A142994 Crystal ball sequence for the lattice C_5.

1, 51, 501, 2471, 8361, 22363, 50973, 103503, 192593, 334723, 550725, 866295, 1312505, 1926315, 2751085, 3837087, 5242017, 7031507, 9279637, 12069447, 15493449, 19654139, 24664509, 30648559, 37741809, 46091811, 55858661, 67215511, 80349081

Offset: 0

Author

Peter Bala, Jul 18 2008

Keywords

• nonn
• easy

Comments

The lattice C_5 consists of all integer lattice points v = (x_1,...,x_5) in Z^5 such that (x_1 + ... + x_5) is even, equipped with the taxicab type norm ||v|| = (1/2) * (|x_1| + ... + |x_5|). The crystal ball sequence of C_5 gives the number of lattice points v in C_5 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].

Partial sums of A019561.

Examples

			a(1) = 51. The origin has norm 0. The 50 lattice points in Z^5 of norm 1 (as defined above) are +-2*e_i, 1 <= i <= 5 and (+- e_i +- e_j), 1 <= i < j <= 5, where e_1, ... , e_5 denotes the standard basis of Z^5. These 50 vectors form a root system of type C_5. Hence the sequence begins 1, 1 + 50 = 51, ... .
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000
• R. Bacher, P. de la Harpe, and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, Annales de l'Institut Fourier, Tome 49 (1999) no. 3, pp. 727-762.
• R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Row 5 of A142992. Cf. A019561, A063496, A142993.

Programs

Magma

[(2*n+1)*(32*n^4+64*n^3+88*n^2+56*n+15)/15: n in [0..30]]; // Vincenzo Librandi, Dec 16 2015

Maple

a := n -> (2*n+1)*(32*n^4+64*n^3+88*n^2+56*n+15)/15: seq(a(n), n = 0..20)

Mathematica

CoefficientList[Series[(1 + 45 x + 210 x^2 + 210 x^3 + 45 x^4 + x^5)/(1 - x)^6, {x, 0, 33}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1},{1, 51, 501, 2471, 8361, 22363}, 25] (* Vincenzo Librandi, Dec 16 2015 *)

Python

A142994_list, m = [], [512, -768, 352, -48, 2, 1]
for _ in range(10**2):
    A142994_list.append(m[-1])
    for i in range(5):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

Formula

a(n) = (2*n + 1)*(32*n^4 + 64*n^3 + 88*n^2 + 56*n + 15)/15.

a(n) = Sum_{k = 0..5} binomial(10, 2*k)*binomial(n+k, 5).

a(n) = Sum_{k = 0..5} binomial(10, 2*k+1)*binomial(n+k+1/2, 5).

O.g.f.: (1 + 45*x + 210*x^2 + 210*x^3 + 45*x^4 + x^5)/(1 - x)^6 = 1/(1 - x) * T(5, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.

Sum_{n >= 1} 1/(n*a(n-1)*a(n)) = 2*log(2) - 41/30.

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), for n > 5. - Vincenzo Librandi, Dec 16 2015

From Peter Bala, Mar 11 2024: (Start)

Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(51 - 3/(59 - 60/(75 - 315/(99 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*5^2))))).

E.g.f.: exp(x)*(1 + 50*x + 400*x^2/2! + 1120*x^3/3! + 1280*x^4/4! + 512*x^5/5!).

Note that -T(10, i*sqrt(x)) = 1 + 50*x + 400*x^2 + 1120*x^3 + 1280*x^4 + 512*x^5. See A008310. (End)