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A324672
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Starting at n, a(n) is the sign of the farthest point from zero 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. In case of a tie, a(n)=0.
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1
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0, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, 1, 1, 1, 1, -1, 1, 1, -1, -1, 1, -1, -1, -1, -1, 1, -1, 1, 1, -1, -1, 1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, -1, -1, -1, 1, -1, -1, -1, 1, 1, 1, 1
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OFFSET
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0
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COMMENTS
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The only known value for which there is a tie (up to n=1000) happens at n=89.
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LINKS
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EXAMPLE
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For n=2, the points visited are 2,1,-1,-4,0. The farthest point from zero is -4, and sgn(-4) = -1, thus a(2) = -1.
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PROG
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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}
def sgn(x):
if x:
return x//abs(x)
return 0
#Actual sequence
def a(n):
d = trip(n)
mx=max(d['spots'])
mn=min(d['spots'])
return sgn(mx+mn)
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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