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A027989 a(n) = self-convolution of row n of array T given by A027926. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

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).

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)

MATHEMATICA

Table[((5+4n)Fibonacci[1+2n]-(1+2n)Fibonacci[2n])/5, {n, 0, 20}] [Emanuele Munarini, Apr 07 2011]

PROG

(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 */

CROSSREFS

Cf. A027926, A054142, A172991, A188648.

Sequence in context: A062454 A121523 A115240 * A096483 A093043 A061566

Adjacent sequences:  A027986 A027987 A027988 * A027990 A027991 A027992

KEYWORD

nonn

AUTHOR

Clark Kimberling

STATUS

approved

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Last modified January 24 09:01 EST 2022. Contains 350534 sequences. (Running on oeis4.)