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A055417
Number of points in N^n of norm <= 2.
3
1, 3, 6, 11, 20, 36, 63, 106, 171, 265, 396, 573, 806, 1106, 1485, 1956, 2533, 3231, 4066, 5055, 6216, 7568, 9131, 10926, 12975, 15301, 17928, 20881, 24186, 27870, 31961, 36488, 41481, 46971, 52990, 59571, 66748, 74556, 83031, 92210, 102131, 112833, 124356
OFFSET
0,2
COMMENTS
Binomial transform of [1, 0, 2, -1, 2, -1, 1, -1, 1, -1, 1, ...]. - Gary W. Adamson, Mar 12 2009
LINKS
FORMULA
a(n) = (n^3 - 3*n^2 + 14*n + 24)*(n+1)/24. Proof: The coordinates of such a point are a permutation of one of the vectors (0, ..., 0), (0, ..., 0, 1), (0, ..., 0, 2), (0, ..., 0, 1, 1), (0, ..., 0, 1, 1, 1), or (0, ..., 0, 1, 1, 1, 1), so the number of points is 1 + n + n + binomial(n,2) + binomial(n,3) + binomial(n,4). - Formula conjectured by Frank Ellermann, Mar 16 2002 and explained by Michael Somos, Apr 25 2003
G.f.: (1-2*x+x^2+x^3)/(1-x)^5. - Michael Somos, Apr 25 2003
EXAMPLE
{(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 1, 0), (0, 1, 1), (0, 2, 0), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1), (2, 0, 0)} are all the points in N^3 of norm <= 2 so a(3)=11.
MATHEMATICA
CoefficientList[Series[(-z^3 - z^2 + 2*z - 1)/(z - 1)^5, {z, 0, 100}], z] (* and *) Table[(n^4 - 6*n^3 + 23 n^2 + 6*n)/24, {n, 1, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 17 2011 *)
PROG
(PARI) a(n)=(n^3-3*n^2+14*n+24)*(n+1)/24
CROSSREFS
Row n=2 of A302998.
Sequence in context: A208851 A182845 A265076 * A018918 A077855 A054887
KEYWORD
nonn
STATUS
approved