OFFSET
1,2
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 9,11,16,21,32,37,40,47 are 2,1,2,1,2,1,2,1, respectively.
Conjecture: every column is eventually increasing.
EXAMPLE
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,...)
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Jan 15 2025
STATUS
approved