The following remarks about the C_3 lattice assume the sequence offset is 0.
Partial sums of
A010006. So this sequence is the crystal ball sequence for the C_3 lattice - row 3 of
A142992. The lattice C_3 consists of all integer lattice points v = (a,b,c) in Z^3 such that a + b + c is even, equipped with the taxicab type norm ||v|| = (1/2) * (|a| + |b| + |c|).
The crystal ball sequence of C_3 gives the number of lattice points v in C_3 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].
For example, a(1) = 19 because the origin has norm 0 and the 18 lattice points in Z^3 of norm 1 (as defined above) are +-(2,0,0), +-(0,2,0), +-(0,0,2), +-(1,1,0), +-(1,0,1), +-(0,1,1), +-(1,-1,0), +-(1,0,-1) and +-(0,1,-1). These 18 vectors form a root system of type C_3.
O.g.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4 = x/(1 - x) * T(3, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
2*log(2) = 4/3 + Sum_{n >= 1} 1/(n*a(n)*a(n+1)). (End)
a(n+1) = (1/Pi) * Integral_{x=0..Pi} (sin((n+1/2)*x)/sin(x/2))^4. -
Yalcin Aktar, Nov 02 2011, corrected by
R. J. Mathar, Dec 01 2011
G.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4.
E.g.f.: (-3 + 6*x + 24*x^2 + 16*x^3)*exp(x)/3 + 1. (End)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k+1)) = 1/(19 - 3/(27 - 60/(43 - 315/(67 - ... -n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*3^2))))).
E.g.f.: exp(x)*(1 + 18*x + 48*x^2/2! + 32*x^3/3!). Note that -T(6, i*sqrt(x)) = 1 + 18*x + 48*x^2 + 32*x^3, where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. See
A008310. (End)
