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%I #5 Mar 11 2019 20:45:16
%S 0,0,0,0,0,0,0,-50,0,0,0,0,-15,-14,0,0,0,-145,0,0,-6,0,0,0,0,-34,0,0,
%T 0,-62,-2,-1,-59,-58,-57,0,0,-5,-1,-3,-2,-1,0,0,-26,0,-21,-23,-22,-21,
%U -20,-19,-18,0,0,0,-44,-43,-42,-1,-40,-39,-38,-37,-56,0,0
%N Starting at n, a(n) is the maximal negative position from which a spot must be revisited on the next move, or zero if no such positions exist, according to the following rules. On the k-th 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.
%e For n=41, the points visited are 41, 40, 38, 35, 31, 26, 20, 13, 5, -4, 6, -5, 7, -6, 8, -7, 9, -8, 10, -9, 11, -10, 12, -11, -35, -60, -34, -61, -33, -62, -32, -1, -33, 0. The only position from which we are forced to revisit a spot is -1 which forces a return to -33. As this is the only position and it is negative, it is the maximum negative position and thus a(41)=-1.
%o (Python)
%o #Sequences A324660-A324692 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['stuck'] if i<0])
%Y Cf. A228474, A324660-A324692.
%K sign
%O 0,8
%A _David Nacin_, Mar 10 2019
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