

A063496


a(n) = (2*n1)*(8*n^28*n+3)/3.


48



1, 19, 85, 231, 489, 891, 1469, 2255, 3281, 4579, 6181, 8119, 10425, 13131, 16269, 19871, 23969, 28595, 33781, 39559, 45961, 53019, 60765, 69231, 78449, 88451, 99269, 110935, 123481, 136939, 151341, 166719, 183105, 200531, 219029
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OFFSET

1,2


COMMENTS

Number of potential flows in a 2 X 2 matrix with integer velocities in n..n, i.e., number of 2 X 2 matrices with adjacent elements differing by no more than n, counting matrices differing by a constant only once.  R. H. Hardin, Feb 27 2002
Number of ordered quadruples (a,b,c,d), (n1)<= a,b,c,d<=n1, such that a+b+c+d=0.  Benoit Cloitre, Jun 14 2003
If Y and Z are 2blocks of a (2n+1)set X then a(n1) is the number of 5subsets of X intersecting both Y and Z.  Milan Janjic, Oct 28 2007
Equals binomial transform of [1, 18, 48, 32, 0, 0, 0,...].  Gary W. Adamson, Jul 19 2008


LINKS

Harry J. Smith, Table of n, a(n) for n=1..1000
R. Bacher, P. de la Harpe and B. Venkov, Series de croissance et series d'Ehrhart associees aux reseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 11371142.
Milan Janjic, Two Enumerative Functions
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,6,4,1).


FORMULA

From Peter Bala, Jul 18 2008: (Start)
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)/(1x)^4 = x/(1x) * T(3,(1+x)/(1x)), where T(n,x) denotes the Chebyshev polynomial of the first kind.
2*log(2) = 4/3 + sum {n = 1..inf} 1/(n*a(n)*a(n+1)). (End)
a(n+1) = integral( (sin((n+1/2)x)/sin(x/2))^4, x=0..Pi)/Pi.  Yalcin Aktar, Nov 02 2011, corrected by R. J. Mathar, Dec 01 2011
From G. C. Greubel, Dec 01 2017: (Start)
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)


MAPLE

A063496:=n>(2*n1)*(8*n^28*n+3)/3; seq(A063496(n), n=1..40); # Wesley Ivan Hurt, May 09 2014


MATHEMATICA

Table[(2*n  1)*(8*n^2  8*n + 3)/3, {n, 40}] (* Wesley Ivan Hurt, May 09 2014 *)
LinearRecurrence[{4, 6, 4, 1}, {1, 19, 85, 231}, 30] (* G. C. Greubel, Dec 01 2017 *)


PROG

(PARI) { for (n=1, 1000, write("b063496.txt", n, " ", (2*n  1)*(8*n^2  8*n + 3)/3) ) } \\ Harry J. Smith, Aug 23 2009
(PARI) x='x+O('x^30); Vec(serlaplace((3+6*x+24*x^2+16*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017
(MAGMA) [(2*n1)*(8*n^28*n+3)/3: n in [1..40]]; // Wesley Ivan Hurt, May 09 2014


CROSSREFS

1/12*t*(2*n^33*n^2+n)+2*n1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Cf. A003215, A010006, A142992, A142993, A142994.
Sequence in context: A209369 A288749 A039609 * A027848 A183623 A039454
Adjacent sequences: A063493 A063494 A063495 * A063497 A063498 A063499


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Aug 01 2001


STATUS

approved



