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Starting at n, a(n) is the number of times we move away from zero from a positive position 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.
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%I #5 Mar 11 2019 20:46:20

%S 0,0,0,0,1,2,0,331,0,1,0,5,17,18,0,0,24,146798,0,1,33885,0,2,0,1,

%T 175429,0,1,0,102,2658893,137573,137573,137574,137575,0,0,18,49239,

%U 49239,45,0,0,1,169,0,17,11,12,7,8,40,3,0,1,0,149,149,149,92857,92857

%N Starting at n, a(n) is the number of times we move away from zero from a positive position 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=11, the points visited are 11, 10, 8, 5, 1, -4, 2, -5, 3, -6, 4, -7, -19, -32, -18, -3, 13, 30, 12, 31, 51, 72, 50, 27, 51, 26, 0. Of these, we move away from zero at -7, -19, 13, 12, 31, 51, and 27. Five of these are positive and thus a(11)=5.

%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 sgn(x):

%o if x:

%o return x//abs(x)

%o return 0

%o #Actual sequence

%o def a(n):

%o d = trip(n)

%o return sum(1 for i in d['flee'] if i > 0)

%Y Cf. A228474, A324660-A324692.

%K nonn

%O 0,6

%A _David Nacin_, Mar 10 2019