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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).
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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