A324666 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.
0, 1, 2, 2, 7, 9, 3, 786, 6, 8, 4, 18, 50, 52, 8, 5, 71, 336258, 10, 12, 74949, 6, 21, 13, 15, 438113, 12, 14, 7, 245, 6219115, 299928, 299928, 299930, 299932, 14, 8, 59, 103544, 103544, 125, 16, 16, 18, 423, 9, 62, 48, 50, 37, 39, 106, 28, 18, 20, 10, 363
Offset: 0
Keywords
Examples
For n=2, the points visited are 2,1,-1,-4,0. As exactly two of these are positive, we have a(2)=2.
Links
- David Nacin, A324666(n)/A228474(n)
Programs
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Python
#Sequences A324660-A324692 generated by manipulating this trip function #spots - positions in order with possible repetition #flee - positions from which we move away from zero with possible repetition #stuck - positions from which we move to a spot already visited with possible repetition def trip(n): stucklist = list() spotsvisited = [n] leavingspots = list() turn = 0 forbidden = {n} while n != 0: turn += 1 sign = n // abs(n) st = sign * turn if n - st not in forbidden: n = n - st else: leavingspots.append(n) if n + st in forbidden: stucklist.append(n) n = n + st spotsvisited.append(n) forbidden.add(n) return {'stuck':stucklist, 'spots':spotsvisited, 'turns':turn, 'flee':leavingspots} #Actual sequence def a(n): d = trip(n) return sum(1 for i in d['spots'] if i > 0)