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%I #7 Dec 19 2023 12:51:38
%S 1,0,2,-1,4,-3,-2,8,-7,-6,-4,3,16,-15,-14,-12,-8,5,6,7,32,-31,-30,-28,
%T -24,-16,-5,9,10,12,13,14,15,64,-63,-62,-60,-56,-48,-32,-13,-11,-10,
%U -9,17,18,20,24,25,26,28,29,30,31,128,-127,-126,-124,-120,-112
%N Irregular triangular array of all the integers, each exactly once, ordered as in Comments.
%C 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.
%H Clark Kimberling, <a href="/A243610/b243610.txt">Table of n, a(n) for n = 1..4000</a>
%H Danielle Cox and Karyn McLellan, <a href="https://www.fq.math.ca/Papers1/55-2/CoxMcLellan021717.pdf">A Problem on Generation Sets Containing Fibonacci Numbers</a>, Fibonacci Quart. 55 (2017), no. 2, 105-113.
%e First 7 rows of the array:
%e 1
%e 0 .... 2
%e -1 ... 4
%e -3 ... -2 ... 8
%e -7 ... -6 ... -4 ... 3 .... 16
%e -15 .. -14 .. -12 .. -8 ... 5 .... 6 ... 7 .. 32
%e -31 .. -30 .. -28 .. -24 .. -16 .. -5 .. 9 .. 10 . 12 . 13 . 14 . 15 . 64
%t z = 12; g[1] = {1}; f1[x_] := 2 x; f2[x_] := 1 - x; h[1] = g[1];
%t b[n_] := b[n] = DeleteDuplicates[Union[f1[g[n - 1]], f2[g[n - 1]]]];
%t h[n_] := h[n] = Union[h[n - 1], g[n - 1]];
%t g[n_] := g[n] = Complement [b[n], Intersection[b[n], h[n]]]
%t u = Table[g[n], {n, 1, 12}]
%t v = Flatten[u]
%Y Cf. A243571, A000045.
%K easy,tabf,sign
%O 1,3
%A _Clark Kimberling_, Jun 08 2014
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