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A324676 Starting at n, a(n) is the maximal negative 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. 0
0, 0, 0, 0, 0, 0, 0, -50, 0, 0, 0, 0, -15, -14, 0, 0, 0, -145, 0, 0, -6, 0, 0, 0, 0, -34, 0, 0, 0, -62, -2, -1, -59, -58, -57, 0, 0, -5, -1, -3, -2, -1, 0, 0, -26, 0, -21, -23, -22, -21, -20, -19, -18, 0, 0, 0, -44, -43, -42, -1, -40, -39, -38, -37, -56, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,8

LINKS

Table of n, a(n) for n=0..66.

EXAMPLE

For n=41, the points visited are 41, 40, 38, 35, 31, 26, 20, 13, 5, -4, 6, -5, 7, -6, 8, -7, 9, -8, 10, -9, 11, -10, 12, -11, -35, -60, -34, -61, -33, -62, -32, -1, -33, 0.  The only position from which we are forced to revisit a spot is -1 which forces a return to -33.  As this is the only position and it is negative, it is the maximum negative position and thus a(41)=-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 maxorzero(x):

    if x:

        return max(x)

    return 0

#Actual sequence

def a(n):

    d=trip(n)

    return maxorzero([i for i in d['stuck'] if i<0])

CROSSREFS

Cf. A228474, A324660-A324692.

Sequence in context: A013843 A303146 A023936 * A015066 A022078 A337330

Adjacent sequences:  A324673 A324674 A324675 * A324677 A324678 A324679

KEYWORD

sign

AUTHOR

David Nacin, Mar 10 2019

STATUS

approved

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Last modified November 29 21:32 EST 2021. Contains 349416 sequences. (Running on oeis4.)