A233695 - OEIS

%I #37 Nov 06 2018 03:59:39

%S 10,18,30,56,109,219,450,933,1946,4071,8516,17823,37310,78112,163551,

%T 342461,717083,1501509,3144031,6583341,13784976

%N a(n) gives the 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 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 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

%H Jack W Grahl, <a href="/A233695/a233695.hs.txt">Haskell code to generate this sequence</a>

%e 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.

%t 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 t2 = Flatten[Position[x, _?(IntegerQ[#] && Negative[#] &)]] (* A233695 *)

%t t = Union[t1, t2]  (* A233696 *)

%t (* _Peter J. C. Moses_, Dec 21 2013 *)

%Y Cf. A233694, A233696, A232559, A226130, A232723, A226080.

%K nonn,more

%O 1,1

%A _Clark Kimberling_, Dec 19 2013

%E More terms by _Jack W Grahl_, Aug 10 2018