A379047 - OEIS

A379047

Rectangular array read by descending antidiagonals: the Type 2 runlength index array of A000002 (the Kolakoski sequence); see Comments.

1, 3, 2, 5, 4, 8, 6, 11, 28, 13, 7, 16, 35, 80, 53, 9, 18, 48, 121, 217, 112, 10, 22, 62, 135, 449, 332, 305, 12, 26, 67, 175, 472, 1478, 1451, 296, 14, 31, 89, 203, 513, 1974, 1947, 1358, 1331, 15, 38, 94, 244, 812, 2101, 2683, 1920, 1827, 964, 17, 40, 107

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OFFSET

1,2

COMMENTS

We begin with a definition of Type 2 runlength array, V(s), of any sequence s for which all the runs referred to have finite length:

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

(row 0) = s

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

For n>=2,

(row n) = sequence of last 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 2 runlength index array, The runlength index array, VI(s) is now contructed from V(s) in two steps:

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

(2) Then VI(s) results by retaining only n in V*.

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

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

***

Examples: (1) if s is monotonic, then VI(s) has just one row, the positive integers, A000027.

(2) if s = A010060 (Thue-Morse sequence), then VI(s) has exactly two rows: A003159 and A036554. The type 1 runlength index array of s also has exactly two rows: A285385 and A072989.

LINKS

Table of n, a(n) for n=1..58.

EXAMPLE

Corner:

1 3 5 6 7 9 10 12 14 15 17 19

2 4 11 16 18 22 26 31 38 40 44 51

8 28 35 48 62 67 89 94 107 130 150 157

13 80 121 135 175 203 244 359 417 458 499 540

53 217 449 472 513 812 879 1069 1272 1511 1725 1786

112 332 1478 1974 2101 2423 2710 3282 3638 3715 3950 4145

305 1451 1947 2683 2883 3605 3706 3827 4528 4749 4963 5076

296 1358 1920 2590 2850 3542 5745 6400 7103 7567 7796 8346

1331 1827 2491 2805 3437 5652 6373 7769 8265 9315 11508 11738

Using s = A000002 as an example, we have for V*(s):

(row 1) = ((1,1), (3,2), (5,1), (6,2), (7,1), (9,2), (10,1), (12,2), (14,1),...)

c(1) = ((2,2), (4,1), (8,2), (11,2), (13,1), (16,1), (18,2), (22,1), (26,2), ...)

(row 2) = ((2,2), (4,1), (11,2), (16,1), (18,2), (22,1), (26,2), (31,1), (35,2), ...)

c(2) = (8,2), (13,1), (28,1), ...)

(row 3) = (8,2), (28,1),

so that VI(s) has

(row 1) = (1,3,5,6,7,9,10,12, ...)

(row 2) = (2,4,11,16,18,22,26, ...)

(row 3) = (8,28,35,48,62,67,...)

MATHEMATICA

r[seq_] := seq[[Flatten[Position[Append[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, A379046 (Type 1 array).