

A191437


Dispersion of ([nx+n+x2]), where x=(golden ratio) and [ ]=floor, by antidiagonals.


1



1, 3, 2, 8, 5, 4, 21, 13, 11, 6, 55, 34, 29, 16, 7, 144, 89, 76, 42, 18, 9, 377, 233, 199, 110, 47, 24, 10, 987, 610, 521, 288, 123, 63, 26, 12, 2584, 1597, 1364, 754, 322, 165, 68, 32, 14, 6765, 4181, 3571, 1974, 843, 432, 178, 84, 37, 15, 17711, 10946
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OFFSET

1,2


COMMENTS

Rows 1 and 2: Fibonacci numbers. Rows 3 and 5: Lucas numbers. Row n satisfies the recurrence x(n)=3*x(n1)x(n2).
Background discussion: Suppose that s is an increasing sequence of positive integers, that the complement t of s is infinite, and that t(1)=1. The dispersion of s is the array D whose nth row is (t(n), s(t(n)), s(s(t(n)), s(s(s(t(n)))), ...). Every positive integer occurs exactly once in D, so that, as a sequence, D is a permutation of the positive integers. The sequence u given by u(n)=(number of the row of D that contains n) is a fractal sequence. Examples:


LINKS



EXAMPLE

Northwest corner:
1....3....8....21...55
2....5....13...34...89
4....11...29...76...199
6....16...42...110..288
7....18...47...123..322


MATHEMATICA

(* Program generates the dispersion array T of increasing sequence f[n] *)
r = 40; r1 = 12; c = 40; c1 = 12; x = GoldenRatio;
f[n_] := Floor[n*x+n+x2] (* complement of column 1 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]]
Flatten[Table[t[k, n  k + 1], {n, 1, c1}, {k, 1, n}]] (* A191437 sequence *)


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KEYWORD



AUTHOR



STATUS

approved



