OFFSET
1,2
COMMENTS
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 n-th 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, as in A248514.
REFERENCES
Clark Kimberling, Fractal sequences and interspersions, Ars Combinatoria 45 (1997) 157-168.
LINKS
Clark Kimberling, Antidiagonals n = 1..60, flattened
Clark Kimberling, Interspersions and dispersions, Proceedings of the American Mathematical Society, 117 (1993) 313-321.
EXAMPLE
Northwest corner:
1 ... 2 ... 3 ... 5 ... 9 .... 17 ... 33
4 ... 8 ... 15 .. 29 .. 57 ... 113 .. 225
6 ... 12 .. 23 .. 45 .. 89 ... 177 .. 353
7 ... 14 .. 27 .. 53 .. 105 .. 209 .. 417
10 .. 20 .. 39 .. 77 .. 153 .. 305 .. 609
MATHEMATICA
r = 40; r1 = 10; (* r = # rows of T, r1 = # rows to show *);
c = 40; c1 = 12; (* c = # cols of T, c1 = # cols to show *);
x = GoldenRatio;
s[n_] := s[n] = If[n < 1, 0, 2 n - Mod[Total[IntegerDigits[n - 1, 2]], 2]];
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]; rows = {NestList[s, 1, c]};
Do[rows = Append[rows, NestList[s, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]]; TableForm[Table[t[i, j], {i, 1, r1}, {j, 1, c1}]]
u = Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A248513 *)
row[i_] := row[i] = Table[t[i, j], {j, 1, c}]
f[n_] := Select[Range[r], MemberQ[row[#], n] &]
v = Flatten[Table[f[n], {n, 1, 200}]] (* A248514 *)
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Oct 08 2014
STATUS
approved