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%I #38 Jul 10 2022 16:43:15
%S 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,
%T 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,
%U 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
%N a(n,k) is the number of ways to choose integers i,j from {0,1,...,k} such that the inequalities |i-j| <= n <= i+j are satisfied.
%C The rows are of length 1, 3, 5, 7, ...
%C 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,
%C {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}.
%C Basis elements of the tensor product space are given by
%C psi^{l,l'}_{p,q} = Sum_{m,m'} C^{ll'p}_{mm'q} Y_{l,m}Y'_{l'm'}
%C 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.
%C Sum_{k=0..2*n+1} a(n,k)*(2*k+1) = (n+1)^4. - _L. Edson Jeffery_, Oct 29 2012
%C 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
%H Michael De Vlieger, <a href="/A176850/b176850.txt">Table of n, a(n) for n = 0..10200</a> (rows n = 0..100, flattened)
%H Eliahu Cohen, Tobias Hansen, and Nissan Itzhaki, <a href="http://arxiv.org/abs/1511.06623">From Entanglement Witness to Generalized Catalan Numbers</a>, arXiv:1511.06623 [quant-ph], 2015.
%F 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.
%e Triangle begins
%e 1;
%e 2, 3, 1;
%e 3, 6, 6, 3, 1;
%e 4, 9, 11, 10, 6, 3, 1;
%e 5, 12, 16, 17, 15, 10, 6, 3, 1;
%e 6, 15, 21, 24, 24, 21, 15, 10, 6, 3, 1;
%e 7, 18, 26, 31, 33, 32, 28, 21, 15, 10, 6, 3, 1;
%e 8, 21, 31, 38, 42, 43, 41, 36, 28, 21, 15, 10, 6, 3, 1;
%e 9, 24, 36, 45, 51, 54, 54, 51, 45, 36, 28, 21, 15, 10, 6, 3, 1;
%e 10, 27, 41, 52, 60, 65, 67, 66, 62, 55, 45, 36, 28, 21, 15, 10, 6, 3, 1;
%p 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;
%t 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 *)
%Y Cf. A005917.
%K tabf,nonn,easy
%O 0,2
%A _Sean Murray_, Apr 27 2010
%E Edited by _Sean Murray_, Oct 05 2011
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