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%I #38 Feb 05 2023 09:18:41
%S 1,3,15,105,903,8778,92235,1023165,11821953,141061206,1727926291,
%T 21634600078,275950576450,3576315994020,46995014634435,
%U 625082431593285,8403885851894445,114069364107664350,1561609592248119645,21543838447412548410,299299110959202973710
%N Expected value of trace(O)^(2n), where O is a 4 X 4 orthogonal matrix randomly selected according to Haar measure.
%C The corresponding sequences for 2 X 2, 3 X 3, and 5 X 5 matrices are A001700, A099251, and A247304.
%C 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
%H MathOverflow, <a href="http://mathoverflow.net/questions/180110/moments-of-the-trace-of-orthogonal-matrices">Moments of the trace of orthogonal matrices</a>
%F In the MathOverflow link, Nathaniel Johnston conjectures a(n) = A000108(n)*(A000108(n)+1)/2. - _Robert Israel_, Jan 17 2020
%p 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);
%t 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;
%t Table[Print[n, " ", a[n]]; a[n], {n, 1, 21}] (* _Jean-François Alcover_, Feb 05 2023 *)
%Y Cf. A000108, A001700, A099251, A247304, A247306.
%K nonn
%O 1,2
%A _Nathaniel Johnston_, Sep 05 2014
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