OFFSET
1,1
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.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
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).
EXAMPLE
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.
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Dec 01 2013
STATUS
approved