0,2

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 232.

Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

T. D. Noe, Table of n, a(n) for n=0..200

Index entries for sequences related to linear recurrences with constant coefficients

Tanya Khovanova, Recursive Sequences

R. Knott, Mathematics of the Fibonacci Series

a(n)=sum(k=0, n, binomial(n, k)*F(k)*2^k) - Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 25 2003

a(n) = 4*a(n-1) + a(n-2); a(-1) = 2, a(0) = 0. a(n) = 2*A001076(n). a(n) = (F(n+1))^3 + (F(n))^3 - (F(n-1))^3. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 11 2004

a(n)=Sum(C(n, 2k+1)5^k 2^(n-2k), k=0, .., Floor[(n-1)/2]) - Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004

a(n)=sum(k=0, n, F(n+k)*binomial(n, k)) - Benoit Cloitre (benoit7848c(AT)orange.fr), May 15 2005

O.g.f.: -2*x/(-1+4*x+x^2). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 06 2008

a(n)=second binomial transform of (2,4,10,20,50,100,250). This is 2* (1,2,5,10,25,50,125) or 5^n (offset 0) *2 for the odd numbers or *4 for the even. The sequences are interpolated. Also a(n)=2*((2+sqrt5)^n-(2-sqrt5)^n)/sqrt20 offset 1. [From Al Hakanson (hawkuu(AT)gmail.com), May 02 2009]

(Mupad) numlib::fibonacci(3*n) $ n = 0..30; - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 09 2008

Table[Fibonacci[3n], {n, 0, 23}] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Apr 07 2006

(Other) sage: [fibonacci(3*n) for n in xrange(0, 24)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 15 2009]

Cf. A000045, A001076.

Equals 2*A001076. First differences of A099919. Third column of array A102310.

Sequence in context: A111643 A000163 A117616 this_sequence A113440 A034999 A067336

Adjacent sequences: A014442 A014443 A014444 this_sequence A014446 A014447 A014448

nonn,easy,nice

Mohammad K. Azarian (ma3(AT)evansville.edu)

More terms from Jud McCranie, j.mccranie(AT)comcast.net.

One more term from Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Apr 07 2006