OFFSET
1,2
COMMENTS
Decree that row 1 is (1), row 2 is (2, 3), and row 3 is (4, 6, 9). Let r(n) = A001590(n+2), so that r(r) = r(n-1) + r(n-2) + r(n-3) with r(1) =1, r(2) = 2, r(3) = 3. Row n of the array, for n >= 4, consists of the numbers, in increasing order, defined as follows: all 3*x from x in row n-1, together with all 1 + 3*x from x in row n-2, together with all 2 + 3*x from x in row n-3. Thus, the number of numbers in row n is r(n), a tribonacci number. Every positive integer occurs exactly once in the array, so that the resulting sequence is a permutation of the positive integers.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1500
EXAMPLE
First 5 rows of the array:
1
2 ... 3
4 ... 6 ... 9
5 ... 7 ... 10 .. 12 .. 18 .. 27
8 ... 11 .. 13 .. 15 .. 19 .. 21 .. 28 .. 30 .. 36 .. 54 .. 81
MATHEMATICA
z = 10; g[1] = {1}; f1[x_] := x + 1; f2[x_] := 3 x; h[1] = g[1];
b[n_] := b[n] = DeleteDuplicates[Union[f1[g[n - 1]], f2[g[n - 1]]]];
h[n_] := h[n] = Union[h[n - 1], g[n - 1]];
g[n_] := g[n] = Complement [b[n], Intersection[b[n], h[n]]]
u = Table[g[n], {n, 1, z}]; v = Flatten[u] (* A243572 *)
CROSSREFS
KEYWORD
nonn,easy,tabf
AUTHOR
Clark Kimberling, Jun 07 2014
STATUS
approved