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, y-1), (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.
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
Offset: 0
Keywords
Examples
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.
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..10000
- Rémy Sigrist, Representation of the configuration for n = 100000 (blue pixels correspond to sites with one grain, red pixels to sites with two grains)
Programs
-
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..#s-1]))) }
Formula
a(3*n) + 1 = a(3*n + 1) = a(3*n + 2).
Comments