A272391 Degeneracies of entanglement witness eigenstates for 2n spin 5/2 irreducible representations.

1, 1, 6, 111, 2666, 70146, 1949156, 56267133, 1670963202, 50720602314, 1566629938776, 49080774275121, 1555873464248076, 49814409137161480, 1608523756282054800, 52323002586904505427, 1712956041168844662002

Offset: 0

Author

Gheorghe Coserea, Apr 28 2016

Keywords

• nonn

Links

• Gheorghe Coserea, Table of n, a(n) for n = 0..200
• Hacène Belbachir, Oussama Igueroufa, Combinatorial interpretation of bisnomial coefficients and Generalized Catalan numbers, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 47-54.
• Eliahu Cohen, Tobias Hansen, Nissan Itzhaki, From Entanglement Witness to Generalized Catalan Numbers, arXiv:1511.06623 [quant-ph], 2015.
• Thomas Curtright, Thomas Van Kortryk, and Cosmas Zachos, Spin Multiplicities, hal-01345527, 2016.

Crossrefs

For spin S = 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4, 9/2, 5 we get A000108, A005043, A264607, A007043, this sequence, A264608, A272392, A272393, A272394, A272395.

Programs

Mathematica

a[n_] := 2/Pi * 2^(2 * n) * Integrate[Sqrt[1 - t] * ((4 * t - 1)(4 * t - 3))^(2 * n) * Sqrt[t]^(2 * n - 1), {t, 0, 1}] (* Thomas Curtright, Jun 23 2016 *)
a[n_] := c[0, 2 n, 5/2] - c[1, 2 n, 5/2]; c[j_, n_, s_]:= Sum[(-1)^k * Binomial[n, k] * Binomial[j - (2 * s + 1) * k + n + n * s - 1, j - (2 * s + 1) * k + n * s], {k, 0, Min[n, Floor[(j + n * s)/(2 * s + 1)]]}]; Table[a[n], {n, 0, 20}] (* Thomas Curtright, Jul 26 2016 *)

PARI

N = 34; S = 5/2;
M = matrix(N+1, N*numerator(S)+1);
Mget(n, j) =  { M[1 + n, 1 + j*denominator(S)] };
Mset(n, j, v) = { M[1 + n, 1 + j*denominator(S)] = v };
Minit() = {
  my(step = 1/denominator(S));
  Mset(0, 0, 1);
  for (n = 1, N, forstep (j = 0, n*S, step,
     my(acc = 0);
     for (k = abs(j-S), min(j+S, (n-1)*S), acc += Mget(n-1, k));
     Mset(n, j, acc)));
};
Minit();
vector(1 + N\denominator(S), n, Mget((n-1)*denominator(S),0))

Formula

a(n) ~ (6*sqrt(210)/1225)*6^(2*n)/(sqrt(Pi)*(2*n)^(3/2)) * (1-123/(280n)+O(1/n^2)). - Thomas Curtright, Jun 17 2016, updated Jul 26 2016

Recurrence: 5*n*(5*n - 8)*(5*n - 3)*(5*n - 2)*(5*n - 1)*(5*n + 1)*(7*n - 16)*(7*n - 10)*(7*n - 9)*a(n) = 6*(2*n - 1)*(5*n - 8)*(7*n - 16)*(499359*n^6 - 2314137*n^5 + 4264709*n^4 - 3984323*n^3 + 1983172*n^2 - 496780*n + 48720)*a(n-1) - 864*(n-1)*(2*n - 3)*(2*n - 1)*(7*n - 2)*(25480*n^5 - 160398*n^4 + 375142*n^3 - 401079*n^2 + 192819*n - 33500)*a(n-2) + 31104*(n-2)*(n-1)*(2*n - 5)*(2*n - 3)*(2*n - 1)*(5*n - 3)*(7*n - 9)*(7*n - 3)*(7*n - 2)*a(n-3). - Vaclav Kotesovec, Jun 24 2016

a(n) = (1/Pi)*int((sin(6x)/sin(x))^(2n)*(sin(x))^2,x,0,2 Pi). - Thomas Curtright, Jun 24 2016