A095922 Dimension of invariants of n-th tensor power of 5-dimensional irreducible representation of B_2.

1, 0, 1, 0, 3, 1, 15, 15, 105, 190, 945, 2410, 10263, 31890, 127699, 444458, 1751685, 6518736, 25807445, 100152288, 401449271, 1602902055, 6519160851, 26580508625, 109656966853, 454524861846, 1899821492925, 7982263725826, 33757439931675

Offset: 0

Author

Alec Mihailovs (alec(AT)mihailovs.com), Jul 11 2004

Keywords

• easy
• nice
• nonn

Comments

The analogous sequence for G_2 is A059710.

Examples

			a(2)=1 because SO(5) has unique (up to multiplication by a constant) invariant in V ⊗ V - the quadratic form x^2+y^2+z^2+u^2+v^2.
		

References

• Alec Mihailovs, A Combinatorial Approach to Representations of Lie Groups and Algebras, Springer-Verlag New York (2004).

Links

• T. D. Noe, Table of n, a(n) for n = 0..500
• G. Lachaud, On the distribution of the trace in the unitary symplectic group and the distribution of Frobenius, arXiv preprint arXiv:1506.06482, 2015.

Crossrefs

Cf. A000108, A059710, A247304.

Programs

Maple

ca:=n->binomial(n+n,n)/(n+1); a:=n->add(ca(i)*ca(i+1)*binomial(n,2*i),i=0..floor(n/2))- add(ca(i)^2*binomial(n,2*i-1),i=0..floor((n+1)/2)); seq(a(n),n=0..40);
A095922:=rsolve({(n+3)*(n+4)*A(n)=3*(n-1)*(n+2)*A(n-1)+(n-1)*(13*n+4)*A(n-2)-15*(n-1)*(n-2)*A(n-3),A(0)=1,A(1)=0,A(2)=1},A(n),makeproc);

Mathematica

t = {0, 1, 0}; Do[AppendTo[ t, (3 (n - 1) (n + 2) t[[n - 1]] + (n - 1) (13 n + 4) t[[n - 2]] - 15 (n - 1) (n - 2) t[[n - 3]])/((n + 3) (n + 4))], {n, 4, 25}]; t = Join[{1}, t] (* T. D. Noe, Apr 11 2014 *)
a[n_] := -n*HypergeometricPFQ[{3/2, 1/2 - n/2, 1 - n/2}, {3, 3}, 16] + HypergeometricPFQ[{3/2, 1/2 - n/2, -n/2}, {2, 3}, 16]; Table[a[n], {n, 0, 28}] (* Jean-François Alcover, Oct 03 2016 *)

Formula

a(n) =sum(A000108(i)*A000108(i+1)*binomial(n, 2*i), i=0..floor(n/2)) - sum(A000108(i)^2*binomial(n, 2*i-1), i=0..floor((n+1)/2)); exponential generating function = exp(t)*b(t) where b(t) is the exponential generating function of the sequence B(n) = (-1)^n*A000108(floor((n+1)/2))*A000108(floor(n/2+1)).

a(0)=1, a(1)=0, a(2)=1 and (n+3)(n+4)a(n)=3(n-1)(n+2)a(n-1)+(n-1)(13n+4)a(n-2)-15(n-1)(n-2)a(n-3) for n>2.

a(n) ~ 3 * 5^(n+5) / (128*Pi*n^5). - Vaclav Kotesovec, Oct 03 2016