A355080 - OEIS

%I #40 Jun 22 2022 20:41:30

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

%T 2,3,5,6,4,7,7,1,7,2,6,7,5,3,4,7,8,8,6,8,1,5,8,7,2,3,8,6,4,7,9,9,8,9,

%U 5,1,9,2,6,8,9,7,3,4,5,9,10,10,8,10,1,7,10,6

%N Start with the positive integers. Term by term from left to right insert a copy of the current term x m steps further, where m is the number of times x has appeared.

%C We start with the sequence of positive integers 1, 2, 3, ... . We process number by number from left to right. If the number is found the first time in the sequence, we will place a copy of it directly after, by shifting the remaining part of the sequence to the right. If we have already seen this number m times, we will place a copy of this number after skipping m numbers on the right.

%C We could use the term ordinal instead of number here, as the numerical value itself is not important, only the ordering.

%C This sequence was inspired by sequence A354223 from _Tamas Sandor Nagy_, which shares the idea to start with a predefined sequence and to insert copies ahead of element-wise evaluation.

%C A mysterious constant C:

%C The indices where this sequence reaches the next greater number for the first time are roughly approximated by a parabola: a(floor(b+(1/C)*n^2)) = 1, 2, 3, ... .

%C a(k) approximates round(sqrt(C*k)) if we choose for k the indices where a new number appears the first time in this sequence.

%C For each number in this sequence the indices of the appearance can be roughly approximated by some polynomial b+(1/C)*n^2, where b is some individual constant for each number, but C always appears to be the same constant, known thus far to be 1.1738... . The author used the value of sqrt(2/u), where u is Soldner's constant, with very good results, but there is yet not any evidence known that Soldner's constant has any relation to this sequence. Can we estimate C more accurately? Can we find an expression or series to describe C?

%C _Tamas Sandor Nagy_ noticed that the value 1 appears exactly once between the first appearances of any two consecutive record values. He further noticed that if we break this sequence up into an irregular triangle in which each record value starts a new row, we will observe columns (A000124) which show a progression with the row number. See example section for details.

%C The mean value of the rows mentioned above as a function of the row index r is approximately r/(Pi*log(2)^2) - 1/2.

%H Thomas Scheuerle, <a href="/A355080/b355080.txt">Table of n, a(n) for n = 1..10000</a>

%H Thomas Scheuerle, <a href="/A355080/a355080.png">Scatter plot of the first 10000 values</a>

%H Thomas Scheuerle, <a href="/A355080/a355080_2.png">y = k - (1/C)*(x-0.283)^2, where k is the least k such that a(k) = x</a>. For C we chose sqrt(2/u) with u = 1.45... (A070769).

%H Thomas Scheuerle, <a href="/A355080/a355080_3.png">y = k - (1/C)*(x-1.275)^2, where k is such that a(k) = 1 for the x-th appearance</a>. For C we chose sqrt(2/u) with u = 1.45... (A070769).

%e Step-by-step development of the sequence is as follows; the asterisk marks the actual term that will be processed:

%e * 1 was previously seen 0 times -> insert directly after.

%e 1,2,3,4,5,6,7,8,9,10

%e 1,1,2,3,4,5,6,7,8,9,10

%e   * 1 was previously seen once -> insert one later.

%e 1,1,2,3,4,5,6,7,8,9,10

%e 1,1,2,1,3,4,5,6,7,8,9,10

%e     * 2 was previously seen 0 times -> insert directly after.

%e 1,1,2,1,3,4,5,6,7,8,9,10

%e 1,1,2,2,1,3,4,5,6,7,8,9,10

%e       * 2 was previously seen once -> insert one later.

%e 1,1,2,2,1,3,4,5,6,7,8,9,10

%e 1,1,2,2,1,2,3,4,5,6,7,8,9,10

%e         * 1 was previously seen twice -> insert two later.

%e 1,1,2,2,1,2,3,4,5,6,7,8,9,10

%e 1,1,2,2,1,2,3,1,4,5,6,7,8,9,10

%e .

%e This sequence written as an irregular triangle:

%e *  *

%e 1, 1     *

%e 2, 2, 1, 2

%e 3, 3, 1, 3, 2     *

%e 4, 4, 3, 4, 1, 2, 4, 3

%e 5, 5, 1, 5, 4, 2, 5, 3        *

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

%e Each column below an asterisk shows a linear progression.

%o (MATLAB)

%o function a = A355080( max_n )

%o     a = 1:max_n;

%o     for n = 1:max_n

%o         j = length(find(a(1:n) == a(n)));

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

%o     end

%o     a = a(1:max_n);

%o end

%Y Cf. A000124, A354223.

%K nonn,hear

%O 1,3

%A _Thomas Scheuerle_, Jun 18 2022