OFFSET
0,1
COMMENTS
The following conjecture, if not already well-known, is probably easy to prove: a(n) = 3a(n-1)-a(n-2)-2(-1)^n, for n=4,5,6,... . (This has been verified up to n=1000.)
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint arXiv:1206.4864 [math.CO], 2012.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).
FORMULA
a(n) = 2*A061646(n+1) = 4*F(n+1)^2-2*(-1)^(n+1). - Emeric Deutsch, Apr 04 2007; Gary Detlefs, Nov 27 2010
a(n) = 2*(F(n)^2+F(n+1)^2+F(n)*F(n+1)). - Emeric Deutsch, Apr 04 2007
G.f.: 2(1+x-x^2)/((1+x)(1-3x+x^2)). - R. J. Mathar, Nov 25 2008
EXAMPLE
a(2)=14 because F(2)^2+F(3)^2+F(4)^2=1+4+9=14.
MAPLE
with(combinat): a:=n->fibonacci(n)^2+fibonacci(n+1)^2+fibonacci(n+2)^2: seq(a(n), n=0..32); # Emeric Deutsch, Apr 04 2007
A000045 := proc(n) combinat[fibonacci](n) ; end: A127546 := proc(n) add( A000045(i+1)^2, i=n..n+2) ; end: for n from 1 to 33 do printf("%d, ", A127546(n)) ; od ; # R. J. Mathar, Apr 03 2007
with(combinat): seq(4*fibonacci(n+1)^2-2*(-1)^n, n=0..29)
MATHEMATICA
Total/@(Partition[Fibonacci[Range[0, 30]], 3, 1]^2) (* Harvey P. Dale, Oct 20 2011 *)
PROG
(PARI) for(n=0, 10, print1(4*fibonacci(n+1)^2-2*(-1)^n, ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Simone Severini, Apr 01 2007
EXTENSIONS
STATUS
approved