A243610 Irregular triangular array of all the integers, each exactly once, ordered as in Comments.

1, 0, 2, -1, 4, -3, -2, 8, -7, -6, -4, 3, 16, -15, -14, -12, -8, 5, 6, 7, 32, -31, -30, -28, -24, -16, -5, 9, 10, 12, 13, 14, 15, 64, -63, -62, -60, -56, -48, -32, -13, -11, -10, -9, 17, 18, 20, 24, 25, 26, 28, 29, 30, 31, 128, -127, -126, -124, -120, -112

Offset: 1

Author

Clark Kimberling, Jun 08 2014

Keywords

• easy
• tabf
• sign

Comments

Let F = A000045 (the Fibonacci numbers). To construct the array, decree the first 4 rows as in the Example. Thereafter, row n consists of F(n) numbers in increasing order, generated as follows: the F(n-1) numbers 2*x from x in row n-1, together with the F(n-2) numbers 1 - 2*x from numbers x in row n-2. For n >= 3, row n consists of F(n-1) negative integers and F(n-2) positive integers; also, row n consists of F(n-1) even integers and F(n-2) odd integers. Conjecture: Every row contains F(k) or -F(k) for some k.

Examples

			First 7 rows of the array:
1
0 .... 2
-1 ... 4
-3 ... -2 ... 8
-7 ... -6 ... -4 ... 3 .... 16
-15 .. -14 .. -12 .. -8 ... 5 .... 6 ... 7 .. 32
-31 .. -30 .. -28 .. -24 .. -16 .. -5 .. 9 .. 10 . 12 . 13 . 14 . 15 . 64
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..4000
• Danielle Cox and Karyn McLellan, A Problem on Generation Sets Containing Fibonacci Numbers, Fibonacci Quart. 55 (2017), no. 2, 105-113.

Crossrefs

Cf. A243571, A000045.

Programs

Mathematica

z = 12; g[1] = {1}; f1[x_] := 2 x; f2[x_] := 1 - 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, 12}]
v = Flatten[u]