A280921 Degree of SO(n,C), the special orthogonal group, as an algebraic variety.
2, 8, 40, 384, 4768, 111616, 3433600, 196968448, 14994641408, 2112561610752, 397713919469568, 137785594909556736, 64120367727755108352, 54666180849611078369280, 62864933930402036994048000, 131959858152100309567348408320, 374913851106401853810511580364800, 1938349609799484523235647407112847360, 13603397258157549964912652571654029312000
Offset: 2
Keywords
Examples
For n = 4 we have a(4) = 2^3*det({6,1},{1,1}) = 2^3*(6-1) = 40.
Links
- M. Brandt, D. Bruce, T. Brysiewicz, R. Krone, E. Robeva, The degree of SO(n), arXiv:1701.03200 [math.AG], 2017
Programs
-
Mathematica
a[n_] := 2^(n-1) Det[Table[Binomial[2n-2i-2j, n-2i], {i, n/2}, {j, n/2}]]; Table[a[n], {n, 2, 20}] (* Jean-François Alcover, Aug 12 2018 *) -
PARI
a(n) = 2^(n-1)*matdet(matrix(n\2,n\2,i,j,binomial(2*n-2*i-2*j,n-2*i))); \\ Michel Marcus, Jan 14 2017
Formula
a(n) = 2^(n-1)*det(binomial(2n-2i-2j, n-2i))_{i,j=1..floor(n/2)}.
a(2*n+1) = A280922(n) * 2^(2*n).
Let M_n be the n X n matrix M_n(i, j) = binomial(2*i+2*j-2, 2*i-1) = A103328(i+j-1, i-1); then a(2*n+1) = 2^(2*n)*det(M_n).
Let M_n be the n X n matrix M_n(i,j) = binomial(2*i+2*j-4, 2*i-2) = A086645(i+j-2, i-1); then a(2*n) = 2^(2*n-1)*det(M_n).