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

0,1

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

Cf. A061646.

Sequence in context: A275208 A000634 A006654 * A192484 A217861 A188492

Adjacent sequences: A127543 A127544 A127545 * A127547 A127548 A127549

Keyword

nonn

Author

Simone Severini, Apr 01 2007

Extensions

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

Status

approved