login
Number of facets of the Alternating Sign Matrix polytope ASM(n).
3

%I #94 Feb 14 2024 15:00:28

%S 1,1,2,8,20,40,68,104,148,200,260,328,404,488,580,680,788,904,1028,

%T 1160,1300,1448,1604,1768,1940,2120,2308,2504,2708,2920,3140,3368,

%U 3604,3848,4100,4360,4628,4904,5188,5480,5780,6088,6404,6728,7060,7400,7748,8104

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

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

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

%H Jessica Striker, <a href="http://arXiv.org/abs/0705.0998">The alternating sign matrix polytope</a>, arXiv:0705.0998 [math.CO], 2007-2009.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

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

%F From _Harvey P. Dale_, Mar 05 2012: (Start)

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

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

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

%o (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

%Y Cf. A005130 (number of vertices).

%K easy,nonn

%O 0,3

%A _Jonathan Vos Post_, May 09 2007

%E More terms from _Harvey P. Dale_, Mar 05 2012

%E Initial 3 terms and formulas corrected by _Ludovic Schwob_, Feb 14 2024