%I #5 Mar 11 2019 20:45:37
%S 0,0,0,0,0,0,0,3442,0,0,0,27,28,29,0,0,84,2368556,0,0,638385,0,0,0,0,
%T 3642394,0,0,0,464,43807680,2117461,2117462,2117463,2117464,0,0,0,
%U 562491,562492,179,0,0,1,1007,0,104,79,80,0,0,190,0,0,0,0,1107,1108
%N Starting at n, a(n) is the maximum positive 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=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. The only position from which we are forced to revisit a spot is 27, which forces a return to 51. Since this is the only time this happens it is also the maximal positive point for which this happens, thus a(11)=27.
%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 nonn
%O 0,8
%A _David Nacin_, Mar 10 2019
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