A138354 Central moment sequence of tr(A^4) in USp(4).

1, 0, 3, 1, 21, 26, 215, 498, 2821, 9040, 43695, 165375, 752785, 3101970, 13881803, 59837183, 267860685, 1184749704, 5337504263, 23996776941, 108964583121, 495544446410, 2267450194443, 10402298479276, 47926692348121

Offset: 0

Author

Andrew V. Sutherland, Mar 16 2008, Mar 31 2008

Keywords

• nonn

Comments

Binomial transform of A018224.

If A is a random matrix in the compact group USp(4) (4 X 4 complex are unitary and symplectic), then a(n)=E[(tr(A^4)+1)^n] is the n-th central moment of the trace of A^4, since E[tr(A^4)] = -1 (see A018224).

Examples

			a(3) = 1 because E[(tr(A^4)+1)^3] = 1.
a(3) = 1*A018224(0) + 3*A018224(1) + 3*A018224(2) + 1*A018224(1) = 1*1 + 3*(-1) + 3*4 + 1*(-9) = 1.
		

Links

• Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.

Crossrefs

Cf. A018224.

Programs

Mathematica

a18224[n_] := Binomial[n, Floor[n/2]]^2;
a[n_] := Sum[(-1)^i Binomial[n, i] a18224[i], {i, 0, n}];
Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Aug 13 2018 *)

Formula

a(n) = (1/2)*Integral_{x=0..Pi,y=0..Pi}(2cos(4x)+2cos(4y)+1)^n*(2cos(x)-2cos(y))^2*(2/Pi*sin^2(x))*(2/Pi*sin^2(y))dxdy.

a(n) = Sum_{i=0..n} (-1)^i binomial(n,i)*A018224(i). [corrected by Jean-François Alcover, Aug 13 2018]

a(n) = (5*A201805(n) - A201805(n+1))/4. - Mark van Hoeij, Nov 29 2024