A242364 Irregular triangular array of all the integers ordered as in Comments.

1, 0, 2, -1, 4, -3, -2, 3, 8, -7, -6, -4, 5, 6, 16, -15, -14, -12, -8, -5, 7, 9, 10, 12, 32, -31, -30, -28, -24, -16, -11, -10, -9, 13, 14, 15, 17, 18, 20, 24, 64, -63, -62, -60, -56, -48, -32, -23, -22, -20, -19, -18, -17, -13, 11, 25, 26, 28, 29, 30, 31

Offset: 1

Author

Clark Kimberling, Jun 11 2014

Keywords

• easy
• tabf
• sign

Comments

Let f1(x) = 2x, f2(x) = 1-x, f3(x) = 2-x, g(1) = (1), and g(n) = union(f1(g(n-1)), f2(g(n-1)),f3(g(n-1))) for n >1. Let T be the array whose n-th row consists of the numbers in g(n) arranged in increasing order. It is easy to prove that every integer occurs exactly once in T. Conjectures: (1) |g(n)| = 2*F(n-1) for n >=2, where F = A000045 (the Fibonacci numbers), and exactly half of the numbers in g(n) are positive; (2) the number of even numbers in g(n) is 2*F(n-2) and the number of odd numbers is 2*F(n-3).

Examples

			First 6 rows of the array:
1
0 .... 2
-1 ... 4
-3 ... -2 .... 3 .... 8
-7 ... -6 ... -4 .... 5 .... 6 .... 16
-15 .. -14 .. -12 .. -8 .... -5 ... 7 ... 9 ... 10 ... 12 ... 32
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..3000

Crossrefs

Cf. A242365, A243735, A243736, A242448, A000045.

Programs

Mathematica

z = 12; g[1] = {1}; f1[x_] := 2 x; f2[x_] := 1 - x; f3[x_] := 2 - x; h[1] = g[1]; b[n_] := b[n] = DeleteDuplicates[Union[f1[g[n - 1]], f2[g[n - 1]], f3[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, 9}]
u1 = Flatten[u]  (* A242364 *)
v = Table[Reverse[Drop[g[n], Fibonacci[n - 1]]], {n, 1, z}]
v1 = Flatten[v]  (* A242365 *)
w1 = Table[Apply[Plus, g[n]], {n, 1, 20}]   (* A243735 *)
w2 = Table[Apply[Plus, v[[n]]], {n, 1, 10}] (* A243736 *)