cp's OEIS Frontend

A233694 Position of n in the sequence (or tree) S generated in order by these rules: 0 is in S; if x is in S then x + 1 is in S; if nonzero x is in S then 1/x is in S; if x is in S, then i*x is in S; where duplicates are deleted as they occur.

%I A233694 #25 Aug 26 2018 00:55:50
%S A233694 1,2,3,5,11,23,49,102,212,443,926,1939,4064,8509,17816,37303,78105,
%T A233694 163544,342454,717076,1501502,3144024,6583334,13784969
%N A233694 Position of n in the sequence (or tree) S generated in order by these rules: 0 is in S; if x is in S then x + 1 is in S; if nonzero x is in S then 1/x is in S; if x is in S, then i*x is in S; where duplicates are deleted as they occur.
%C A233694 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.
%C A233694 The differences of this sequence give the number of elements in each level of the tree. This means that d(n) = a(n) - a(n-1) is at least 1, and is bounded by 3*d(n-1), since there are three times as many elements in each level, before we exclude repetitions. - _Jack W Grahl_, Aug 10 2018
%e A233694 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. The positions of the nonnegative integers are 1, 2, 3, 5, 11.
%t A233694 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 *)
%t A233694 t2 = Flatten[Position[x, _?(IntegerQ[#] && Negative[#] &)]]  (* A233695 *)
%t A233694 t = Union[t1, t2]  (* A233696 *)
%t A233694 (* _Peter J. C. Moses_, Dec 21 2013 *)
%Y A233694 Cf. A233695, A233696, A232559, A226130, A232723, A226080.
%K A233694 nonn,more
%O A233694 0,2
%A A233694 _Clark Kimberling_, Dec 19 2013
%E A233694 More terms from _Jack W Grahl_, Aug 10 2018