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%I #27 Jan 06 2013 02:48:58
%S 0,1,3,6,10,15,21,27,34,42,51,61,72,84,96,109,123,138,154,171,189,208,
%T 227,247,268,290,313,337,362,387,413,440,468,497,527,558,590,622,655,
%U 689,724,760,797,835,873,912,952,993,1035,1078,1122,1167,1212
%N Sum{floor(sqrt(3)*k/2) : 1<=k<=n}
%C Partial sums of A171970.
%C Comment from _R. J. Mathar_, Dec 02 2012 (Start):
%C a(n-1) is the number of unit squares regularly packed into the isosceles triangle of edge length n.
%C The triangle may be aligned with the Cartesian axes by putting its bottom edge on the horizontal axis, so its vertices are at (x,y) = (0,0), (n,0) and (n/2,sqrt(3)*n/2), see A010527.
%C The area inside the triangle is sqrt(3)*n^2/4 = A120011*n^2. There is an obvious upper limit of floor(sqrt(3)*n^2/4) = A171971(n) to the count of non-overlapping unit squares inside this triangle.
%C Regular packing: We place the first row of unit squares so they touch the bottom edge of the triangle. Their number is limited by the length of the horizontal section of the line y=1 inside the triangle, n-2*y/sqrt(3), which touches all of these first-row squares at their top.
%C The number of unit squares in the next row, between y=1 and y=2, is limited by the length of the horizontal section of the line y=2 inside the triangle, n-2*y/sqrt(3). Continuing, in row y=1, 2, ... we insert floor(n-2*y/sqrt(3)) unit squares, all with the same orientation.
%C The total number of squares is sum_{ y=1, 2, ..., floor(n*sqrt(3)/2) } floor( n-2*y/sqrt(3) ), and resummation yields, up to an index shift, this sequence here.
%C (End)
%t r = Sqrt[3]/2;
%t c[k_] := Sum[Floor[j*r], {j, 1, k}];
%t Table[c[k], {k, 1, 90}]
%o (PARI) a(n)=sum(k=1,n,sqrtint(3*k^2\4)) \\ _Charles R Greathouse IV_, Jan 06 2013
%Y Cf. A171970.
%K nonn
%O 1,3
%A _Clark Kimberling_, Aug 17 2011
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