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A055417
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Number of points in N^n of norm <= 2.
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Binomial transform of [1, 0, 2, -1, 2, -1, 1, -1, 1, -1, 1,...]. [From Gary W. Adamson, Mar 12 2009]
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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
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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.
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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}] (* From Vladimir Joseph Stephan Orlovsky, Jul 17 2011 *)
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PROG
| (PARI) a(n)=(n^3-3*n^2+14*n+24)*(n+1)/24
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CROSSREFS
| Sequence in context: A180086 A116365 A182845 * A018918 A077855 A054887
Adjacent sequences: A055414 A055415 A055416 * A055418 A055419 A055420
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KEYWORD
| nonn
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AUTHOR
| David W. Wilson (davidwwilson(AT)comcast.net)
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