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Degeneracies of entanglement witness eigenstates for n spin 5 irreducible representations.
7

%I #70 Jun 17 2017 14:27:57

%S 1,0,1,1,11,76,671,5916,54131,504316,4779291,45898975,445798221,

%T 4371237794,43213522209,430241859971,4310236148075,43417944574136,

%U 439495074016427,4468208369691396,45605656313488271,467140985042718910

%N Degeneracies of entanglement witness eigenstates for n spin 5 irreducible representations.

%C The Mathematica formula for a(n) as the difference of two generalized binomial coefficients is adapted from the Appendix of the Mendonça link. - _Thomas Curtright_, Jul 27 2016

%H Gheorghe Coserea, <a href="/A272395/b272395.txt">Table of n, a(n) for n = 0..401</a>

%H Eliahu Cohen, Tobias Hansen, Nissan Itzhaki, <a href="http://arxiv.org/abs/1511.06623">From Entanglement Witness to Generalized Catalan Numbers</a>, arXiv:1511.06623 [quant-ph], 2015.

%H T. L. Curtright, T. S. Van Kortryk, and C. K. Zachos, <a href="https://hal.archives-ouvertes.fr/hal-01345527">Spin Multiplicities</a>, hal-01345527, 2016.

%H Vaclav Kotesovec, <a href="/A272395/a272395.txt">Recurrence (of order 6)</a>

%H J. R. G. Mendonça, <a href="http://arxiv.org/abs/1207.4106">Exact eigenspectrum of the symmetric simple exclusion process on the complete, complete bipartite and related graphs</a>, Journal of Physics A: Mathematical and Theoretical 46:29 (2013) 295001. arXiv:1207.4106 [cond-mat.stat-mech], 2012-2013.

%F a(n)=(1/Pi)*int((sin(11x)/sin(x))^n*(sin(x))^2,x,0,2Pi). - _Thomas Curtright_, Jun 24 2016

%F a(n) ~ (1/20)^(3/2)*11^n/(sqrt(Pi)*n^(3/2))(1-63/(80n)+O(1/n^2)). - _Thomas Curtright_, Jul 26 2016

%t a[n_]:= 2/Pi*Integrate[Sqrt[(1-t)/t]*(1024t^5-2304t^4+1792t^3-560t^2 +60t-1)^n, {t, 0, 1}] (* _Thomas Curtright_, Jun 24 2016 *)

%t a[n_]:= c[0, n, 5]-c[1, n, 5]

%t 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, Floor[(j + n*s)/(2*s + 1)]}] (* _Thomas Curtright_, Jul 26 2016 *)

%t a[n_]:= mult[0, n, 5]

%t mult[j_,n_,s_]:=Sum[(-1)^(k+1)*Binomial[n,k]*Binomial[n*s+j-(2*s+1)*k+n- 1,n*s+j-(2*s+1)*k+1],{k,0,Floor[(n*s+j+1)/(2*s+1)]}] (* _Thomas Curtright_, Jun 14 2017 *)

%o (PARI)

%o N = 22; S = 5;

%o M = matrix(N+1, N*numerator(S)+1);

%o Mget(n, j) = { M[1 + n, 1 + j*denominator(S)] };

%o Mset(n, j, v) = { M[1 + n, 1 + j*denominator(S)] = v };

%o Minit() = {

%o my(step = 1/denominator(S));

%o Mset(0, 0, 1);

%o for (n = 1, N, forstep (j = 0, n*S, step,

%o my(acc = 0);

%o for (k = abs(j-S), min(j+S, (n-1)*S), acc += Mget(n-1, k));

%o Mset(n, j, acc)));

%o };

%o Minit();

%o vector(1 + N\denominator(S), n, Mget((n-1)*denominator(S),0))

%o (PARI) c(j, n) = sum(k=0, min((j + 5*n)\11, n), (-1)^k*binomial(n, k)*binomial(j - 11*k + n + 5*n - 1, j - 11*k + n*5))

%o a(n)=c(0, n)-c(1, n) \\ _Charles R Greathouse IV_, Jul 28 2016; adapted from Curtright's Mathematica code

%Y For spin S = 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4, 9/2, 5 we get A000108, A005043, A264607, A007043, A272391, A264608, A272392, A272393, A272394, this sequence.

%K nonn

%O 0,5

%A _Gheorghe Coserea_, Apr 28 2016