Fractal erasure

Hello SeqFans and MathFun,

Consider the monotonic sequence S hereunder:

S= 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,110,111,112,113,114,115,116,117,118,119,1110,1111,1112,1113,1114,1115,...

Mark in yellow the first digit of each integer:

S= 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,110,111,112,113,114,115,116,117,118,119,1110,1111,1112,1113,1114,1115,...

Erase all yellow digits (and get S):

=                     0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 110, 111, 112, 113, 114, 115,...

One sees that =S.

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Lets modify S in order to build U.

We will add a constraint on the erased digits and ask them to reproduce the succession of the digits of the sequence itself (this forces U not to start with zero and not to remain monotonic):

U=1,2,3,4,5,6,7,8,9,11,12,13,24,15,36,27,48,19,511,312,613,224,715,436,827,148,919,5511,1312,1613,3224,1715,2436,6827,1148,3919,...

=                   1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 24, 15, 36, 27, 48, 19, 511, 312, 613, 224, 715, 436, 827, 148, 919...

U=1,2,3,4,5,6,7,8,9,1  1  1  2  1  3  2  4  1  5   3   6   2   7   4   8   1   9   5    1    1    3    1    2    6    1    3 ...

Could it be said that U is twice fractal -- a double fractal? One gets the same string concatenating the digits of U, U or U...

[U is the starting sequence, U is what you erase (the leftmost digit of every integer), and U what is left]

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Instead of 9 different digits, like in U, we could have used 8, or 7, or 6, or... 2 digits only, in order to produce a similar double fractal:

V= 1,2,11,12,111,212,1111,1212,11111,21212,111111,221212,1111111,1221212,11111111,11221212,111111111,211221212,1111111111,2211221212,

V=      1  2  11  12  111  212  1111  1212  11111  21212  111111  221212  1111111  1221212  11111111  11221212  111111111  211221212

V= 1,2,1  1  1   2   1    1    1     2     1      2      1       1       1        1        1         2         1          2

[V is the starting sequence, V is what you erase (the leftmost digit of every integer), and V what is left. V uses only two different digits: 1 and 2. Here also V=V=V (if the digits of V are properly concatenated)]

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Lets build T in the same way we have built S -- but instead of erasing the first digit of every integer (the leftmost one), we will erase the last digit (the rightmost one):

T= 1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100,200,300,400,500,600,700,800,900,1000,2000,3000,4000,5000,6000,7000,...

If you erase all yellow digits youll get T which is equal to T --> =T.

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As we did before with S, we can add a constraint on the erased digits of T -- and ask them to reproduce the succession of the digits of the sequence itself (the new sequence W wont start with zero either -- but remains monotonic):

W=1,2,3,4,5,6,7,8,9,11,21,32,41,53,62,74,81,95,113,216,322,417,534,628,741,819,955,1131,2161,3223,4172,5341,6286,7413,8192,9552,11314,...

W=                  1  2  3  4  5  6  7  8  9  11  21  32  41  53  62  74  81  95  113  216  322  417  534  628  741  819  955  1131...

W=1,2,3,4,5,6,7,8,9, 1  1  2  1  3  2  4  1  5   3   6   2   7   4   8   1   9   5    1    1    3    2    1    6    3    2    2     4...

[W is the starting sequence, W is what you erase (the rightmost digit of every integer), and W what is left. W=W=W (if the digits of W are properly concatenated)]

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Instead of 9 different digits, like in W, we could have used 8, or 7, or 6, or... 2 digits only, in order to produce a similar double fractal:

X= 1,2,11,21,112,211,1121,2111,11212,21112,112121,211121,1121211,2111211,11212112,21112111,112121122,211121111,1121211221,2111211111,

X=     1  2  11  21  112  211  1121  2111  11212  21112  112121  211121  1121211  2111211  11212112  21112111  112121122  211121111

X= 1,2, 1  1   2   1    1    1     2     2      1      1       1       1        2        1         2         1          1          1

[X is the starting sequence, X is what you erase (the leftmost digit of every integer), and X what is left. X uses only two different digits: 1 and 2. Here also we have X=X=X (if the digits of X are properly concatenated)]

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U, V, W and X will soon appear in the OEIS.

Best,

Ι.

__________

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