A379047 - OEIS

%I #11 Dec 20 2024 12:04:11

%S 1,3,2,5,4,8,6,11,28,13,7,16,35,80,53,9,18,48,121,217,112,10,22,62,

%T 135,449,332,305,12,26,67,175,472,1478,1451,296,14,31,89,203,513,1974,

%U 1947,1358,1331,15,38,94,244,812,2101,2683,1920,1827,964,17,40,107

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

%C 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:

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

%C (row 0) = s

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

%C For n>=2,

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

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

%C ***

%C The corresponding Type 2 runlength index array,  The runlength index array, VI(s) is now contructed from V(s) in two steps:

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

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

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

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

%C ***

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

%C (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.

%e Corner:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

%e so that VI(s) has

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

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

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

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

%t 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 *)

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

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

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

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

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

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

%t Grid[t]  (* array *)

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

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

%t (* _Peter J. C. Moses_, Dec 04 2024 *)

%Y Cf. A000002, A379046 (Type 1 array).

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Dec 16 2024