OFFSET
1,3
COMMENTS
Same as A001224 except that there a(2)=2 not 1. - N. J. A. Sloane, Mar 30 2015
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1001
A. R. Ashrafi, J. Azarija, K. Fathalikhani, S. Klavzar, et al., Orbits of Fibonacci and Lucas cubes, dihedral transformations, and asymmetric strings, 2014.
R. J. Mathar, Paving rectangular regions with rectangular tiles, ..., arXiv:1311.6135 [math.CO], Table 9.
W. E. Patten (proposer) and S. W. Golomb (solver), Problem E1470, "Covering a 2Xn rectangle with dominoes", Amer. Math. Monthly, 69 (1962), 61-62.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,0,-1,-1).
FORMULA
If F(n) is the n-th Fibonacci number, then a(2n) = (F(2n) + F(n+1))/2 and a(2n+1) = (F(2n+1) + F(n))/2 for n > 1.
G.f.: -x*(x^7 + x^6 + x^5 + 2*x^4 - x^3 + x^2 - 1) / ((x^2 + x - 1)*(x^4 + x^2 - 1)). - Colin Barker, Dec 15 2012
EXAMPLE
a(3)=2 because of the configurations |= and |||.
MAPLE
with(combinat); F:=fibonacci;
f:=proc(n) option remember;
if n=2 then 1 # change this to 2 to get A001224
elif (n mod 2) = 0 then (F(n+1)+F(n/2+2))/2;
else (F(n+1)+F((n+1)/2))/2; fi; end;
[seq(f(n), n=1..50)];
MATHEMATICA
CoefficientList[Series[-(x^7 + x^6 + x^5 + 2 x^4 - x^3 + x^2 - 1) / ((x^2 + x - 1) (x^4 + x^2 - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 22 2014 *)
LinearRecurrence[{1, 2, -1, 0, -1, -1}, {1, 1, 2, 4, 5, 9, 12, 21}, 40] (* Harvey P. Dale, Mar 13 2024 *)
PROG
(Magma) [n eq 1 select 1 else (1/2)*(Fibonacci(n+2)+Fibonacci(Floor((n-(-1)^n)/2)+2)): n in [0..40]]; // Vincenzo Librandi, Nov 22 2014
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Thomas Ward, Mar 27 2001
EXTENSIONS
Edited by N. J. A. Sloane, Mar 30 2015
STATUS
approved