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%I #15 Mar 29 2023 09:20:24
%S -1,1378,-1,595,66,36,153,758,1185,78,1732,171,2510,2094,1407,253,630,
%T 210,780,2385,1326,300,1225,990,2800,406,3267,4333,4124,528,4309,741,
%U 5951,666,2701,903,30418,820,3321,1081,4186,990,8299,2775,4560,1176,4753,39951,5778
%N 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.
%C 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.
%o (Python)
%o # reformatted by _R. J. Mathar_, 2023-03-29
%o class A352731():
%o def __init__(self,n) :
%o self.n = n
%o self.KM=[(n, 1), (1, n), (-1, n), (-n, 1), (-n, -1), (-1, -n), (1, -n), (n, -1)]
%o @staticmethod
%o def _idx(loc):
%o i, j = loc
%o return (i+j-1)*(i+j-2)//2 + j
%o def _next_move(self,loc, visited):
%o i, j = loc
%o moves = [(i+io, j+jo) for io, jo in self.KM if i+io>0 and j+jo>0]
%o available = [m for m in moves if m not in visited]
%o return min(available, default=None, key=lambda x: A352731._idx(x))
%o def _aseq(self):
%o locs = [[], []]
%o loc, s, turn, alst = [(1, 1), (1, 1)], {(1, 1)}, 0, [1]
%o m = self._next_move(loc[turn], s)
%o while m != None:
%o loc[turn], s, turn, alst = m, s|{m}, 0 , alst + [A352731._idx(m)]
%o locs[turn] += [loc[turn]]
%o m = self._next_move(loc[turn], s)
%o if len(s)%100000 == 0:
%o print(self.n,'{steps} moves in'.format(steps = len(s)))
%o return alst
%o def at(self,n) :
%o if n == 1 or n == 3:
%o return -1
%o else:
%o return self._aseq()[-1]
%o for n in range(1,40):
%o a352731 = A352731(n)
%o print(a352731.at(n))
%Y Cf. A323469, A323471, A352730.
%K sign
%O 1,2
%A _Andrew Smith_, Mar 30 2022
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