|
%I #5 Mar 11 2019 20:46:14
%S 0,0,1,0,8,8,0,409,2,2,0,2,31,31,2,0,15,201444,3,3,25388,0,10,5,5,
%T 347717,3,3,0,75,874341,103653,103652,103652,103652,3,0,10,52376,
%U 52375,48,5,4,4,251,0,32,23,23,16,16,44,10,4,4,0,113,112,111,129093
%N Starting at n, a(n) is the number of times we move away from zero from a negative 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. Two of these are negative and thus a(11)=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 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 0,5
%A _David Nacin_, Mar 10 2019
|
