A246860 Expected value of trace(O)^(2n), where O is a 4 X 4 orthogonal matrix randomly selected according to Haar measure.

1, 3, 15, 105, 903, 8778, 92235, 1023165, 11821953, 141061206, 1727926291, 21634600078, 275950576450, 3576315994020, 46995014634435, 625082431593285, 8403885851894445, 114069364107664350, 1561609592248119645, 21543838447412548410, 299299110959202973710

Offset

1,2

Comments

The corresponding sequences for 2 X 2, 3 X 3, and 5 X 5 matrices are A001700, A099251, and A247304.

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

Table of n, a(n) for n=1..21.

MathOverflow, Moments of the trace of orthogonal matrices

Formula

In the MathOverflow link, Nathaniel Johnston conjectures a(n) = A000108(n)*(A000108(n)+1)/2. - Robert Israel, Jan 17 2020

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;
Table[Print[n, " ", a[n]]; a[n], {n, 1, 21}] (* Jean-François Alcover, Feb 05 2023 *)

Crossrefs

Cf. A000108, A001700, A099251, A247304, A247306.

Sequence in context: A067546 A015682 A291744 * A357596 A249014 A258498

Adjacent sequences: A246857 A246858 A246859 * A246861 A246862 A246863

Keyword

nonn

Author

Nathaniel Johnston, Sep 05 2014

Status

approved