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

0,2

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.

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).

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)

Example

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, ... .

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 *)

Prog

(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
(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

Crossrefs

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

Sequence in context: A128511 A347922 A355418 * A251932 A319542 A173804

Adjacent sequences: A142991 A142992 A142993 * A142995 A142996 A142997

Keyword

nonn,easy

Author

Peter Bala, Jul 18 2008

Status

approved