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A142994
Crystal ball sequence for the lattice C_5.
4
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
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.
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
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 18 2008
STATUS
approved