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A324682
Starting at n, a(n) is the sign of the closest nonzero point to zero visited for which the next move is a step away from zero, 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. In the case of a tie, or if the set of points is empty, we set a(n)=0.
0
0, 0, -1, 0, -1, 1, 0, -1, -1, 1, 0, -1, -1, 1, -1, 0, 1, 0, -1, 1, 1, 0, -1, -1, 1, 0, -1, 1, 0, -1, -1, -1, -1, 1, 1, -1, 0, -1, -1, -1, -1, -1, -1, 1, 1, 0, -1, -1, 1, -1, 1, 1, -1, -1, 1, 0, -1, -1, -1, -1, -1, 1, 1, 1, -1, -1, 0, -1, 1, 0, -1, -1, -1
OFFSET
0,1
EXAMPLE
For n=9, the points visited are 9, 8, 6, 3, -1, 4, -2, 5, -3, -12, -22, -11, 1, 14, 0. The three times moves are made away from zero happen at -3, -12 and 1. The closest of these is to zero is 1 and thus a(9) = sgn(1) = 1.
PROG
(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}
def sgn(x):
if x:
return x//abs(x)
return 0
def maxorzero(x):
if x:
return max(x)
return 0
def minorzero(x):
if x:
return min(x)
return 0
#Actual sequence
def a(n):
d = trip(n)
neg=maxorzero([i for i in d['flee'] if i < 0])
pos=minorzero([i for i in d['flee'] if i > 0])
if neg and not pos:
return -1
return -sgn(neg+pos)
CROSSREFS
KEYWORD
sign
AUTHOR
David Nacin, Mar 10 2019
STATUS
approved