%I #14 May 30 2024 16:00:05
%S 0,1,2,2,7,9,3,786,6,8,4,18,50,52,8,5,71,336258,10,12,74949,6,21,13,
%T 15,438113,12,14,7,245,6219115,299928,299928,299930,299932,14,8,59,
%U 103544,103544,125,16,16,18,423,9,62,48,50,37,39,106,28,18,20,10,363
%N Starting at n, a(n) is the total number of positive positions visited 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 from zero instead.
%H David Nacin, <a href="/A324666/a324666.png">A324666(n)/A228474(n)</a>
%e For n=2, the points visited are 2,1,-1,-4,0. As exactly two of these are positive, we have a(2)=2.
%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 #Actual sequence
%o def a(n):
%o d = trip(n)
%o return sum(1 for i in d['spots'] if i > 0)
%Y Cf. A228474, A324660-A324692. Equals A228474 - A324665.
%K nonn
%O 0,3
%A _David Nacin_, Mar 10 2019