A sequence embedding itself,
twice
For n=1 to infinity,
underline the [a(n)+n]th term not yet underlined in S.
The sequence is such
that:
- the
underlined terms re-build S
- the
not underlined terms re-build S too.
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,2,2,2,...
Let’s check if this is
true:
----->Step 1, we compute [a(n)+n] for n=1.
This is [a(1)+1]
as a(1) means “the first integer of S”, we
have a(1)=3
so
[a(1)+1] = [3 + 1] = 4
We now underline the
4th term not yet underlined in S:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,...
----->Step 2, we compute [a(n)+n] for n=2.
This is [a(2)+2]
as a(2) means “the second integer of S”,
we have a(2)=2
so
[a(2)+2] = [2 + 2] = 4
We underline the 4th
term not yet underlined in S:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,...
----->Step 3, we compute [a(n)+n] for n=3.
This is [a(3)+3]
as a(3) means “the third integer of S”, we
have a(3)=2
so
[a(3)+3] = [2 + 3] = 5
We underline the 5th
term not yet underlined in S:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,...
----->Step 4, we compute [a(n)+n] for n=4.
This is [a(4)+4]
as a(4) means “the fourth integer of S”,
we have a(4)=3
so
[a(4)+4] = [3 + 4] = 7
We underline the 7th
term not yet underlined in S:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,...
----->Step 5, we compute [a(n)+n] for n=5.
This is [a(5)+5]
as a(5) means “the fifth integer of S”, we
have a(5)=2
so
[a(5)+5] = [2 + 5] = 7
We underline the 7th
term not yet underlined in S:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,...
----->Step 6, we compute [a(n)+n] for n=6.
This is [a(6)+6]
as a(6) means “the sixth integer of S”, we
have a(6)=3
so [a(6)+6] = [3
+ 6] = 9
We underline the 9th
term not yet underlined in S:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,...
----->etc.
Let’s call U the sequence of underlined terms and N the non underlined ones; we have:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,...
U = 3,2, 2, 3,2,
3, ...
N = 3,2,2, 3,
2,3, ...
... the three sequences are (by
construction) the same.
Questions:
— Is S chaotic? (this is: can I guess the 2008th term
of S without computing the first
2007 terms?)
— Are the 2’s and 3’s
equally distributed?
Construction method:
Start with a(1)=3,
a(2)=2, a(3)=2 (many other starts fail in producing
interesting sequences; we have tested the starts [1], [2,1],
[3,1],[2,3],[3,2,1]... without success):
S = 3,2,2,...
Now, if we go through
the Step 1 (above), we see that we have to underline the 4th
term not yet underlined of S:
S = 3,2,2,.,...
This term (the first underlined term) *must* be equal to the first term of S (by
definition); so:
S = 3,2,2,3,...
U = 3,
N = 3,2,2,
We now go through
Step 2 -- and see
that we have to underline (again) the 4th term not yet underlined of S
(this is not the same term as before -- but the next
one, obviously):
S = 3,2,2,3,.,...
This second
underlined term *must* be equal to the second term of S (by definition, again); so:
S = 3,2,2,3,2,...
U = 3,2,
N = 3,2,2,
We now go through
Step 3 and see that
we have to underline the 5th term not yet underlined of S:
S = 3,2,2,3,2,.,.,...
This leaves us with a
“hole” (a hole “h” between the last
integer 2 and the underlined spot); how do we fill this hole? Looking at
the triple split of the sequence S gives
the answer:
S = 3,2,2,3,2,h,.,...
U = 3,2,
N = 3,2,2,
The above hole “h”
obviously belongs to the N sequence (as “h” is *not* underlined). We thus have:
S = 3,2,2,3,2,h,.,...
U = 3,2,
N = 3,2,2, h
... and, as N has to be the same as S, we have no choice: h being the 4th term of N *must* be
equal to the 4th term of S --> thus h = 3. We have:
S = 3,2,2,3,2,3,.,...
U = 3,2,
N = 3,2,2, 3
We must determine now
the missing underlined term “u” -- but this is easy, according to the triple split
again; we see that the missing term is also the 3rd term of U:
S = 3,2,2,3,2,3,u,...
U = 3,2, u
N = 3,2,2, 3
... and this 3rd term of U has to be the same as the
3rd term of S; so “u”
must be equal to 2 and we have:
S = 3,2,2,3,2,3,2,...
U = 3,2, 2
N = 3,2,2, 3
No more hole or
missing underlined term has to be fixed; we then proceed to Step 4 -- which says that we
have to underline the 7th term not
yet underlined of S (“step 4” is the same as “add 4 to the 4th term of S and underline accordingly”):
S = 3,2,2,3,2,3,2,.,.,.,...
U = 3,2, 2
N = 3,2,2, 3
We know that the
“holes” in S will be filled thanks
to the N sequence -- and the missing underlined term thanks to the U sequence (both sequences must be equal to S
by definition -- and S grows faster
than U and/or N); we have thus:
S = 3,2,2,3,2,3,2,h,h,u,...
U = 3,2, 2 3
N = 3,2,2, 3 2,3
...and:
S = 3,2,2,3,2,3,2,2,3,3,...
U = 3,2, 2 3
N = 3,2,2, 3 2,3
No more holes or
waiting underscore? We proceed to Step 5 (“add 5 to the 5th
term of S and underline accordingly”) and underline the 7th term not yet underlined of S:
S = 3,2,2,3,2,3,2,2,3,3,.,...
U = 3,2, 2 3
N = 3,2,2, 3 2,3
... we advance in the writing of U to help us fill the
underscore:
S = 3,2,2,3,2,3,2,2,3,3,.,...
U = 3,2, 2
3,2
N = 3,2,2, 3 2,3
... We have:
S = 3,2,2,3,2,3,2,2,3,3,2,...
U = 3,2, 2
3,2
N = 3,2,2, 3 2,3
Step 6 asks to underline the 9th
term not yet underlined of S:
S = 3,2,2,3,2,3,2,2,3,3,2,.,.,.,...
U = 3,2, 2
3,2
N = 3,2,2, 3 2,3
Two holes and one
underscore to fill; we read S and
prolong accordingly U and N:
S = 3,2,2,3,2,3,2,2,3,3,2,.,.,.,...
U = 3,2, 2
3,2 3
N = 3,2,2, 3 2,3
2,2
... gives:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,...
U = 3,2, 2
3,2 3
N = 3,2,2, 3 2,3
2,2
Job done; we proceed
to Step 7 and fill:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,.,....
U = 3,2, 2
3,2 3
N = 3,2,2, 3 2,3
2,2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,.,....
U = 3,2, 2
3,2 3,2
N = 3,2,2, 3 2,3
2,2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,....
U = 3,2, 2
3,2 3,2
N = 3,2,2, 3 2,3
2,2
Step 8:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,.,.,...
U = 3,2, 2
3,2 3,2
N = 3,2,2, 3 2,3
2,2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,.,.,...
U = 3,2, 2
3,2 3,2 2
N = 3,2,2, 3 2,3
2,2 3
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,...
U = 3,2, 2
3,2 3,2 2
N = 3,2,2, 3 2,3
2,2 3
Step 9:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,.,.,.,...
U = 3,2, 2
3,2 3,2 2
N = 3,2,2, 3 2,3
2,2 3
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,.,.,.,...
U = 3,2, 2
3,2 3,2 2
3
N = 3,2,2, 3 2,3
2,2 3
3,2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,...
U = 3,2, 2
3,2 3,2 2
3
N = 3,2,2, 3 2,3
2,2 3 3,2
Step 10:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,.,.,...
U = 3,2, 2
3,2 3,2 2
3
N = 3,2,2, 3 2,3
2,2 3 3,2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,.,.,...
U = 3,2, 2
3,2 3,2 2
3 3
N = 3,2,2, 3 2,3
2,2 3 3,2
2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,...
U = 3,2, 2
3,2 3,2 2
3 3
N = 3,2,2, 3 2,3
2,2 3 3,2 2
Step 11:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,.,...
U = 3,2, 2
3,2 3,2 2
3 3
N = 3,2,2, 3 2,3
2,2 3 3,2 2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,.,...
U = 3,2, 2
3,2 3,2 2
3 3,2
N = 3,2,2, 3 2,3
2,2 3 3,2 2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,2,...
U = 3,2, 2
3,2 3,2 2
3 3,2
N = 3,2,2, 3 2,3
2,2 3 3,2 2
Step 12:
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,2,.,.,...
U = 3,2, 2
3,2 3,2 2
3 3,2
N = 3,2,2, 3 2,3
2,2 3 3,2 2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,2,.,.,...
U = 3,2, 2
3,2 3,2 2
3 3,2 2
N = 3,2,2, 3 2,3
2,2 3 3,2 2 2
S = 3,2,2,3,2,3,2,2,3,3,2,2,2,3,2,3,2,3,2,3,2,3,2,2,2,...
U = 3,2, 2
3,2 3,2 2
3 3,2 2
N = 3,2,2, 3 2,3
2,2 3 3,2 2 2
Step 13:
etc.
__________
If we call f(n):[a(n)+n] the generating
function, I think that the generating function of A117943 is f(n):[3n]. I’m working on this -- hopefully there is something fun to stumble
on!
Best,
É.
[a warm thanks to Philippe Lallouet
for his private mail about generating functions and splitting of sequences;
another thanks to Jean-Marc Falcoz for his computer skills and encouraging words]
__________
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