A143010 Crystal ball sequence for the A4 x A4 lattice.

1, 41, 661, 5741, 33001, 142001, 494341, 1465661, 3833941, 9073501, 19789001, 40328641, 77620661, 142282141, 250054001, 423621001, 694880441, 1107728161, 1721435341, 2614694501, 3890418001, 5681377241, 8156775661, 11529853541

Offset: 0

Author

Peter Bala, Jul 22 2008

Keywords

• easy
• nonn

Comments

The A_4 lattice consists of all vectors v = (a,b,c,d,e) in Z^5 such that a+b+c+d+e = 0. The lattice is equipped with the norm ||v|| = 1/2*(|a| + |b| + |c| + |d| + |e|). Pairs of lattice points (v,w) in the product lattice A_4 x A_4 have norm ||(v,w)|| = ||v|| + ||w||. Then the k-th term in the crystal ball sequence for the A_4 x A_4 lattice gives the number of such pairs (v,w) for which ||(v,w)|| is less than or equal to k.

Links

• 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 (9,-36, 84,-126,126,-84,36,-9,1).

Crossrefs

Cf. A143007 (row 4), A143008, A143009, A143011.

Programs

Magma

[5*n*(n+1)*(n^2+n+2)*(7*n^4+14*n^3+77*n^2+70*n+120)/288+1: n in [0..30]]; // Vincenzo Librandi, Dec 16 2015

Maple

p := n -> (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288: seq(p(n), n = 0..24);

Mathematica

LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {1, 41, 661, 5741, 33001, 142001, 494341, 1465661, 3833941}, 25] (* Vincenzo Librandi, Dec 16 2015 *)

Python

A143010_list, m = [], [4900, -14700, 17500, -10500, 3340, -540, 40, 0, 1]
for _ in range(10**2):
    A143010_list.append(m[-1])
    for i in range(8):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

Formula

a(n) = (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288 = 5*n*(n + 1)*(n^2 + n + 2)*(7*n^4 + 14*n^3 + 77*n^2 + 70*n + 120)/288 + 1.

O.g.f. : 1/(1-x)*[Legendre_P(4,(1+x)/(1-x))]^2.

Apery's constant zeta(3) = (1+1/2^3+1/3^3+1/4^3) + Sum {n = 1..inf} 1/(n^3*a(n-1)*a(n)).

G.f.: (1+16*x+36*x^2+16*x^3+x^4)^2/(1-x)^9. [Colin Barker, Mar 16 2012]

a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n>8. - Vincenzo Librandi, Dec 16 2015