A127546 a(n) = F(n)^2 + F(n+1)^2 + F(n+2)^2, where F(n) denotes the n-th Fibonacci number.

2, 6, 14, 38, 98, 258, 674, 1766, 4622, 12102, 31682, 82946, 217154, 568518, 1488398, 3896678, 10201634, 26708226, 69923042, 183060902, 479259662, 1254718086, 3284894594, 8599965698, 22515002498, 58945041798, 154320122894, 404015326886, 1057725857762

Offset: 0

Author

Simone Severini, Apr 01 2007

Keywords

• nonn

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.)

Examples

			a(2)=14 because F(2)^2+F(3)^2+F(4)^2=1+4+9=14.
		

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).

Crossrefs

Cf. A061646.

Programs

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 *)

PARI

for(n=0,10,print1(4*fibonacci(n+1)^2-2*(-1)^n,", "))

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

Extensions

Edited and extended by R. J. Mathar, Emeric Deutsch and John W. Layman, Apr 09 2007