

A350188


Consider a 2D sandpile model where each site with 3 or more grains, say at location (x, y), topples and transfers one grain of sand to the sites at locations (x+1, y1), (x+1, y) and (x+1, y+1); a(n) gives the number of nonempty sites after stabilization of a configuration starting with n grains at the origin.


2



0, 1, 1, 3, 4, 4, 3, 4, 4, 7, 8, 8, 10, 11, 11, 10, 11, 11, 14, 15, 15, 17, 18, 18, 17, 18, 18, 21, 22, 22, 24, 25, 25, 24, 25, 25, 27, 28, 28, 30, 31, 31, 30, 31, 31, 36, 37, 37, 39, 40, 40, 39, 40, 40, 39, 40, 40, 42, 43, 43, 42, 43, 43, 45, 46, 46, 48, 49
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OFFSET

0,4


COMMENTS

Sites containing 0, 1 or 2 grains are stable.
After stabilization, there are:
 2*a(n)  n sites with one grain,
 n  a(n) sites with two grains.


LINKS



FORMULA

a(3*n) + 1 = a(3*n + 1) = a(3*n + 2).


EXAMPLE

For n = 10 :
 the model evolves (for example) as follows:
1 1
3 . 2 . 2 1
10 > 1 3 > 1 . 3 > 1 . . 1
3 . 2 . 2 1
1 1
 there are 8 nonempty sites in the stabilized configuration,
 so a(10) = 8.


PROG

(PARI) a(n) = { my (s=[n], v=0); for (k=1, oo, if (vecmax(s)==0, return (v), v += sum(k=1, #s, s[k]%3>0); s \= 3; s = concat([ s , [0], [0]]) + concat([[0], s , [0]]) + concat([[0], [0], s ]); while (#s>2 && s[1]==0, s = s[2..#s1]))) }


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



