A379404 -id:A379404 - 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 *)

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

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