A188623 Number of reachable configurations in a chip-firing game on a triangle starting with n chips on one vertex.
1, 2, 2, 5, 7, 8, 12, 15, 17, 22, 26, 29, 35, 40, 44, 51, 57, 62, 70, 77, 83, 92, 100, 107, 117, 126, 134, 145, 155, 164, 176, 187, 197, 210, 222, 233, 247, 260, 272, 287, 301, 314, 330, 345, 359, 376, 392, 407, 425, 442, 458, 477, 495, 512, 532, 551, 569, 590, 610, 629, 651, 672, 692, 715, 737, 758, 782, 805, 827, 852, 876, 899, 925, 950, 974, 1001
Offset: 1
Examples
For n=4, a(4)=5 because the reachable configurations are: (4, 0, 0), (2, 1, 1), (0, 2, 2), (1, 0, 3), (3, 0, 1).
Links
- Yifan Xie, Table of n, a(n) for n = 1..10000
- J. Schneider, Enumeration and Quasipolynomiality of Chip-Firing Configurations, arXiv:1104.0279 [math.CO], 2011.
- Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).
Programs
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Mathematica
Table[(n (n + 3) - 4 (-1)^Floor[2 n/3 + 1/3] - 2)/6, {n, 1, 80}] (* Bruno Berselli, Feb 03 2016 *) -
Sage
[(n*(n+3)-4*(-1)^floor(2*n/3+1/3)-2)/6 for n in (1..80)] # Bruno Berselli, Feb 03 2016
Formula
a(3*k) = (3*k^2 + 3*k - 2)/2,
a(3*k+1) = (3*k^2 + 5*k + 2)/2,
a(3*k+2) = (3*k^2 + 7*k + 4)/2.
G.f.: x*(1 - x^2 + 2*x^3 - x^4)/((1 + x + x^2)*(1 - x)^3). - Bruno Berselli, Feb 03 2016
a(n) = (n*(n + 3) - 4*(-1)^floor(2*n/3 + 1/3) - 2)/6. - Bruno Berselli, Feb 03 2016
Comments