A232867 Positions of the negative integers in the sequence (or tree) of complex numbers generated by these rules: 0 is in S, and if x is in S, then x + 1 and i*x are in S, where duplicates are deleted as they occur.

8, 12, 19, 30, 45, 64, 87, 114, 145, 180, 219, 262, 309, 360, 415, 474, 537, 604, 675, 750, 829, 912, 999, 1090, 1185, 1284, 1387, 1494, 1605, 1720, 1839, 1962, 2089, 2220, 2355, 2494, 2637, 2784, 2935, 3090, 3249, 3412, 3579, 3750, 3925, 4104, 4287, 4474

Offset: 1

Author

Clark Kimberling, Dec 01 2013

Keywords

• nonn
• easy

Comments

Let S be the sequence (or tree) of complex numbers defined by these rules: 0 is in S, and if x is in S, then x + 1, and i*x are in S. Deleting duplicates as they occur, the generations of S are given by g(1) = (0), g(2) = (1), g(3) = (2,i), g(4) = (3, 2i, 1+i, -1), ... Concatenating these gives 0, 1, 2, i, 3, 2*i, 1 + i, -1, 4, 3*i, 1 + 2*i, -2, 2 + i, -1 + i, -i, 5, ... It appears that if c and d are integers, than the positions of c*n+d*i, for n>=0, comprise a linear recurrence sequence with signature beginning with 3, -3, 1, following for zero or more 0's.

Examples

			Each x begets x + 1, and i*x, but if either these has already occurred it is deleted.  Thus, 0 begets (1); then 1 begets (2,i,); then 2 begets 3 and 2*i, and i begets 1 + i and -1, so that g(4) = (3, 2*i, 1 + i, -1), etc.; a(1) = 8 because -1 occurs in the 8th position of S.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A232559, A232866, A232868.

Programs

Mathematica

x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, x + 1, I*x}]]], {40}]; x;
t1 = Flatten[Table[Position[x, n], {n, 0, 30}]]    (* A232866 *)
t2 = Flatten[Table[Position[x, -n], {n, 1, 30}]]  (* A232867 *)
Union[t1, t2]  (* A232868 *)

Formula

a(n+1) = 2*n^2 + n + 9 for n >= 1 (conjectured).

G.f.: (-8 + 12 x - 7 x^2 - x^3)/(x -1)^3 (conjectured).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 5 (conjectured).