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

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

Author

Clark Kimberling, Jan 15 2025

Keywords

• nonn
• tabl

Comments

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.

Examples

			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, ...)
		

Crossrefs

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

Programs

Mathematica

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 *)