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%I #16 Mar 11 2022 07:42:51
%S 0,1,1,3,3,3,3,5,5,5,5,7,7,7,7,9,9,9,9,9,9,9,9,11,11,11,11,13,13,13,
%T 13,15,15,15,15,15,15,15,15,15,15,15,15,17,17,17,17,19,19,19,19,19,19,
%U 19,19,19,19,19,19,21,21,21,21,23,23,23,23,23,23,23,23
%N Consider a 2D sandpile model where each site with 2 or more grains, say at location (x, y), topples and transfers one grain of sand to the sites at locations (x+1, y) and (x, y+1). Let S(n) be the configuration after stabilization of a configuration with n grains at the origin. a(n) = Max_{ (x,y) in S(n) } (x+y).
%C Sites containing 0 or 1 grain are stable. S(n) contains n elements.
%H Rémy Sigrist, <a href="/A352226/b352226.txt">Table of n, a(n) for n = 1..10000</a>
%H Rémy Sigrist, <a href="/A352226/a352226.png">Representation of the configuration for n = 100000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Abelian_sandpile_model#Sandpile_models_on_directed_graphs">Sandpile models on directed graphs</a>
%e For n = 15:
%e - S(15) corresponds to the following configuration:
%e 4| X X X
%e 3|X X X
%e 2|X X X
%e 1|X X
%e 0|X X X X
%e +---------
%e 0 1 2 3 4
%e - x+y is maximized for (x,y) = (4,3) and (3,4),
%e - so a(15) = 3+4 = 7.
%o (PARI) a(n) = { my (s=[n]); for (k=-1, oo, if (vecmax(s)==0, return (k), s \= 2; s = concat(0, s) + concat(s, 0); if (#s>2 && s[1]==0, s = s[2..#s-1]))) }
%Y Cf. A180230, A349990, A350188, A351783.
%K nonn
%O 1,4
%A _Rémy Sigrist_, Mar 08 2022