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%I #10 Aug 13 2018 09:04:40
%S 1,0,2,1,11,16,95,232,1085,3460,14820,54275,227095,895688,3756688,
%T 15462293,65586405,277342336,1192038266,5136760581,22357937431,
%U 97730561480,430177280197,1901975209706,8454151507801,37734802709796
%N Central moment sequence of tr(A^2) in USp(4).
%C If A is a random matrix in the compact group USp(4) (4 X 4 complex matrices which are unitary and symplectic), then a(n)=E[(tr(A^2)+1)^n] is the n-th central moment of the trace of A^2, since E[tr(A^2)] = -1 (see A138350).
%H Kiran S. Kedlaya and Andrew V. Sutherland, <a href="https://arxiv.org/abs/0803.4462">Hyperelliptic curves, L-polynomials and random matrices</a>, arXiv:0803.4462 [math.NT], 2008-2010.
%H Kiran S. Kedlaya and Andrew V. Sutherland, <a href="https://dspace.mit.edu/handle/1721.1/64701">Hyperelliptic curves, L-polynomials and random matrices</a>, in Arithmetic, Geometry, Cryptography, and Coding Theory: International Conference, November 5-9, 2007, CIRM, Marseilles, France. Gilles Lachaud, Christophe Ritzenthaler, Michael A. Tsfasman, editors. 2009. (Contemporary Mathematics ; v.487)..
%F a(n) = (1/2)Integral_{x=0..Pi,y=0..Pi}(2cos(2x)+2cos(2y)+1)^n(2cos(x)-2cos(y))^2(2/Pi*sin^2(x))(2/Pi*sin^2(y))dxdy.
%F a(n) = Sum_{i=0..n} binomial(n,i)*A138350(i).
%e a(4) = 11 because E[((tr(A^2)+1)^4] = 11 for a random matrix A in USp(4).
%e a(4) = 1*A138350(0)+4*A138350(1)+6*A138350(2)+4*A138350(3)+1*A138350(4)
%e = 1*1 + 4*(-1) + 6*3 + 4*(-6) + 1*20 = 11.
%t a126120[n_] := If[EvenQ[n], CatalanNumber[n/2], 0];
%t a138364[n_] := If[EvenQ[n], 0, Binomial[n, Floor[n/2]], 0];
%t a138350[n_] := a126120[n] a138364[n+1] - a138364[n] a126120[n+1];
%t a[n_] := Sum[Binomial[n, i] a138350[i], {i, 0, n}];
%t Table[a[n], {n, 0, 25}] (* _Jean-François Alcover_, Aug 13 2018 *)
%Y Cf. A138350.
%K nonn
%O 0,3
%A _Andrew V. Sutherland_, Mar 16 2008, Mar 31 2008
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