|
|
A246860
|
|
Expected value of trace(O)^(2n), where O is a 4 X 4 orthogonal matrix randomly selected according to Haar measure.
|
|
4
|
|
|
1, 3, 15, 105, 903, 8778, 92235, 1023165, 11821953, 141061206, 1727926291, 21634600078, 275950576450, 3576315994020, 46995014634435, 625082431593285, 8403885851894445, 114069364107664350, 1561609592248119645, 21543838447412548410, 299299110959202973710
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
a(n) is the number of triangulations with middle chord of an 2n+2-gon modulo the cyclic action. So a(n) = A000108(n)^2 - A000107(A000108(n)-1). The first part A000108(n)^2 means the cartes of two n+2-gons separated by the middle chord, second part is the duplicated joins need to be removed. - Yuchun Ji, Aug 11 2020
|
|
LINKS
|
|
|
FORMULA
|
|
|
MAPLE
|
A246860 := proc (n) return (1/8)*integrate(integrate((cos(x)-cos(y))^2*(2*cos(x)+2*cos(y))^(2*n), y = 0 .. 2*Pi), x = 0 .. 2*Pi)/Pi^2+(1/2)*integrate((1-cos(z)^2)*(2*cos(z))^(2*n), z = 0 .. 2*Pi)/Pi end proc; seq(A246860(n), n = 1 .. 21);
|
|
MATHEMATICA
|
a[n_] := a[n] = (1/8)*Integrate[Integrate[(Cos[x] - Cos[y])^2 * (2 Cos[x] + 2 Cos[y])^(2 n), {y, 0, 2 Pi}], {x, 0, 2 Pi}]/ Pi^2 + (1/2)*Integrate[(1 - Cos[z]^2)*(2 Cos[z])^(2 n), {z, 0, 2 Pi}]/Pi;
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|