A352731 On a diagonally numbered square grid, with labels starting at 1, this is the number of the last cell that a (1,n) leaper reaches before getting trapped when moving to the lowest available unvisited square, or -1 if it never gets trapped.

-1, 1378, -1, 595, 66, 36, 153, 758, 1185, 78, 1732, 171, 2510, 2094, 1407, 253, 630, 210, 780, 2385, 1326, 300, 1225, 990, 2800, 406, 3267, 4333, 4124, 528, 4309, 741, 5951, 666, 2701, 903, 30418, 820, 3321, 1081, 4186, 990, 8299, 2775, 4560, 1176, 4753, 39951, 5778

Offset

1,2

Comments

A (1,2) leaper is a chess knight. (1,1) and (1,3) leapers both never get trapped. This is understandable for the (1,1) leaper but not so much for the (1,3) which does get trapped on the spirally numbered board (see A323471). Once the (1,3) leaper reaches 39 it then performs the same set of 4 moves repeatedly, meaning that it never gets trapped.

Links

Table of n, a(n) for n=1..49.

Prog

(Python)
# reformatted by R. J. Mathar, 2023-03-29
class A352731():
    def __init__(self, n) :
        self.n = n
        self.KM=[(n, 1), (1, n), (-1, n), (-n, 1), (-n, -1), (-1, -n), (1, -n), (n, -1)]
    @staticmethod
    def _idx(loc):
        i, j = loc
        return (i+j-1)*(i+j-2)//2 + j
    def _next_move(self, loc, visited):
        i, j = loc
        moves = [(i+io, j+jo) for io, jo in self.KM if i+io>0 and j+jo>0]
        available = [m for m in moves if m not in visited]
        return min(available, default=None, key=lambda x: A352731._idx(x))
    def _aseq(self):
        locs = [[], []]
        loc, s, turn, alst = [(1, 1), (1, 1)], {(1, 1)}, 0, [1]
        m = self._next_move(loc[turn], s)
        while m != None:
            loc[turn], s, turn, alst = m, s|{m}, 0 , alst + [A352731._idx(m)]
            locs[turn] += [loc[turn]]
            m = self._next_move(loc[turn], s)
            if len(s)%100000 == 0:
                print(self.n, '{steps} moves in'.format(steps = len(s)))
        return alst
    def at(self, n) :
        if n == 1 or n == 3:
            return -1
        else:
            return self._aseq()[-1]
for n in range(1, 40):
    a352731 = A352731(n)
    print(a352731.at(n))

Crossrefs

Cf. A323469, A323471, A352730.

Sequence in context: A345515 A345768 A323815 * A307422 A256627 A260368

Adjacent sequences: A352728 A352729 A352730 * A352732 A352733 A352734

Keyword

sign

Author

Andrew Smith, Mar 30 2022

Status

approved