A128445 Number of facets of the Alternating Sign Matrix polytope ASM(n).

1, 1, 2, 8, 20, 40, 68, 104, 148, 200, 260, 328, 404, 488, 580, 680, 788, 904, 1028, 1160, 1300, 1448, 1604, 1768, 1940, 2120, 2308, 2504, 2708, 2920, 3140, 3368, 3604, 3848, 4100, 4360, 4628, 4904, 5188, 5480, 5780, 6088, 6404, 6728, 7060, 7400, 7748, 8104

Offset: 0

Author

Jonathan Vos Post, May 09 2007

Keywords

• easy
• nonn

Comments

The number of vertices (Bressoud) is Product_{j=0..n-1}(3j+1)!/(n+j)!.

References

• D. M. Bressoud, Proofs and confirmations: the story of the alternating sign matrix conjecture, MAA Spectrum, 1999.

Links

• Jessica Striker, The alternating sign matrix polytope, arXiv:0705.0998 [math.CO], 2007-2009.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A005130 (number of vertices).

Programs

Mathematica

Table[4((n-2)^2+1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{20,8,4},50] (* Harvey P. Dale, Mar 05 2012 *)

PARI

a(n)=([0,1,0; 0,0,1; 1,-3,3]^n*[20;8;4])[1,1] \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 4*((n-2)^2 + 1) for n >= 3.

From Harvey P. Dale, Mar 05 2012: (Start)

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n > 5.

G.f.: (2*x^5+x^4+4*x^3+2*x^2-4*x+1)/(1-x)^3. (End)

Extensions

More terms from Harvey P. Dale, Mar 05 2012

Initial 3 terms and formulas corrected by Ludovic Schwob, Feb 14 2024