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A055417 Number of points in N^n of norm <= 2.

%I A055417 #22 Aug 30 2025 12:39:32
%S A055417 1,3,6,11,20,36,63,106,171,265,396,573,806,1106,1485,1956,2533,3231,
%T A055417 4066,5055,6216,7568,9131,10926,12975,15301,17928,20881,24186,27870,
%U A055417 31961,36488,41481,46971,52990,59571,66748,74556,83031,92210,102131,112833,124356
%N A055417 Number of points in N^n of norm <= 2.
%C A055417 Binomial transform of [1, 0, 2, -1, 2, -1, 1, -1, 1, -1, 1, ...]. - _Gary W. Adamson_, Mar 12 2009
%H A055417 Andrew Howroyd, <a href="/A055417/b055417.txt">Table of n, a(n) for n = 0..1000</a>
%H A055417 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A055417 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
%F A055417 G.f.: (1-2*x+x^2+x^3)/(1-x)^5. - _Michael Somos_, Apr 25 2003
%e A055417 {(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.
%t A055417 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 *)
%t A055417 LinearRecurrence[{5,-10,10,-5,1},{1,3,6,11,20},50] (* _Harvey P. Dale_, Aug 30 2025 *)
%o A055417 (PARI) a(n)=(n^3-3*n^2+14*n+24)*(n+1)/24
%Y A055417 Row n=2 of A302998.
%K A055417 nonn,changed
%O A055417 0,2
%A A055417 _David W. Wilson_