1,2

Decree that row 1 is (1) and row 2 is (2). For n >= 3, row n consists of numbers in increasing order generated as follows: 2*x for each x in row n-1 together with 1+2*x for each x in row n-2. It is easy to prove that row n consists of F(n) numbers, where F = A000045 (the Fibonacci numbers), and that every positive integer occurs exactly once. Row n has F(n-1) even numbers and F(n-2) odd numbers.

The least and greatest numbers in row n are A083329(n-1) and 2^(n-1), for n >= 1.

For n>=2, row n consists of numbers in increasing order generated as follows: 2*x for each odd x and x+1 and 2*x for each even x in row n-1. - Ya-Ping Lu, Oct 15 2025

Clark Kimberling, Table of n, a(n) for n = 1..1500

First 6 rows of the array:

1

2

3 ... 4

5 ... 6 ... 8

7 ... 9 ... 10 .. 12 .. 16

11 .. 13 .. 14 .. 17 .. 18 .. 20 .. 24 .. 32

z = 10; g[1] = {1}; f1[x_] := x + 1; f2[x_] := 2 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] (* A243571 *)

(Python)

def NextRow(n): return [n+1, 2*n] if n%2 == 0 else [2*n]

R = [1]

for _ in range(9):

print(*R, sep = ', ', end = ', '); L = []

for i in range(len(R)): c = NextRow(R[i]); {L.append(c[j]) for j in range(len(c))}

L.sort(); R = L # Ya-Ping Lu, Oct 15 2025

Cf. A232559, A243572, A243573, A000045.

Cf. A052955 for the first element in each row.

Sequence in context: A362178 A358122 A334111 * A385818 A215366 A333483

Adjacent sequences: A243568 A243569 A243570 * A243572 A243573 A243574

nonn,easy,tabf

Clark Kimberling, Jun 07 2014

approved