%I
%S 0,0,1,0,2,1,0,1,2,1,0,3,2,1,1,0,2,1,3,2,1,0,1,3,2,
%T 1,2,1,0,6,1,1,3,2,1,1,0,4,1,1,1,1,3,2,1,0,1,2,1,
%U 3,2,1,1,2,1,0,1,1,6,1,4,3,2,1,1,1,0,3
%N Starting at n, a(n) is the maximum negative point visited, or zero if no negative points are visited, according to the following rules. On the kth step (k=1,2,3,...) move a distance of k in the direction of zero. If the number landed on has been landed on before, move a distance of k away from zero instead.
%e For n=2, the points visited are 2,1,1,4,0. As 1 is the larger of the two negative values, a(2) = 1.
%o (Python)
%o #Sequences A324660A324692 generated by manipulating this trip function
%o #spots  positions in order with possible repetition
%o #flee  positions from which we move away from zero with possible repetition
%o #stuck  positions from which we move to a spot already visited with possible repetition
%o def trip(n):
%o stucklist = list()
%o spotsvisited = [n]
%o leavingspots = list()
%o turn = 0
%o forbidden = {n}
%o while n != 0:
%o turn += 1
%o sign = n // abs(n)
%o st = sign * turn
%o if n  st not in forbidden:
%o n = n  st
%o else:
%o leavingspots.append(n)
%o if n + st in forbidden:
%o stucklist.append(n)
%o n = n + st
%o spotsvisited.append(n)
%o forbidden.add(n)
%o return {'stuck':stucklist, 'spots':spotsvisited,
%o 'turns':turn, 'flee':leavingspots}
%o def maxorzero(x):
%o if x:
%o return max(x)
%o return 0
%o #Actual sequence
%o def a(n):
%o d = trip(n)
%o return maxorzero([i for i in d['spots'] if i < 0])
%Y Cf. A228474, A324660A324692.
%K sign
%O 0,5
%A _David Nacin_, Mar 10 2019
