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):
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=S.
----------
Lets
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,...
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...
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]
----------
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)]
--------------------
Lets
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 youll get T
which is equal to T --> T=T.
----------
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)]
----------
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)]
----------
U, V, W and X will soon appear in the OEIS.
Best,
Ι.
__________
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