OFFSET
1,1
COMMENTS
It can be proved using the division algorithm for Gaussian integers that S is the set of Gaussian rational numbers: (b + c*i)/d, where b,c,d are integers and d is not 0.
Empirically, it appears that a(n) = A233694(n+2) + 7 for n > 2. It seems clear that positive integers appear for the first time at the start of a new level of the tree. If this is always the case, then the row starting with n will be followed by a row starting n+1, 1/n, ni, followed by a row starting n+2, 1/(n+1), (n+1)i, 1+1/n, n+1, i/(n+1), 1+ni, -i/n, -n. It may be possible to show that of these 9 values, only n+1 has ever appeared before. If so, then -n will always appear exactly 7 places after n + 2 in the sequence. - Jack W Grahl, Aug 10 2018
LINKS
EXAMPLE
The first 16 numbers generated are as follows: 0, 1, 2, i, 3, 1/2, 2 i, 1 + i, -i, -1, 4, 1/3, 3 i, 3/2, i/2, 1 + 2 i. -1 appears in the 10th place, so a(1) = 10.
MATHEMATICA
Off[Power::infy]; x = {0}; Do[x = DeleteDuplicates[Flatten[Transpose[{x, x + 1, 1/x, I*x} /. ComplexInfinity -> 0]]], {18}]; On[Power::infy]; t1 = Flatten[Position[x, _?(IntegerQ[#] && NonNegative[#] &)]] (*A233694*)
t2 = Flatten[Position[x, _?(IntegerQ[#] && Negative[#] &)]] (* A233695 *)
t = Union[t1, t2] (* A233696 *)
(* Peter J. C. Moses, Dec 21 2013 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Clark Kimberling, Dec 19 2013
EXTENSIONS
More terms by Jack W Grahl, Aug 10 2018
STATUS
approved