cp's OEIS Frontend

A369370 Numerator of the maximum expected number of steps of a random walk on the cells of the triangular lattice before it lands on a mined cell, given that all but n cells are mined.

%I A369370 #11 Jan 28 2024 09:20:35
%S A369370 0,1,3,15,3,156,15,1284,87,642,172,2189,149,15,2865,215,87
%N A369370 Numerator of the maximum expected number of steps of a random walk on the cells of the triangular lattice before it lands on a mined cell, given that all but n cells are mined.
%C A369370 For all n <= 16 except n = 7, 8, and 15, the optimal placement of the mine-free cells is unique up to rotations and reflections of the lattice (leaving the starting cell fixed). The three exceptional cases all have two optimal placements. For n = 7, the two optimal placements have the same underlying graph, but that is not the case for n = 8 and n = 15. See linked illustration.
%H A369370 Pontus von Brömssen, <a href="/A369370/a369370.svg">Illustration of the optimal mine-free cells for n = 1..16</a>. (The random walk starts at the black dot.)
%H A369370 Pontus von Brömssen, <a href="https://oeis.org/plot2a?name1=A369370&amp;name2=A369371&amp;tform1=untransformed&amp;tform2=untransformed&amp;shift=0&amp;radiop1=ratio&amp;drawpoints=true">Plot of a(n)/A369371(n) vs n</a>, using Plot2.
%e A369370 For n = 0, the random walk stops before it can take any step, so a(0) = 0.
%e A369370 For n = 1, only the starting cell can be swept, so the random walk always stops after 1 step and a(1) = 1.
%e A369370 For n = 2, we can sweep the starting cell and one adjacent cell. The random walk then has probability 1/3 of surviving at each step, which implies that the expected number of steps is 3/2, so a(2) = 3. (The number of steps follows a geometric distribution.)
%e A369370 For n = 3, the best strategy is to sweep the starting cell and two of its neighboring cells. Let x be the expected length of the random walk with the given starting cell, and let y be the expected length of a random walk starting at one of the other two cells. By conditioning on the first step, it follows that the equations x = 1 + y*2/3 and y = 1 + x/3 hold, giving x = 15/7 and a(3) = 15.
%e A369370 See linked illustration for optimal solutions for 1 <= n <= 16.
%Y A369370 Cf. A369371 (denominators), A366998 (square lattice), A369368 (hexagonal lattice).
%K A369370 nonn,frac,more
%O A369370 0,3
%A A369370 _Pontus von Brömssen_, Jan 24 2024