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A324680 Starting at n, a(n) is the largest distance from zero among all positions 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. 2
0, 0, 0, 0, 0, 0, 0, 3442, 0, 0, 0, 27, 140, 139, 0, 0, 84, 3072845, 0, 0, 638385, 0, 0, 0, 0, 4869724, 0, 0, 0, 464, 43807680, 2117461, 2117462, 2117463, 2117464, 0, 0, 24, 696919, 696918, 179, 1, 0, 1, 1920, 0, 148, 86, 85, 84, 83, 190, 63, 0, 0, 0, 1107 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,8

LINKS

Table of n, a(n) for n=0..56.

David Nacin, A324680

David Nacin, A324680(n)/A228474(n)

EXAMPLE

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 has the largest distance from zero, thus a(11)=27.

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([abs(i) for i in d['stuck']])

CROSSREFS

Cf. A228474, A324660-A324692.

Sequence in context: A254093 A254086 A324679 * A273341 A107537 A260500

Adjacent sequences:  A324677 A324678 A324679 * A324681 A324682 A324683

KEYWORD

nonn

AUTHOR

David Nacin, Mar 10 2019

STATUS

approved

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Last modified October 20 13:41 EDT 2019. Contains 328257 sequences. (Running on oeis4.)