%I
%S 6,12,1620,2160,2551500,3061800,33756345000,38578680000,
%T 4060381958325000,4511535509250000,3168740859543387253125000,
%U 3456808210410967912500000,34159303730702924635072148437500
%N Denominator of the rational coefficient in the main term in the dynamical analog of Mertens's theorem for a full ndimensional shift, n >= 2.
%C a(n) for n >= 2 may be defined as follows. For a full ndimensional shift, let M(N) = Sum_{L} O(L)/exp(h[L]) where the sum is over subgroups L of finite index in Z^n, O(L) is the number of points with stabilizer L and exp(h) is the number of symbols.
%C Then M(N) is asymptotic to a rational times a power of Pi times a product of values of the zeta function at odd integers and a(n) is the denominator of that rational.
%H Vaclav Kotesovec, <a href="/A159282/b159282.txt">Table of n, a(n) for n = 2..63</a>
%H R. Miles and T. Ward, <a href="https://doi.org/10.1090/S0002993908096494">Orbitcounting for nilpotent group shifts</a>, Proc. Amer. Math. Soc. 137 (2009), 14991507.
%F By Perron's formula, M(N) = residue(z=n1, zeta(z+1)...zeta(zn+2)N^z) = (a(n)/b(n))*N^(d1)*Pi^(floor(n/2)*(floor(n/2)+1)*Product_{j=1..floor((n1)/2)} zeta(2*j+1).
%e For n=3, using the formula in terms of residues, we have residue(zeta(z1)*zeta(z)*zeta(z+1)*N^z/z,z=2) = (1/12)*zeta(3)*Pi^2N^2, so a(3)=12.
%p residue(product(Zeta(zj),j=1..(n2))*N^z/z,z=n1) # generates an expression from which a(n) can be read off
%t Denominator[Table[Residue[Product[Zeta[z  j], {j, 1, n2}]/z, {z, n1}], {n, 2, 14}]] (* _Vaclav Kotesovec_, Sep 05 2019 *)
%Y This is the denominator of a rational sequence whose numerator is A159283.
%K easy,frac,nonn
%O 2,1
%A Thomas Ward (t.ward(AT)uea.ac.uk), Apr 08 2009
