A194082 Sum{floor(sqrt(3)*k/2) : 1<=k<=n}.

0, 1, 3, 6, 10, 15, 21, 27, 34, 42, 51, 61, 72, 84, 96, 109, 123, 138, 154, 171, 189, 208, 227, 247, 268, 290, 313, 337, 362, 387, 413, 440, 468, 497, 527, 558, 590, 622, 655, 689, 724, 760, 797, 835, 873, 912, 952, 993, 1035, 1078, 1122, 1167, 1212

Offset: 1

Author

Clark Kimberling, Aug 17 2011

Keywords

• nonn

Comments

Partial sums of A171970.

Comment from R. J. Mathar, Dec 02 2012 (Start):

a(n-1) is the number of unit squares regularly packed into the isosceles triangle of edge length n.

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.

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.

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.

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.

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.

(End)

Crossrefs

Cf. A171970.

Programs

Mathematica

r = Sqrt[3]/2;
c[k_] := Sum[Floor[j*r], {j, 1, k}];
Table[c[k], {k, 1, 90}]

PARI

a(n)=sum(k=1,n,sqrtint(3*k^2\4)) \\ Charles R Greathouse IV, Jan 06 2013