A379046 -id:A379046 - cp's OEIS frontend

A379401 Rectangular array, read by descending antidiagonals: the Type 1 runlength index array of A039701 (primes mod 3); see Comments.

1, 2, 10, 3, 12, 17, 4, 16, 22, 56, 5, 19, 33, 75, 57, 6, 24, 38, 97, 134, 98, 7, 37, 41, 115, 165, 274, 109, 8, 40, 48, 162, 181, 299, 275, 166, 9, 47, 55, 180, 220, 466, 318, 276, 241, 11, 52, 68, 201, 273, 554, 467, 363, 279, 256, 13, 59, 92, 264, 294

Offset: 1

Clark Kimberling, Jan 15 2025

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.
***
Regarding the present array, each row of U(s) splits an increasing sequence of primes according to remainder modulo 3; e.g., in row 2, the remainders of primes in positions 10,12,16,19,24,37 are 2,1,2,1,2,1,2,1, respectively.

			Corner:
      1    2    3     4    5      6     7     8     9    11    13    14
     10   12   16    19   24     37    40    47    52    59    72    74
     17   22   33    38   41     48    55    68    92   101   104   112
     56   75   97   115   162   180   201   264   293   328   359   440
     57  134  165   181   220   273   294   341   360   451   545   623
     98  274  299   466   554   624   661   742   786   836   898   941
    109  275  318   467   555   631   704   749   823   839   903  1046
    166  276  363   500   600   758   824   856   912  1059  1176  1212
    241  279  364   505   601   861   913  1076  1177  1229  1258  1368
    256  510  608   866   964  1077  1180  1533  1645  2006  2156  2215
    421  521  709  1088  1181  2007  2163  2248  2551  2690  2919  3138
    424  522  710  1089  1184  2008  2174  2785  2920  3141  3466  3938
Starting with s = A039701, we have for U*(s):
(row 1) = ((1,1), (2,0), (3,2), (4,2), (5,2), (6,1), (7,2), (8,1), (9,2), ...)
c(1) = ((10,2), (12,1), (16,2), (17,2), (14,1), (17,1), (19,1), (22,1), (24,2), ...)
(row 2) = ((10,2), (12,1), (16,2), (19,1), (24,2), (23,1), (27,2), (29,1), (36,2), ...)
c(2) = ((17,2), (22,1), (33,2), ...)
(row 3) = ((17,2), (22,1), ...)
so that UI(s) has
(row 1) = (1,2,3,4,5,6,7,8,9,11,13, ...)
(row 2) = (10,12,16.19,24, ...)
(row 3) = (17,22,33,...)

Cf. A000040, A039701, A379046, A379402, A379403, A379404.

r[seq_] := seq[[Flatten[Position[Prepend[Differences[seq[[All, 1]]], 1], _?(# != 0 &)]], 2]];  (* Type 1 *)
row[0] = Mod[Prime[Range[4000]], 3];(* A039701 *)
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 *)

A379402 Rectangular array, read by descending antidiagonals: the Type 2 runlength index array of A039701 (primes mod 3); see Comments.

Original entry on oeis.org

1, 2, 9, 3, 11, 15, 4, 16, 18, 54, 5, 21, 23, 58, 91, 6, 32, 36, 102, 110, 205, 7, 37, 39, 129, 160, 272, 194, 8, 40, 46, 161, 167, 419, 271, 139, 10, 47, 55, 174, 238, 499, 416, 260, 86, 12, 56, 73, 245, 273, 597, 496, 359, 257, 357, 13, 67, 96, 274, 292

Offset: 1

Clark Kimberling, Jan 15 2025

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.
***
Regarding the present array, each row of U(s) splits a sequence of primes according to remainder modulo 3; e.g., in row 2, the remainders of primes in positions 9,11,16,21,32,37,40,47 are 2,1,2,1,2,1,2,1, respectively.
Conjecture: every column is eventually increasing.

			Corner:
      1    2    3    4      5     6     7     8    10    12    13    14
      9   11   16   21     32    37    40    47    56    67    71    74
     15   18   23   36     39    46    55    73    96    99   107   111
     54   58  102  129    161   174   245   274   311   326   423   515
     91  110  160  167    238   273   292   321   420   508   598   621
    205  272  419  499    597   618   703   733   813   835   896   932
    194  271  416  496    576   617   702   730   776   834   989  1128
    139  260  359  489    699   713   771   831   988  1127  1173  1190
     86  257  358  464    698   830   987  1124  1164  1185  1251  1298
    357  461  697  829    942  1107  1412  1498  1717  2059  2138  2179
    356  438  889  1062  1714  2046  2137  2176  2551  2820  2927  3270
    291  437  882  1055  1711  2033  2550  2741  2926  3269  3699  3918
Starting with s = A039701, we have for U*(s):
(row 1) = ((1,1), (2,0), (3,2), (4,2), (5,2), (6,1), (7,2), (8,1), (10,2), ...)
c(1) = ((9,2), (11,1), (15,2), (16,2), (18,1), (21,1), (23,1), (32,2), ...)
(row 2) = ((9,2), (11,1), (16,2), (21,1), (36,1), ...)
c(2) = ((15,2), (37,1), ...)
(row 3) = ((15,2), (18,1), (23,2), ...)
so that UI(s) has
(row 1) = (1,2,3,4,5,6,7,8,10,12,13, ...)
(row 2) = (9,11,16.21,32, ...)
(row 3) = (15,18,23,...)

Cf. A000040, A039701, A379046, A379401, A379403, A379404.

r[seq_] := seq[[Flatten[Position[Append[Differences[seq[[All, 1]]], 1], _?(# != 0 &)]], 2]];  (* Type 2 *)
row[0] = Mod[Prime[Range[4000]], 3];(* A039701 *)
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 *)

A379403 Rectangular array, read by descending antidiagonals: the Type 1 runlength index array of A039702 (primes mod 4); see Comments.

Original entry on oeis.org

1, 2, 5, 3, 7, 20, 4, 9, 26, 23, 6, 13, 39, 71, 48, 8, 15, 60, 93, 80, 49, 10, 25, 76, 137, 94, 89, 96, 11, 28, 79, 156, 140, 95, 204, 133, 12, 30, 92, 187, 157, 199, 241, 356, 242, 14, 32, 113, 230, 198, 236, 271, 512, 457, 243, 16, 45, 118, 260, 233, 268

Offset: 1

Clark Kimberling, Jan 15 2025

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.
***
Regarding the present array, each row of U(s) splits an increasing sequence of primes according to remainder modulo 3; e.g., in row 2, the remainders of primes in positions 10,12,16,19,24,37 are 2,1,2,1,2,1,2,1, respectively.

			Corner:
    1    2     3     4     6     8    10    11    12    14    16    17
    5    7     9    13    15    25    28    30    32    45    47    51
   20   26    39    60    76    79    92   113   118   123   132   136
   23   71    93   137   156   187   230   260   283   296   318   326
   48   80    94   140   157   198   233   265   286   343   377   382
   49   89    95   199   236   268   472   595   635   702   732   755
   96  204   241   271   473   600   642   841   899   956  1120  1279
  133  356   512   601   643   844   906   961  1129  1402  1440  1482
  242  457   549   869   921   962  1220  1403  1567  1910  1946  2097
  243  460   566   870  1223  1406  1570  1917  1947  2102  2336  2655
  248  991  1242  1483  1745  2103  2367  2664  2981  3322  3440  3953
  249  992  1247  1484  1750  2118  2368  2667  3042  3323  3455  3956
Starting with s = A039702, we have for U*(s):
(row 1) = ((1,2), (2,3), (3,1), (4,3), (6,1), (8,3), (10,1), (11,3), ...)
c(1) = ((5,3), (7,1), (9,3), (13,1), (15,3), (20,3), (23,3), (25,1), (26,1), ...)
(row 2) = ((5,3), (7,1), (9,3), (13,1), (15,3), (25,1), (28,3), (30,1), (32,3), ...)
c(2) = ((20,3), (23,3), (26,1), ...)
(row 3) = ((20,3), (26,1), ...)
so that UI(s) has
(row 1) = (1,2,3,4,5,6,8,10,11, ...)
(row 2) = (5,7,9,13,15,25, ...)
(row 3) = (20,26,...)

Cf. A039702, A379046, A379401, A379402, A379404.

r[seq_] := seq[[Flatten[Position[Prepend[Differences[seq[[All, 1]]], 1], _?(# != 0 &)]], 2]];
row[0] = Mod[Prime[Range[4000]], 4];(* A039702 *)
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]
w[n_, k_] := t[[n]][[k]];
Table[w[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten
(* Peter J. C. Moses, Dec 04 2024 *)

A379404 Rectangular array, by descending antidiagonals: the Type 2 runlength index array of A039702 (primes mod 4); see Comments.

Original entry on oeis.org

1, 2, 4, 3, 6, 19, 5, 8, 24, 46, 7, 12, 47, 78, 31, 9, 22, 65, 128, 77, 14, 10, 25, 72, 135, 93, 50, 91, 11, 27, 87, 154, 134, 92, 168, 239, 13, 29, 94, 197, 153, 183, 240, 337, 232, 15, 38, 97, 247, 196, 241, 400, 540, 254, 229, 16, 44, 114, 264, 246, 435

Offset: 1

Clark Kimberling, Jan 15 2025

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.
***
Regarding the present array, each row of U(s) splits a sequence of primes according to remainder modulo 3; e.g., in row 2, the remainders of primes in positions 4,6,8,12,22,25,27,29 are 3,1,3,1,3,1,3,1, respectively.
Conjecture: every column is eventually increasing.

			Corner:
    1     2      3      5      7      9    10      11     13     15     16     17
    4     6      8     12     22     25    27      29     38     44     48     59
   19    24     47     65     72     87    94      97    114    121    131    136
   46    78    128    135    154    197   247     264    281    287    303    319
   31    77     93    134    153    196   246     263    280    338    363    378
   14    50     92    183    241    435   546     574    675    691    724    744
   91   168    240    400    543    571   758     834    887   1041   1240   1261
  239   337    540    568    707    833   886    1002   1381   1397   1407   1501
  232   254    674    824    885    987   1380   1500   1811   1883   1976   2280
  229   251    669    986   1377   1481   1802   1882   1971   2271   2444   2911
  626   983   1376   1480   1944   2240   2439   2910   3179   3295   3710   3939
  619   982   1333   1469   1943   2239   2366   2909   3178   3294   3701   3892
Starting with s = A039702, we have for U*(s):
(row 1) = ((1,2), (2,3), (3,1), (4,3), (5,3), (7,1), (9,3), (10,1), ...)
c(1) = ((4,3), (6,1), (8,3), (12,1), (14,3), (19,3), (22,3), (24,1), (25,1), ...)
(row 2) = ((4,3), (6,1), (8,3), (12,1), (22,3), (25,1), (27,3), (29,1) ...)
c(2) = ((14,3), (19,3), (24,1), ...)
(row 3) = ((19,3), (24,1), ...)
so that UI(s) has
(row 1) = (1,2,3,5,7,9,10,11,13, ...)
(row 2) = (4,6,8,12,22,25, ...)
(row 3) = (19,24,47, ...)

Cf. A039702, A379046, A379401, A379402, A379403.

r[seq_] := seq[[Flatten[Position[Append[Differences[seq[[All, 1]]], 1], _?(# != 0 &)]], 2]];  (* Type 2 *)
row[0] = Mod[Prime[Range[4000]], 4];(* A039701 *)
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 *)

A379047 Rectangular array read by descending antidiagonals: the Type 2 runlength index array of A000002 (the Kolakoski sequence); see Comments.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Dec 16 2024

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.

			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,...)

Cf. A000002, A379046 (Type 1 array).

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 *)

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

Original entry on oeis.org

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

Clark Kimberling, Jan 11 2025

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.

			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,...)

Cf. A000002, A379046, A378142.

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 *)

cp's OEIS Frontend

A379401 Rectangular array, read by descending antidiagonals: the Type 1 runlength index array of A039701 (primes mod 3); see Comments.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A379402 Rectangular array, read by descending antidiagonals: the Type 2 runlength index array of A039701 (primes mod 3); see Comments.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A379403 Rectangular array, read by descending antidiagonals: the Type 1 runlength index array of A039702 (primes mod 4); see Comments.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A379404 Rectangular array, by descending antidiagonals: the Type 2 runlength index array of A039702 (primes mod 4); see Comments.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A379047 Rectangular array read by descending antidiagonals: the Type 2 runlength index array of A000002 (the Kolakoski sequence); see Comments.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica