A055417 Number of points in N^n of norm <= 2.
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
Examples
{(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.
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Crossrefs
Row n=2 of A302998.
Programs
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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}] (* Vladimir Joseph Stephan Orlovsky, Jul 17 2011 *) LinearRecurrence[{5,-10,10,-5,1},{1,3,6,11,20},50] (* Harvey P. Dale, Aug 30 2025 *) -
PARI
a(n)=(n^3-3*n^2+14*n+24)*(n+1)/24
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
Comments