OFFSET
0,2
COMMENTS
The rows are of length 1, 3, 5, 7, ...
a(n,k) is also the number of independent rank n tensor operators to appear in the tensor product of two spaces each spanned by k+1 tensor operators of ranks 0 to k,
{Y_{l,m},l=0,1,...,k, m:-l,-l+1,...,l} times {Y'_{l'm'}, l'=0,1,...,k, m':-l,-l+1,...,l}.
Basis elements of the tensor product space are given by
psi^{l,l'}_{p,q} = Sum_{m,m'} C^{ll'p}_{mm'q} Y_{l,m}Y'_{l'm'}
for all l,l' = 0,1,...,k and where p = |l-l'|, |l-l'|+1, ..., l+l' is the rank, q=-p, -p+1,...,p and where C^{ll'p}_{mm'q} are the Clebsch-Gordon coefficients.
Sum_{k=0..2*n+1} a(n,k)*(2*k+1) = (n+1)^4. - L. Edson Jeffery, Oct 29 2012
Sum_{k=0..2*n+1} (a(n,k) - a(n-1,k))*(2*k+1) = n^4 - (n-1)^4 = A005917(n+1), for n > 0. - L. Edson Jeffery, Nov 02 2012
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..10200 (rows n = 0..100, flattened)
Eliahu Cohen, Tobias Hansen, and Nissan Itzhaki, From Entanglement Witness to Generalized Catalan Numbers, arXiv:1511.06623 [quant-ph], 2015.
FORMULA
a(n,k) = -(3/2)*n^2 + 2*k*n + n/2 + k + 1 for n=0,1,...,k, a(n) = (2*k-n+1)*(2*k-n+2)/2 for n = k+1,...,2*k.
EXAMPLE
Triangle begins
1;
2, 3, 1;
3, 6, 6, 3, 1;
4, 9, 11, 10, 6, 3, 1;
5, 12, 16, 17, 15, 10, 6, 3, 1;
6, 15, 21, 24, 24, 21, 15, 10, 6, 3, 1;
7, 18, 26, 31, 33, 32, 28, 21, 15, 10, 6, 3, 1;
8, 21, 31, 38, 42, 43, 41, 36, 28, 21, 15, 10, 6, 3, 1;
9, 24, 36, 45, 51, 54, 54, 51, 45, 36, 28, 21, 15, 10, 6, 3, 1;
10, 27, 41, 52, 60, 65, 67, 66, 62, 55, 45, 36, 28, 21, 15, 10, 6, 3, 1;
MAPLE
Seq:=[]: for k from 0 to 15 do for n from 0 to k do Seq:= [op(Seq), -(3/2)*n^2+2*k*n+(1/2)*n+k+1] end do; for n from k+1 to 2*k do Seq:= [op(Seq), (1/2)*(2*k-n+1)*(2*k-n+2)] end do; end do; Seq;
MATHEMATICA
Table[If[n <= k, -(3/2)*n^2 + 2*k*n + n/2 + k + 1, (2*k - n + 1)*(2*k - n + 2)/2], {k, 0, 8}, {n, 0, 2 k}] // Flatten (* Michael De Vlieger, Jul 10 2022 *)
CROSSREFS
KEYWORD
tabf,nonn,easy
AUTHOR
Sean Murray, Apr 27 2010
EXTENSIONS
Edited by Sean Murray, Oct 05 2011
STATUS
approved