 411, 511, 99, ...

Example

S = 11,

T = ...

To form t(1) we replace the first (yellow) digit "1" of 11 by 11 (which contains exactly 2 copies of "1"):

S = 11,

T = 11

To form t(2) we replace the second (yellow) digit "1" of 11 by 101 (as 11 is already in T; 101 contains exactly 2 copies of "1"):

S = 11,

T = 11, 101

We must extend S with s(2) = 101 as we want S and T to be the same sequence (in the end):

S = 11, 101

T = 11, 101

The next yellow digit in S is a "1" again; we will extend T with t(3) = 110 as 110 is the smallest available integer that contains exactly two digits "1":

S = 11, 101

T = 11, 101, 110,

The next yellow digit in S will be "0", which produces t(4) = 100 (which has exactly 2 copies of 0):

S = 11, 101

T = 11, 101, 110, 100

The next yellow digit in S will be another "1", which produces t(5) = 112 (but not 111 as 111 has three copies of "1" instead of two):

S = 11, 101

T = 11, 101, 110, 100, 112

It is time to extend S with s(3), s(4) and s(5):

S = 11, 101, 110, 100, 112

T = 11, 101, 110, 100, 112

Etc.

_______________________

Jean-Marc Falcoz was quick to extend S/T to 10000 terms and compute the according nice graph. Here are the first 180 terms of S/T:

S/T ={11,101,110,100,112,113,114,200,115,300,400,116,117,22,118,119,33,121,131,44,122,500,600,141,151,55,133,700,800,144,900,1001,161,171,66,181,191,77,202,212,211,311,88,411,511,99,233,303,611,220,711,811,313,911,244,344,1010,221,223,155,1002,1003,166,1004,1005,1012,404,1013,1014,255,1015,355,455,1016,323,330,177,1006,1007,188,1008,1009,1017,414,424,199,1020,1030,1018,1040,1050,1019,1021,266,1031,1041,277,1051,366,466,1061,288,1071,1081,299,1091,377,477,224,1060,225,226,1100,227,228,1102,1103,331,1104,1105,388,488,434,1106,1107,505,1108,1109,399,499,229,332,334,335,1070,336,566,1120,1122,232,242,1080,577,1123,1124,588,1125,1126,337,1127,338,599,1128,1129,252,440,441,339,442,443,1130,1090,1132,1200,262,272,1133,282,292,343,1134,515,525,1135,1300,1400,322,1136,1500,1600}

S/T 10000 terms graph

I realize only now the same boomerang idea could be used to produce the sequences U and V where we look for integers having exactly ONE copy of d (instead of TWO for S/T); Jean-Marc was there again (first 100 terms of U/V and graph):

U/V = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 12, 30, 13, 21, 23, 40, 14, 31, 24, 15, 25, 32, 34, 50, 16, 41, 35, 17, 26, 42, 18, 45, 27, 51, 36, 28, 37, 43, 52, 60, 19, 46, 47, 61, 38, 53, 71, 57, 29, 56, 48, 62, 81, 58, 49, 54, 72, 67, 59, 91, 39, 63, 82, 68, 73, 70, 64, 83, 65, 92, 69, 80, 100, 79, 74, 76, 84, 75, 86, 102, 93, 78, 85, 103, 87, 104, 95, 97, 112, 89, 105, 96, 94, 98, 106, 120, 108, 107, 115}

We see hereunder how the sequence works:

the 9 stays on 9,

the 1 of 10 produces 10,

the 0 of 10 produces 20,

the 2 of 20 produces 12,

the 0 of 20 produces 30,

the 1 of 12 produces 13,

the 2 of 12 produces 21, etc.

U/V 1000 terms graph