A134507 Number of rectangles in a pyramid built with squares. The squares counted in A092498 are excluded.

0, 4, 19, 57, 134, 269, 486, 813, 1281, 1926, 2788, 3910, 5340, 7130, 9335, 12015, 15234, 19059, 23562, 28819, 34909, 41916, 49928, 59036, 69336, 80928, 93915, 108405, 124510, 142345, 162030, 183689, 207449, 233442, 261804, 292674, 326196

Offset: 1

Author

Philippe Lallouet (philip.lallouet(AT)orange.fr), Jan 19 2008

Keywords

• easy
• nonn

Comments

At the first step, the pyramid contains only one unitary square. At each step of rank n we add a row of 2*n-1 squares below the previous pyramid. The sequence is the number of rectangles of any size which can be seen in this pyramid of height n.

..........._.

|..|.........|..|

||......_|__|

..........|..|..|..|

..0.......||__|| 3 rectangles 2X1, 1 rectangle 3X1

Examples

			G.f. = 4*x^2 + 19*x^3 + 57*x^4 + 134*x^5 + 269*x^6 + 486*x^7 + 813*x^8 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-5,6,-4,1).

Crossrefs

Cf. A092498.

Programs

Magma

I:=[0,4,19,57,134,269,486]; [n le 7 select I[n] else 4*Self(n-1)-6*Self(n-2)+5*Self(n-3)-5*Self(n-4)+6*Self(n-5)-4*Self(n-6)+Self(n-7): n in [1..50]]; // Vincenzo Librandi, Mar 01 2014

Mathematica

a[ n_] := SeriesCoefficient[ x^2 (4 + 3 x + 5 x^2) / ((1 - x)^5 (1 + x + x^2)), {x, 0, n}]; (* Michael Somos, Feb 25 2014 *)
a[ n_] := Quotient[ 3 n^4 + 5 n^3 - 3 n^2 - 3 n + 2, 18]; (* Michael Somos, Feb 25 2014 *)
CoefficientList[Series[-x (5 x^2 + 3 x + 4)/((x - 1)^5 (x^2 + x + 1)), {x, 0, 40}], x] (* _Vincenzo Librandi Mar 01 2014 *)

PARI

{a(n) = (3*n^4 + 5*n^3 - 3*n^2 - 3*n + 2) \ 18}; /* Michael Somos, Feb 17 2008 */

Formula

For n == 0 mod 3, a(n) = n*(3*n^3+5*n^2-3*n-3)/18; for n == 1 mod 3, a(n) = (n-1)*(3*n^3+8*n^2+5*n+2)/18; for n == 2 mod 3, a(n) = (3*n^4+5*n^3-3*n^2-3*n+2)/18. [corrected and edited by Michel Marcus, Apr 09 2024]

G.f.: -x^2*(5*x^2+3*x+4)/((x-1)^5*(x^2+x+1)). [Colin Barker, Nov 16 2012]

a(n) = (3*n^4+5*n^3-3*n^2-5*n+6*floor((n+1)/3))/18. - Luce ETIENNE, Jul 31 2015