A354223 - OEIS

%I #42 Jul 18 2022 10:00:38

%S 1,2,3,1,3,4,2,1,2,4,1,4,5,3,2,1,2,3,1,3,5,2,1,2,5,1,5,6,4,3,2,1,2,3,

%T 1,3,4,2,1,2,4,1,4,6,3,2,1,2,3,1,3,6,2,1,2,6,1,6,7,5,4,3,2,1,2,3,1,3,

%U 4,2,1,2,4,1,4,5,3,2,1,2,3,1,3,5,2,1,2,5,1,5,7,4,3,2,1,2,3,1,3,4,2,1,2,4,1,4,7,3,2,1,2,3,1,3,7,2,1,2,7,1,7

%N The initial sequence of the positive integers gradually becomes interspersed with the copies of its terms and so is transformed into this sequence. The method is described in the Comments section.

%C Pre-inked rubberstamps of the positive integers 1, 2, 3, ... are lined up on an infinite paper strip, with sufficient gaps between them. Underneath them already there are their imprints on the paper. We lift and move the first stamp over as many others as is the number of the second one in the line. Then, we press it down on the paper in the gap between the two stamps at that location. We leave it to stand there on its new imprint until its next turn. We successively repeat this with every temporary leading pair: the second stamp will govern the first one, how many others it has to leap over. The ink-printed numbers left uncovered on the paper in front of the receding line of the stamps form the sequence.

%C It appears that for every new term k > 1, runs of 2^(k-1)-1 length are in this sequence, with k being at the beginning and at the end of these runs. Also, within these runs k repeats itself in the self-same manner, only in reverse order. Could this indicate a fractal property?

%C If you take the first differences of this sequence as a new sequence and remove all negative numbers from it, then you will obtain A089309. - _Thomas Scheuerle_, May 20 2022

%H Thomas Scheuerle, <a href="/A354223/b354223.txt">Table of n, a(n) for n = 1..6000</a>

%H Thomas Scheuerle, <a href="/A354223/a354223.png">First differences of a(n) as scatter plot from 1 to 10000.</a>

%F From _Thomas Scheuerle_, May 19 2022: (Start)

%F a(2^(m+1) + n - 1) - a(n) < 2, if n < 2^(m+1) - m. This evaluates to zero for all n with m+1 exceptions. For these exceptions it will become 1. The exceptions are n = {A132045(m), A132045(m)+2^m, A132045(m)+2^m+2^(m-1), A132045(m)+2^m+2^(m-1)+2^(m-2), ..., A132045(m)+2^(m+1)-1}.

%F (1/n)*Sum_{k=1..n} a(k) <= 3. It appears that this arithmetic mean converges to 3 if n approaches infinity.

%F a(1 + Sum_{k=1..n} A090739(k)) = a(1 + A120738(n - 1)) = 1.

%F a(Sum_{k=1..n} (1 + A195986(k))) = 2. (End)

%e The initial line of the stamps:

%e (1) (2) (3)         (4)         (5) ...

%e Stamp (1) leaps over two others as is governed by stamp (2), leaving its imprint 1 in front:

%e 1  (2) (3) (1)     (4)         (5) ...

%e Stamp (2) leaps over three others as stamp (3) dictates. Its imprint 2 remains on the paper:

%e 1   2  (3) (1)     (4) (2)     (5) ...

%e Stamp (3) leaps over only the stamp (1), leaving its imprint 3 behind:

%e 1   2   3  (1) (3) (4) (2)     (5) ...

%e Now stamp (1) leaps again, over three others into the gap between stamps (2) and (5). Its print 1 remains:

%e 1   2   3   1  (3) (4) (2) (1) (5) ...

%e (And so on.)

%o (MATLAB)

%o function a = A354223( max_n )

%o   a = [1:max_n];

%o   for n = 1:max_n

%o       m = a(n);

%o       j = a(n+1);

%o       a = [a(1:n+j) m a(n+j+1:end)];

%o   end

%o   a = a(1:max_n);

%o end % _Thomas Scheuerle_, May 20 2022

%Y Cf. A000027, A089309, A090739, A120738, A132045, A195986.

%K nonn

%O 1,2

%A _Tamas Sandor Nagy_, May 19 2022