A241526 Number of different positions in which a square with side length k, 1 <= k <= n - floor(n/3), can be placed within a bi-symmetric triangle of 1 X 1 squares of height n.

2, 7, 16, 31, 53, 83, 123, 174, 237, 314, 406, 514, 640, 785, 950, 1137, 1347, 1581, 1841, 2128, 2443, 2788, 3164, 3572, 4014, 4491, 5004, 5555, 6145, 6775, 7447, 8162, 8921, 9726, 10578, 11478, 12428, 13429, 14482, 15589, 16751, 17969, 19245, 20580, 21975

Offset

1,1

Links

Christopher Hunt Gribble, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (3,-3,2,-3,3,-1).

Formula

a(n) = sum_{j=0..n-1-floor(n/3)} ((4*n-6*j+1-(-1)^j)/4)*((4*n-6*j+3+(-1)^j)/4).

a(n) = (4*n^3+15*n^2+17*n-6*floor(n/3))/18.

G.f.: x*(x^2+x+2) / ((x-1)^4*(x^2+x+1)). - Colin Barker, Apr 26 2014

Example

The bi-symmetric triangle of 1 X 1 squares of height 5 is:
                   ___
                 _|_|_|_
               _|_|_|_|_|_
             _|_|_|_|_|_|_|_
           _|_|_|_|_|_|_|_|_|_
          |_|_|_|_|_|_|_|_|_|_|
.
No. of positions in which a 1 X 1 square can be placed = 2 + 4 + 6 + 8 + 10 = 30.
No. of positions in which a 2 X 2 square can be placed = 1 + 3 + 5 + 7 = 16.
No. of positions in which a 3 X 3 square can be placed = 2 + 4 = 6.
No. of positions in which a 4 X 4 square can be placed = 1.
Thus, a(5) = 30 + 16 + 6 + 1 = 53.

Maple

a := proc (n::integer)::integer;
       (2/9)*n^3+(5/6)*n^2+(17/18)*n-(1/3)*floor((1/3)*n)
     end proc:
seq(a(n), n = 1..60);

Prog

(PARI) Vec(x*(x^2+x+2)/((x-1)^4*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Apr 26 2014

Crossrefs

Cf. A092498.

Sequence in context: A005581 A064468 A225311 * A074470 A216499 A228189

Adjacent sequences: A241523 A241524 A241525 * A241527 A241528 A241529

Keyword

nonn,easy

Author

Christopher Hunt Gribble and Luce ETIENNE, Apr 24 2014

Status

approved