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

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

Author

Clark Kimberling, Jan 15 2025

Keywords

• tabl
• nonn

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     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,...)
		

Crossrefs

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

Programs

Mathematica

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 *)