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

Author

Christopher Hunt Gribble and Luce ETIENNE, Apr 24 2014

Keywords

• nonn
• easy

Examples

			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.
		

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).

Crossrefs

Cf. A092498.

Programs

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);

PARI

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

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