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

Author

Clark Kimberling, Jan 15 2025

Keywords

• nonn
• tabl

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.

***

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.

Examples

			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,...)
		

Crossrefs

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

Programs

Mathematica

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 *)