A247591 Dimension of invariants of 2n-th tensor power of 6-dimensional irreducible representation of A_3.

1, 1, 3, 16, 126, 1296, 16071, 228514, 3607890, 61891050, 1135871490, 22049362440, 448790912004, 9512960347260, 208858963314735, 4728736078065810, 110006925920592810, 2621619942885055530, 63840054782606886630, 1585094577104979776880, 40054740803371374834780, 1028483346608802276173280

Offset: 0

Author

Paul Zinn-Justin, Sep 20 2014

Keywords

• nonn

Comments

The 6-dimensional representation is the usual representation of SO(6)=A_3.

Examples

			For 2n=6, there are 15 invariants corresponding to all ways of pairing the 6 indices with the metric tensor, plus one invariant which is the completely skew-symmetric 6-index tensor.
		

Crossrefs

Cf. A095922, A059710.

Programs

Mathematica

a[0] = 1; a[1] = 1; a[n_] := a[n] = (4*n*(2*n-1)*(5*n+7)*a[n-1] - 36*(n-1)*(2*n-3)*(2*n-1)*a[n-2]) / ((n+2)*(n+3)^2); Table[a[n], {n, 0, 21}]

PARI

N=66; v=vector(N); v[1]=1; v[2]=1;
for(n=2, N-1, my(t=n+1); v[t] = (-36*(n-1)*(2*n-3)*(2*n-1)*v[t-2] + 4*n*(2*n-1)*(5*n+7)*v[t-1]) / ((n+2)*(n+3)^2) );
v \\ Joerg Arndt, Sep 20 2014

Formula

a(n) = (-36*(n-1)*(2*n-3)*(2*n-1)*a(n-2) + 4*n*(2*n-1)*(5*n+7)*a(n-1)) / ((n+2)*(n+3)^2).

a(n) = (9*(n+1)*A005802(n)-(n+5)*A005802(n+1))*binomial(2*n,n)/(2*(n+1)*(n+2)). - Mark van Hoeij, Nov 12 2023