[seqfan] Re: An idea of sequence of sequences concerning with triangle

[Vladimir Shevelev]

Dear Seqfans,

A class of sequences which I considered let us call M, since the
algorithm begins wth the triangle of 1's corresponding to Mersenne sequence.
It is easy to show that, if we use the same algorithm, beginning with another
(0,1)-triangle corresponding to, say, sequence {x(n)}, 0<=x(n)<=2^n-1,
then there are possible only two cases: 1) we obtain the same class X=M; 
2)  we obtain another class X which is disjoint with M.
Indeed, opration "k" means that every k-th entry of (0,1)-triangle is changed by
opposite, such that k(k(T))=T. If {x(n)} is in M, then we get the corresponding
to it (0,1)-triangle from the triangle of 1's by usng a finite number of operations k_1,k_2,...,k_r.
So, conversely, beginning with the (0,1)-triangle corresponding to sequence {x(n)}, using
these operations, but  in the inverse order k_r,k_{r-1},...,k_1, we get the triangle of 1's corresponding to Mersenne sequence. This means that X=M.
For example, denote by I the class of sequences obtained by the same 
algorithm, but beginning with the (0,1)-triangle coresponding to sequence 1,1,...,1,...:
1
01
001
0001
....,
we conjecture that M and I are disjoint;
Denote by P the class of sequences obtained by the same 
algorithm, but beginning with the (0,1)-triangle coresponding to sequence
0 followed by primes 0,2,3,5,7,...:
0
10
011
0101
00111
....
Most likely, M, I and P are mutually disjoint.

An important question: is there a way to know, whether a given
sequence {c(n)}  belongs or not belongs to a considered class X with 
the initial sequence {x(n)}? If this way exists, then having a representative
of every considered class,  we can partition the whole class of nonnegative
sequences c(n) not exceeding 2^n - 1, n>=1, by disjoint classes of type M.
 I think that, to this question it could be devoted a separate research.

Best regards,
Vladimir


________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 09 November 2014 14:58
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: An idea of sequence of sequences concerning with  triangle

 In addition, note that, if to use sequence {a(n)}, we can consider it arbitrary nonnegative.
Indeed, if a_k=1 ("every first entry is replaced by opposite"), then, on the
k-th step, all 1's (0's) are changed by 0's (1's). Besides, we can agree that,
if a_k=0, then all entries are changed by 0's.
 It is interesting to create a calculator such that, when the user inputs several
nonnegative numbers (e.g., 3,2,4,3,5,1), on the screen there apears
the last sequence and, maybe, the last sequence of row sums (e.g., 20-30 terms). Besides, one can create a picture of several first rows of triangle in the last its
state, where 0's and 1's are represented by white and red small circles respectively.

Best regards,
Vladimir

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 08 November 2014 17:58
To: seqfan at list.seqfan.eu
Subject: [seqfan] An idea of sequence of sequences concerning with triangle

Dear seqfans,

Let us start with triangle of 1's

1
11
111
1111
....

Reading it in binary and converting to decimal, we obtain
Mersenne numbers (A000225): 1,3,7,15,31,63,127,...
Numbering 1's in triangle consecutively, let us change
every second 1's by 0. We get triangle

1
01
010
1010
....
Again converting rows to decimal, we have sequence
1,1,2,10,21,21,42,...
Further, we change every third entry in the second triangle
by opposite (i.e., 1->0, 0->1):
1
00
011
1000
....
and get sequence  1,0,3,8,28,28,56,...
Changing here every fourth entry by opposite , we get
sequence 1,0,7,12,20,62,41,..., etc.
Instead of using the sequence "second,third,fourth,...",
we can use "a_2-th, a_3-th,..." for any sequence {a_n(>1)}.

Best regards,
Vladimir

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