A379536 Rectangular array, read by descending antidiagonals: the Type 1 runlength index array of A378142; see Comments.

1, 6, 2, 7, 12, 3, 11, 14, 18, 4, 13, 17, 21, 25, 5, 16, 20, 24, 39, 28, 8, 19, 23, 36, 55, 40, 29, 9, 22, 35, 50, 72, 56, 41, 30, 10, 26, 49, 71, 92, 73, 61, 42, 31, 15, 27, 52, 87, 103, 93, 78, 62, 45, 32, 33, 34, 54, 102, 124, 104, 94, 79, 65, 46, 47, 166, 37, 58, 113, 135, 125, 105, 97, 84, 66, 99, 179, 618

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..78.

Example

Corner:
     1    6    7    11    13    16    19    22    26    27    34    37
     2   12   14    17    20    23    35    49    52    54    58    60
     3   18   21    24    36    50    71    87   102   113   116   119
     4   25   39    55    72    92   103   124   135   157   170   187
     5   28   40    56    73    93   104   125   136   160   171   188
     8   29   41    61    78    94   105   128   137   161   172   193
     9   30   42    62    79    97   108   129   140   162   173   194
    10   31   45    65    84    98   109   130   141   163   174   197
    15   32   46    66   110   131   142   164   177   198   216   231
    33   47   99   147   165   178   199   248   297   310   333   417
   166  179  232   285   298   311   498   549   564   581   631   750
   618  830  882  1262  1342  1561  1976  3056  3767  4616  5459  6112
Starting with s = A000002, we have for U*(s):
(row 1) = ((1,1), (2,1), (3,1), (4,1), (5,1), (6,0), (7,1), (8,1), (9,1), (10,1), (11,0) ...)
c(1) = ((2,1), (3,1), (4,1), (5,1), (8,1), (9,1), (10,1), (12,0), (14,1), (15,1), ...)
(row 2) = ((2,1), (12,2), (14,1), (17,0), (20,1), (22,0), (34,1), ...)
c(2) = ((3,1), (4,1), (5,1), (8,1), (9,1), (10,1), (15,1), (18,0), ...)
(row 3) = ((3,1), (18,0), (21,1), (24,0), ...)
so that UI(s) has
(row 1) = (1,6,7,11,13,16,19....)
(row 2) = (2,12,14,17,20,23,...)
(row 3) = (3,18,21,24,36,...)

Mathematica

r[seq_] := seq[[Flatten[Position[Prepend[Differences[seq[[All, 1]]], 1], _?(# != 0 &)]], 2]];
z = 8000; r1 = 2^(1/4); s1 = 2^(1/2); t1 = 2^(3/4);
row[0] = Table[Floor[n (r1 + t1)/s1] - Floor[n  r1/s1] - Floor[n  t1/s1], {n, 1, z}];
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]];
zz = 12
p[n_] := Take[m[[n]], zz]
t = Table[p[n], {n, 1, zz}]
Grid[t]  (*array*)
w[n_, k_] := t[[n]][[k]];
Table[w[n - k + 1, k], {n, zz {k, n, 1, -1}] // Flatten  (*sequence*)
(*_Peter J.C.Moses_, Dec 04 2024*)

Crossrefs

Cf. A000002, A379046, A378142.

Sequence in context: A307086 A021090 A368669 * A270138 A177889 A086744

Adjacent sequences: A379533 A379534 A379535 * A379537 A379538 A379539

Keyword

nonn,tabl,new

Author

Clark Kimberling, Jan 11 2025

Status

approved