A099016 a(n) = 3*L(2*n)/5 - (-1)^n/5, where L = A000032.

1, 2, 4, 11, 28, 74, 193, 506, 1324, 3467, 9076, 23762, 62209, 162866, 426388, 1116299, 2922508, 7651226, 20031169, 52442282, 137295676, 359444747, 941038564, 2463670946, 6449974273, 16886251874, 44208781348, 115740092171

Offset

0,2

Comments

Let M = an infinite triangle with (1,2,2,3,3,4,4,...) as the left border and all other columns = (0,1,2,3,4,5,...). Then lim_{n->infinity} M^n = A099016, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 26 2010

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..300 from Vincenzo Librandi)

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Formula

G.f.: (1 - 2*x^2)/((1 + x)*(1 - 3*x + x^2)).

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).

a(n) = 2*F(n)^2 + F(n)*F(n-1) + F(n-1)^2, where F = A000045.

a(n) = 3*((3/2 - sqrt(5)/2)^n + (3/2 + sqrt(5)/2)^n)/5 - (-1)^n/5.

a(n) = A099015(n)/F(n+1).

a(n) = 3*A005248(n)/5 - (-1)^n/5.

a(n) = 3*A000032(2*n)/5 - (-1)^n/5.

a(n) = A061646(n) + F(n)^2.

a(n) = 3*F(n)^2 + (-1)^n.

a(n) = F(n+1)^2 + F(n)*F(n-2). See also A192914, fourth formula. - Bruno Berselli, Feb 15 2017

Maple

with(combinat):seq(3*fibonacci(n)^2+(-1)^n, n= 0..27)

Mathematica

CoefficientList[Series[(1 - 2*x^2)/((1 + x)*(1 - 3*x + x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 31 2017 *)

Prog

(Magma) [3*Lucas(2*n)/5-(-1)^n/5: n in [0..35]]; // Vincenzo Librandi, Jun 09 2011
(Magma) F:=Fibonacci; [F(n+1)^2+F(n)*F(n-2): n in [0..30]]; // Bruno Berselli, Feb 15 2017
(PARI) x='x+O('x^30); Vec((1 - 2*x^2)/((1 + x)*(1 - 3*x + x^2))) \\ G. C. Greubel, Dec 31 2017

Crossrefs

Cf. A000032.

Sequence in context: A024962 A202085 A122423 * A026122 A108629 A007048

Adjacent sequences: A099013 A099014 A099015 * A099017 A099018 A099019

Keyword

nonn,easy

Author

Paul Barry, Sep 22 2004

Status

approved