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A138351
Central moment sequence of tr(A^2) in USp(4).
1
1, 0, 2, 1, 11, 16, 95, 232, 1085, 3460, 14820, 54275, 227095, 895688, 3756688, 15462293, 65586405, 277342336, 1192038266, 5136760581, 22357937431, 97730561480, 430177280197, 1901975209706, 8454151507801, 37734802709796
OFFSET
0,3
COMMENTS
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).
LINKS
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, 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)..
FORMULA
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.
a(n) = Sum_{i=0..n} binomial(n,i)*A138350(i).
EXAMPLE
a(4) = 11 because E[((tr(A^2)+1)^4] = 11 for a random matrix A in USp(4).
a(4) = 1*A138350(0)+4*A138350(1)+6*A138350(2)+4*A138350(3)+1*A138350(4)
= 1*1 + 4*(-1) + 6*3 + 4*(-6) + 1*20 = 11.
MATHEMATICA
a126120[n_] := If[EvenQ[n], CatalanNumber[n/2], 0];
a138364[n_] := If[EvenQ[n], 0, Binomial[n, Floor[n/2]], 0];
a138350[n_] := a126120[n] a138364[n+1] - a138364[n] a126120[n+1];
a[n_] := Sum[Binomial[n, i] a138350[i], {i, 0, n}];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 13 2018 *)
CROSSREFS
Cf. A138350.
Sequence in context: A158354 A055459 A080958 * A120293 A063624 A101851
KEYWORD
nonn
AUTHOR
Andrew V. Sutherland, Mar 16 2008, Mar 31 2008
STATUS
approved