A379046 - OEIS

A379046

Rectangular array read by descending antidiagonals: the Type 1 runlength index array of A000002 (the Kolakoski sequence); see Comments.

1, 2, 3, 4, 5, 12, 6, 9, 17, 32, 7, 14, 39, 66, 93, 8, 19, 52, 125, 134, 257, 10, 23, 57, 154, 318, 351, 378, 11, 27, 71, 194, 512, 639, 702, 471, 13, 29, 84, 216, 553, 1627, 1672, 789, 798, 15, 36, 98, 230, 594, 2141, 2168, 1747, 1960, 825, 16, 41, 111, 309

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

We begin with a definition of Type 1 runlength array, U(s), of a sequence s:

Suppose s is a sequence (finite or infinite), and define rows of U(s) as follows:

(row 0) = s

(row 1) = sequence of 1st terms of runs in (row 0); c(1) = complement of (row 1) in (row 0)

For n>=2,

(row n) = sequence of 1st terms of runs in c(n-1); c(n) = complement of (row n) in (row n-1),

where the process stops if and when c(n) is empty for some n.

***

The corresponding Type 1 runlength index array, UI(s) is now contructed from U(s) in two steps:

(1) Let U*(s) be the array obtaining by repeating the construction of U(s) using (n,s(n)) in place of s(n).

(2) Then UI(s) results by retaining only n in U*.

Thus, loosely speaking, (row n) of UI(s) shows the indices in s of the numbers in (row n) of U(s).

The array UI(s) includes every positive integer exactly once.

LINKS

Table of n, a(n) for n=1..59.

EXAMPLE

Corner:

1 2 4 6 7 8 10 11 13 15 16 18

3 5 9 14 19 23 27 29 36 41 45 49

12 17 39 52 57 71 84 98 111 116 139 161

32 66 125 154 194 216 230 309 430 462 491 526

93 134 318 512 553 594 943 1004 1330 1371 1594 1826

257 351 639 1627 2141 2490 2612 2869 3501 3761 3990 4191

378 702 1672 2168 2896 3564 3806 4017 4218 4935 5054 5418

471 789 1747 2729 2905 3651 4547 6578 6763 7768 7962 8185

798 1960 2756 2932 3660 4574 6659 6936 8368 9370 10296 12393

825 1987 2783 3415 3687 4601 8455 9433 10359 12426 13180 15836

Starting with s = A000002, we have for U*(s):

(row 1) = ((1,1), (2,2), (4,1), (6,2), (7,1), (8,2), (10,1), (11,2), (13,1), ...)

c(1) = ((3,2), (5,1), (9,2), (12,2), (14,1), (17,1), (19,2), (23,1), (27,2), ...)

(row 2) = ((3,2), (5,1), (9,2), (14,1), (19,2), (23,1), (27,2), (29,1), (36,2), ...)

c(2) = (12,2), (17,1), (32,1), (39,2), (52,1), (57,2), (66,2), (71,1), ...)

(row 3) = (12,2), (17,1), (39,2), (52,1), (57,2), (71,1), ...)

so that UI(s) has

(row 1) = (1,2,4,6,7,8,10,11,13,...)

(row 2) = (3,5,9,14,19,...)

(row 3) = (12,17,32,66,...)

MATHEMATICA

r[seq_] := seq[[Flatten[Position[Prepend[Differences[seq[[All, 1]]], 1], _?(# != 0 &)]], 2]]; (* Type 1 *)

row[0] = Prepend[Nest[Flatten[Partition[#, 2] /. {{2, 2} -> {2, 2, 1, 1}, {2, 1} -> {2, 2, 1}, {1, 2} -> {2, 1, 1}, {1, 1} -> {2, 1}}] &, {2, 2}, 24], 1]; (* A000002 *)

row[0] = Transpose[{#, Range[Length[#]]}] &[row[0]];

k = 0; Quiet[While[Head[row[k]] === List, row[k + 1] = row[0][[r[

SortBy[Apply[Complement, Map[row[#] &, Range[0, k]]], #[[2]] &]]]]; k++]];

m = Map[Map[#[[2]] &, row[#]] &, Range[k - 1]];

p[n_] := Take[m[[n]], 12]

t = Table[p[n], {n, 1, 12}]

Grid[t] (* array *)

w[n_, k_] := t[[n]][[k]];

Table[w[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* sequence *)

(* Peter J. C. Moses, Dec 04 2024 *)

CROSSREFS

Cf. A000002, A379047 (Type 2 array).