A379403 -id:A379403 - 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 *)

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 *)

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

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

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

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