OFFSET
1,1
COMMENTS
The relevant graph has rotational symmetry so the number of tilings is a square or twice a square, in this case by the formula for a(n) it is always twice a square.
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..200
M. Ciucu, Enumeration of perfect matchings in graphs with reflective symmetry, J. Combin. Theory Ser. A 77 (1997), no. 1, 67-97.
W. Jockusch, Perfect matchings and perfect squares J. Combin. Theory Ser. A 67 (1994), no. 1, 100-115.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).
FORMULA
a(n) = 2 * F(n)^2 * F(n+1)^2 where F(n) is the n-th Fibonacci number - sequence A000045.
G.f.: -2*x*(1-x+x^2) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Jan 30 2011
EXAMPLE
a(1) = 2 because in this case the set S is the unit square and there is one horizontal tiling and one vertical.
MAPLE
with(combinat): for n from 1 to 40 do printf(`%d, `, 2*fibonacci(n)^2*fibonacci(n+1)^2) od:
PROG
(PARI) { a=1; b=0; c=1; for (n=1, 200, f=a+b; g=b+c; a=b; b=c; c=g; write("b060635.txt", n, " ", 2*f^2*g^2); ) } \\ Harry J. Smith, Jul 08 2009
CROSSREFS
KEYWORD
nonn
AUTHOR
Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 16 2001
EXTENSIONS
More terms from James A. Sellers, Apr 16 2001
STATUS
approved