A027989 a(n) = self-convolution of row n of array T given by A027926.

1, 3, 10, 33, 105, 324, 977, 2895, 8462, 24465, 70101, 199368, 563425, 1583643, 4430290, 12342849, 34262337, 94800780, 261545777, 719697255, 1975722326, 5412138033, 14796520365, 40380240528, 110016825025, 299285288499, 813011578522, 2205652007265, 5976479585817

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

a(n) is the number of all columns in stack polyominoes of perimeter 2n+4. - Emanuele Munarini, Apr 07 2011

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
• Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Crossrefs

Cf. A027926, A054142, A172991, A188648.

Programs

Mathematica

Table[((5+4n)Fibonacci[1+2n]-(1+2n)Fibonacci[2n])/5,{n,0,28}] (* Emanuele Munarini, Apr 07 2011 *)

Maxima

makelist(((5+4*n)*fib(1+2*n)-(1+2*n)*fib(2*n))/5,n,0,20); /* Emanuele Munarini, Apr 07 2011 */

PARI

Vec((1-3*x+3*x^2)/(1-3*x+x^2)^2+O(x^66)) /* Joerg Arndt, Apr 08 2011 */

Formula

a(n) = (2/5)*(n + 1)*F(2*n+3) + (1/5)*F(2*n+2) - (4/5)*(n + 1)*F(2*n), where F(n) = A000045(n). - Ralf Stephan, May 13 2004

From Emanuele Munarini, Apr 07 2011: (Start)

a(n) = ((4*n + 5)*F(2*n+1) - (2*n + 1)*F(2*n))/5, where F(n) = A000045(n).

a(n) = Sum_{k=0..n} binomial(2*n-k, k)*(k + 1).

G.f.: (1 - 3*x + 3*x^2)/(1 - 3*x + x^2)^2.

a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). (End)