Decimation-like sequences
Hello
SeqFan and Math-fun ;
(sorry if this is old hat)
Please
have a look at this sequence:
S=
1
2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 3 2 1 3 2 1 3 2 1 3 4 2
1 3 4 2 1 3 4 2 5 1 3 4 2 5 1 3 4 2 6 5 1 3 4 2 6 5 1 3 7 4 2 6 5 1 3 7 4 2 8 6 5 1 3 7 4 2 8 6 9 5 1 3 7 4 2
8 6 9 1 5 1 3 7 4 2 8 6 9 1 1 5 1 3 7 4 2 8 6 1 9 1 1 5 1 3 7 4 2 1 8 6 1 9 1 1
5 1 3 1 7 4 2 1 8 6 1 9 1 1 1 5 1 3 1 7 4 2 1 1 8 6 1 9 1 1 1 5 1 1 3 1 7 4 2 1
...
Mark
in yellow every 10th
integer:
S=
1
2 3 4 5 6 7 8 9 1 1
1 1 1 1 1 1 1 1 2 1
2 1 2 1 2 1 2 1 3 2
1 3 2 1 3 2 1 3 4 2
1 3 4 2 1 3 4 2 5 1
3 4 2 5 1 3 4 2 6 5
1 3 4 2 6 5 1 3 7 4
2 6 5 1 3 7 4 2 8 6 5 1 3 7 4 2 8 6 9 5 1 3 7 4 2 8 6 9 1 5 1 3 7 4 2 8 6 9 1 1 5 1 3 7 4 2 8 6 1 9 1 1 5 1 3 7 4 2 1 8 6 1 9 1 1 5 1 3 1 7 4 2 1 8 6 1 9 1 1 1 5 1 3 1 7 4 2 1 1 8 6 1 9 1 1 1 5 1 1 3 1 7 4 2 1 ...
Erase
all yellow terms:
E=
1
2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 1 1 2 1 2 1 2 1 2 1 2
1 3 2 1 3 2 1 3 2 1 3 4 2 1 3 4 2 1 3 4 2 5 1 3 4 2 5
1 3 4 2 6 5 1 3 4 2 6 5 1 3 7
4 2 6 5 1 3 7 4 2 8 6 5 1 3 7 4 2 8 6 9 5 1
3 7 4 2 8 6 9 1 5 1 3 7 4 2 8 6 9 1 1 5 1 3 7 4 2 8
6 1 9 1 1 5 1 3 7 4 2 1 8 6 1 9 1 1 5 1 3 1 7 4 2 1 8
6 1 9 1 1 1 5 1 3 1 7 4 2 1 ...
If
you concatenate the result (omit the blanks),
you’ll get the starting sequence:
E=S=
1
2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 3 2 1 3 2 1 3 2 1 3 4 2
1 3 4 2 1 3 4 2 5 1 3 4 2 5 1 3 4 2 6 5 1 3 4 2 6 5 1 3 7 4 2 6 5 1 3 7 4 2 8 6 5 1 3 7 4 2 8 6 9 5 1 3 7 4 2
8 6 9 1 5 1 3 7 4 2 8 6 9 1 1 5 1 3 7 4 2 8 6 1 9 1 1 5 1 3 7 4 2 1 8 6 1 9 1 1
5 1 3 1 7 4 2 1 8 6 1 9 1 1 1 5 1 3 1 7 4 2 1 1 8 6 1 9 1 1 1 5 1 1 3 1 7 4 2 1
...
The
nice thing being that the succession of all yellow terms form also the starting
sequence:
Y=E=S=
.
. . . . . . . . 1 .
. . . . . . . . 2 .
. . . . . . . . 3 .
. . . . . . . . 4 .
. . . . . . . . 5 .
. . . . . . . . 6 .
. . . . . . . . 7 .
. . . . . . . . 8 .
. . . . . . . . 9 .
. . . . . . . . 1 .
. . . . . . . . 1 .
. . . . . . . . 1 .
. . . . . . . . 1 .
. . . . . . . . 1 .
. . . . . . . . 1 .
. . . . . . . . 1 .
. . . . . . . . 1 . . . . . . ...
Could
such sequences be called «super-fractals»?
----------
There
are infinitely many decimation-like
sequences: instead of [1,2,3,4,5,6,7,8,9] the first nine integers could have
been [7,15,8933,42,3001,8,666,0,666], for instance (with or without repeted
terms). This new set would have
produced this 10-super-fractal:
S=
7
15 8933 42 3001 8 666 0 666 7 7 7 7 7 7 7 7 7 7 15 7 15 7 15 7 15 7 15 7 8933 15
7 8933 15 7 8933 15 7 8933 42 15 7 8933 42 15 7 8933 42 15 3001 7 8933 42 15 3001
7 8933 42 15 8 3001 7 8933 42 15 8 3001 7 8933 666 42 15 8 3001 7 8933 666 42 15 0 8 3001 7 8933 666 42
15 0 8 666 3001 7 8933 666 42 15 0 8 666 7 3001 7 8933 666 42 15 0 8 666 7 7 3001
7 8933 666 42 15 0 8 7 666 7 7 3001 7 8933 666 42 15 7 0 8 7 666 7 7 3001 7 8933
7 666 42 15 7 0 8 7 666 7 7 7 3001 7 8933 7 666 42 15 7 7 0 8 7 666 7 7 7 3001 7
7 8933 7 666 42 15 7 ...
----------
Instead
of yellowing every 10th integer, one
could mark every 7th, or 215th, or 3rd term, etc. – producing thus another
infinite bunch of super-fractals;
here is such a super-fractal with every 6th term marked:
S=
1
2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 2 1 3 2 1 4 3 2 1 4 3 5 2 1 4 3 5 1 2 1 4 3 5
1 1 2 1 4 3 1 5 1 1 2 1 1 4 ...
1
2 3 4 5 1 1 1 1 1 1
2 1 2 1 2 1 3 2 1 3 2 1 4 3 2 1 4 3 5 2 1 4 3 5 1 2 1 4 3 5 1 1 2 1 4 3 1 5 1 1 2 1 1 4 ...
----------
The
lowest bound (for
the interval between two yellow marks)
seems to be 2; integers 0 et 1 would then produce (a):
(a)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,1,...
Let’s
check if it works:
1)
mark every 3rd integer of (a) in yellow and get (b):
(a)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,1,...
(b)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,1,...
2)
erase all yellow marks from (b) and get (c):
(b)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,1,...
(c)
0,1, 0,0, 0,1, 0,1, 0,0, 1,0 ,0,0, 0,1, 0,1, 0,0, 0,0, 1,1, 0,1, 0,0, 0,0, 0,0, ...
3)
concatenate all terms of (c) and get (d):
(c)
0,1, 0,0, 0,1, 0,1, 0,0, 1,0 ,0,0, 0,1, 0,1, 0,0, 0,0, 1,1, 0,1, 0,0, 0,0, 0,0, ...
(d)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,...
4)
compare (d) to the starting sequence (a):
(a)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,1,...
(d)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,...
...
same sequences!
5)
write down the successive previously yellow marked (and erased) terms from (b) and get (e):
(b)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,1,...
(e)
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, ...
6)
concatenate all terms of (e) and get (f):
(e)
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, ...
(f)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1, ...
7)
Compare (f) to the starting sequence (a):
(a)
0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,1,...
(f) 0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1, ...
...
same sequences again!
Isn’t
this (a) sequence the mother of all decimation-like super-fractals?
(a)
is not in the OEIS (it is now as A117943). Nor is (g) where all 0’s and 1’s of (a)
have been switched:
(g)
1,0,1,1,1,0,1,0,1,1,0,1,1,1,1,0,1,0,1,1,1,1,0,0,1,0,1,1,1,1,1,1,0,0,0,1,1,0,1,1,1,1,1,1,1,1,1,0,...
What
could be the law ruling the distribution of 0’s and 1’s in (a)? Gilles Sadowski has computed the first
1000 terms of (a) :
(a)=
0 1 0 0 0 1 0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1 0 1 1 0 0 0 1 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 0 1 0 1 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 0 1 0 0 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 0 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 1 0 1 1 0 0 0 0 1 0 1 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 1 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 1 0 0 0 1 0 1 1 0 0 1 1 1 1 0 1 0 0 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 0 1 1 1 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 0 1 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 1 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 ...
What
else can be said about decimation-like
sequences and/or super-fractals?
Here
is the building method used for the 10-super-fractal
which opens this page:
1)
choose de decimation-step d (here d=10 --> we will cancel all terms
whose index is a multiple of 10)
2)
choose (d-1) integers at random (they
will form the starting "seed" --> here 1,2,3,4,5,6,7,8,9)
3)
draw line (1):
(1) .........X.........X.........X.........X.........X.........X...
[this line (1) alternates 9 empty simple boxes and 1 "super-box" : a dot (.) stands
for a simple box, an "X"
for a super-box]
4)
draw the "punched" line (2) under line (1):
(1) .........X.........X.........X.........X.........X.........X...
(2) ......... ......... ......... .........
......... ......... ...
(this line is the same as the previous one -- but *without*
the "X" super-boxes)
5)
draw the continuous (S) line, which will be the "sum" of lines (1)
and (2):
(1)
.........X.........X.........X.........X.........X.........X...
(2)
......... ......... ......... ......... ......... ......... ...
(S)
...............................................................
6)
Fill the first 9 empty single boxes of (2) with the chosen "seed" :
(1)
.........X.........X.........X.........X.........X.........X...
(2)
123456789 ......... ......... ......... ......... ......... ...
(S)
...............................................................
Those
9 integers are also the first 9 X's:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 ......... ......... ......... ......... ......... ...
(S)
...............................................................
Start
to "sum" (1) and (2) [to "sum" is to "push" (1)
into (2) which gives S]:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 ......... ......... ......... ......... ......... ...
(S)
1234567891.....................................................
Now
comes the construction trick: line (2) must always be filled by copying S's
succession of integers, *but* one can never write *under* the "X's".
So we will write, to the right of the last 9 of (2), the "1" which
ends S for the moment:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 1........ ......... ......... ......... ......... ...
(S)
1234567891.....................................................
We
see now that the succession of terms in (2) is the same as the succession of
terms in S [this is because we cannot use for writing purpose the
"hole" which is under the "1" of (1)]
We
"sum" again:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 1........ ......... ......... ......... ......... ...
(S)
12345678911....................................................
The
last "1" which appears on line S forces the writing of a
corresponding "1" behind the final "1" of line (2) -- because
line (2) always copies with a little delay the succession of terms in S:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 11....... ......... ......... ......... ......... ...
(S)
12345678911....................................................
We
"sum":
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 11....... ......... ......... ......... ......... ...
(S)
123456789111...................................................
We
copy S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111...... ......... ......... ......... ......... ...
(S)
123456789111...................................................
We
"sum":
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111...... ......... ......... ......... ......... ...
(S)
1234567891111..................................................
We
copy S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 1111..... ......... ......... ......... ......... ...
(S)
1234567891111..................................................
...
and so on -- till:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 ......... ......... ......... ......... ...
(S)
123456789111111111.............................................
We
"sum":
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 ......... ......... ......... ......... ...
(S)
1234567891111111111............................................
We
copy S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 1........ ......... ......... ......... ...
(S)
1234567891111111111............................................
We
"sum":
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 1........ ......... ......... ......... ...
(S)
123456789111111111121..........................................
We
copy the last *2* integers of S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121...... ......... ......... ......... ...
(S)
123456789111111111121..........................................
We
"sum":
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121...... ......... ......... ......... ...
(S)
12345678911111111112121........................................
We
copy the last *2* integers of S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 12121.... ......... ......... ......... ...
(S)
12345678911111111112121........................................
We
"sum" and copy till:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 21....... ......... ......... ...
(S)
12345678911111111112121212121..................................
We
"sum" again:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 21....... ......... ......... ...
(S)
12345678911111111112121212121321...............................
We
copy the last *3* integers of S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 21312.... ......... ......... ...
(S)
12345678911111111112121212121321...............................
We
"sum" and copy, till:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 21....... ......... ...
(S)
12345678911111111112121212121321321321.........................
We
"sum":
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 21....... ......... ...
(S)
123456789111111111121212121213213213213421.....................
We
copy the last *4* integers of S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 213421... ......... ...
(S)
123456789111111111121212121213213213213421.....................
We
"sum" and copy, till:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 213421342 1........ ...
(S)
1234567891111111111212121212132132132134213421.................
We
"sum" again:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 213421342 1........ ...
(S)
123456789111111111121212121213213213213421342134251............
We
copy the last *5* integers of S in (2):
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 213421342 134251... ...
(S)
123456789111111111121212121213213213213421342134251............
We
"sum" and copy till:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 213421342 134251342 51.
(S)
12345678911111111112121212121321321321342134213425134251.......
We
"sum" again:
(1)
.........1.........2.........3.........4.........5.........6...
(2)
123456789 111111111 121212121 213213213 213421342 134251342 51.
(S)
12345678911111111112121212121321321321342134213425134251342651.
...
etc. This should be clear now : line S is the sequence we were looking
for.
[One
must *not* forget that line (1) builds up step by step too: line (1) is also a
copy of S! Thus after the "seed" 1,2,3,4,5,6,7,8,9
the next "X" value is *not* 10 but 1! (because the 10th term of S is
"1")]
We
then get at the end (terms computed by Gilles
Sadowski):
S=
1
2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 3 2 1 3 2 1 3 2 1 3 4 2
1 3 4 2 1 3 4 2 5 1 3 4 2 5 1 3 4 2 6 5 1 3 4 2 6 5 1 3 7 4 2 6 5 1 3 7 4 2 8 6
5 1 3 7 4 2 8 6 9 5 1 3 7 4 2 8 6 9 1 5 1 3 7 4 2 8 6 9 1 1 5 1 3 7 4 2 8 6 1 9
1 1 5 1 3 7 4 2 1 8 6 1 9 1 1 5 1 3 1 7 4 2 1 8 6 1 9 1 1 1 5 1 3 1 7 4 2 1 1 8
6 1 9 1 1 1 5 1 1 3 1 7 4 2 1 1 8 6 1 1 9 1 1 1 5 1 1 3 1 1 7 4 2 1 1 8 6 1 2 1
9 1 1 1 5 1 1 3 1 1 1 7 4 2 1 1 8 6 2 1 2 1 9 1 1 1 5 1 1 1 3 1 1 1 7 4 2 1 2 1
8 6 2 1 2 1 9 1 1 1 1 5 1 1 1 3 1 1 2 1 7 4 2 1 2 1 8 6 1 2 1 2 1 9 1 1 1 1 2 5
1 1 1 3 1 1 2 1 1 7 4 2 1 2 1 8 6 1 3 2 1 2 1 9 1 1 1 1 2 2 5 1 1 1 3 1 1 2 1 1
1 7 4 2 1 2 1 8 3 6 1 3 2 1 2 1 9 1 2 1 1 1 2 2 5 1 1 1 1 3 1 1 2 1 1 1 7 4 3 2
1 2 1 8 3 6 1 3 2 2 1 2 1 9 1 2 1 1 1 1 2 2 5 1 1 1 1 3 3 1 1 2 1 1 1 7 4 3 4 2
1 2 1 8 3 6 1 3 2 2 2 1 2 1 9 1 2 1 1 1 1 1 2 2 5 1 1 1 3 1 3 3 1 1 2 1 1 1 4 7
4 3 4 2 1 2 1 8 2 3 6 1 3 2 2 2 1 2 1 1 9 1 2 1 1 1 1 1 3 2 2 5 1 1 1 3 1 3 4 3
1 1 2 1 1 1 4 7 2 4 3 4 2 1 2 1 8 2 5 3 6 1 3 2 2 2 1 2 1 1 1 9 1 2 1 1 1 1 3 1
3 2 2 5 1 1 1 3 4 1 3 4 3 1 1 2 1 1 2 1 4 7 2 4 3 4 2 1 5 2 1 8 2 5 3 6 1 3 1 2
2 2 1 2 1 1 1 9 3 1 2 1 1 1 1 3 1 3 4 2 2 5 1 1 1 3 4 1 2 3 4 3 1 1 2 1 1 2 6 1
4 7 2 4 3 4 2 1 5 5 2 1 8 2 5 3 6 1 1 3 1 2 2 2 1 2 1 1 3 1 9 3 1 2 1 1 1 1 4 3
1 3 4 2 2 5 1 1 2 1 3 4 1 2 3 4 3 1 6 1 2 1 1 2 6 1 4 7 5 2 4 3 4 2 1 5 5 2 1 1
8 2 5 3 6 1 1 3 3 1 2 2 2 1 2 1 1 3 7 1 9 3 1 2 1 1 1 1 4 4 3 1 3 4 2 2 5 1 2 1
2 1 3 4 1 2 3 4 6 3 1 6 1 2 1 1 2 6 5 1 4 7 5 2 4 3 4 2 1 1 5 5 2 1 1 8 2 5 3 3
6 1 1 3 3 1 2 2 7 2 1 2 1 1 3 7 1 9 4 3 1 2 1 1 1 1 4 4 2 3 1 3 4 2 2 5 1 2 8 1
2 1 3 4 1 2 3 4 6 6 3 1 6 1 2 1 1 2 5 6 5 1 4 7 5 2 4 3 1 4 2 1 1 5 5 2 1 1 3 8
2 5 3 3 6 1 1 3 7 3 1 2 2 7 2 1 2 1 4 1 3 7 1 9 4 3 1 2 2 1 1 1 1 4 4 2 3 1 8 3
4 2 2 5 1 2 8 1 6 2 1 3 4 1 2 3 4 6 9 6 3 1 6 1 2 1 1 2 5 5 6 5 1 4 7 5 2 4 1 3
1 4 2 1 1 5 5 2 3 1 1 3 8 2 5 3 3 6 7 1 1 3 7 3 1 2 2 7 4 2 1 2 1 4 1 3 7 1 2 9
4 3 1 2 2 1 1 1 8 1 4 4 2 3 1 8 3 4 6 2 2 5 1 2 8 1 6 2 9 1 3 4 1 2 3 4 6 9 1
...
The
(simple) algorithm used says:
a)
put the seed at the beginning of (2)
b)
copy the seed in the successive X's of line (1)
c)
"Sum"
d)
copy the "sum" so far in the proper empty boxes of (2) and (1)
e)
loop in (c)
This
produces the three "equivalent" lines (1), (2) and S.
____________________
This
e-mail (in
french), was sent to me by Jacques Tramu on May 1st,
2006 :
Je suis en train de regarder combien il
faut en moyenne d'itérations pour calculer le ne terme pour n allant
de 10 à 10**100 (1 googol).
Voir la courbe ci-joint pour k = 1000.
C'est remarquablement linéaire [en abscisse, log10(n) ; en vert le nombre
max d'itérations].
Éric : pour k = 10, voici le 100.........0000000e terme (cent zéros), et les 10 suivants :
1
2 1 1
1 7 2
1 1 7 7
à suivre.
J.T.
____________________
This
e-mail was sent to me the same day by Kerry
Mitchell :
Hi Eric,
I've been playing with your decimation sequences--very
interesting!
Attached is an image I created to simultaneously show a 2-integer, a 3-integer, and a 4-integer sequence.
Each of the 576 (24 x 24) squares is one of 24 symbols
for the 2 x 3 x 4 = 24 possible values of the 3 sequences.
Kerry
Beautiful,
Kerry, thanks !
___________________
Note of May 10th,
2006
I’ve
noticed yesterday that more super-fractals
can be created by increasing the size of the « X » super-boxes (see above). Here is an example for a super-box of size 3 which follows an
empty 2-single box:
(1) ..010..100..110..000..110..110..000..011..000..110..110..
(2)
01 01
00 11 00
00 11 01
10 00 00
11
(S)
010100110000110110000011000110110000101110000001100011011...
So
S must be understood like this:
S=
0,1,0,1,0,0,1,1,0,0,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,0,0,1,1,0,1,1,0,0,0,0,1,0,1,1,1,0,0,0,0,0,0,1,1,0,0,0,1,1,0,1,1,...
«Keep
two integers, underline the next three; keep two integers again, underline the
next three; keep two again, underline three; etc. The concatenation of all kept
integers is the sequence itself – and so is the concatenation of all underlined
integers, too»
S
could be seen as a «keep 2/underline 3» sequence. Here is a «keep 3/underline
2» sequence:
(1)
...01...20...10...10...20...10...21...00...10...10...21...
(2)
012 010
102 010 210
010 102 102
001 010 102
211
(S)
0120101020102100101021020010101022110200001100101010221211...
S=
0,1,2,0,1,0,1,0,2,0,1,0,2,1,0,0,1,0,1,0,2,1,0,2,0,0,1,0,1,0,1,0,2,2,1,1,0,2,0,0,0,0,1,1,0,0,1,0,1,0,1,0,2,2,1,2,1,1,...
[So A117943, the
«mother of all super-fractals», is simply a «keep 2/underline 1» sequence; and
the decimation sequence which opens
this page a «keep 9/underline 1» sequence]