A022126 Fibonacci sequence beginning 3, 16.

3, 16, 19, 35, 54, 89, 143, 232, 375, 607, 982, 1589, 2571, 4160, 6731, 10891, 17622, 28513, 46135, 74648, 120783, 195431, 316214, 511645, 827859, 1339504, 2167363, 3506867, 5674230, 9181097, 14855327, 24036424, 38891751, 62928175, 101819926, 164748101

Offset

0,1

Comments

a(n) is the sum of eleven consecutive Fibonacci numbers. a(n) = F(n-6) + F(n-5) + ... + F(n+4), where F(n) = A000045(n), extended so that F(-1) = 1, F(-2) = -1, F(-3) = 2, etc. using the rule F(n-1) = F(n+1) - F(n). - Graeme McRae, Apr 24 2014

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Tanya Khovanova, Recursive Sequences

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

Formula

a(0) = 3, a(1) = 16, a(n) = a(n - 2) + a(n - 1).

G.f.: (3 + 13x)/(1 - x - x^2). - Philippe Deléham, Nov 19 2008

From Colin Barker, Jun 03 2016: (Start)

a(n) = (2^(-1 - n)*((1 - sqrt(5))^n*(-29 + 3*sqrt(5)) + (1 + sqrt(5))^n*(29 + 3*sqrt(5))))/sqrt(5).

a(n) = a(n-1) + a(n-2) for n > 1.

(End)

E.g.f.: (29*sqrt(5)*sinh(sqrt(5)*x/2) + 15*cosh(sqrt(5)*x/2))*exp(x/2)/5. - Ilya Gutkovskiy, Jun 03 2016

a(n) = Fibonacci(n+6) - Fibonacci(n-5). - Greg Dresden, Austen Sharrett, and Aoife Chow, Mar 03 2022

Maple

seq(coeff(series((3+13*x)/(1-x-x^2), x, n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Sep 29 2018

Mathematica

CoefficientList[Series[(3 + 13 x)/(1 - x - x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Apr 25 2014 *)
LinearRecurrence[{1, 1}, {3, 16}, 50] (* Alonso del Arte, Sep 29 2018 *)

Prog

(PARI) Vec((3+13*x)/(1-x-x^2) + O(x^100)) \\ Colin Barker, Jun 03 2016
(GAP) a:=[3, 16];; for n in [3..40] do a[n]:=a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Sep 29 2018

Crossrefs

Cf. A000045.

Sequence in context: A219508 A032922 A103655 * A127006 A042075 A342716

Adjacent sequences: A022123 A022124 A022125 * A022127 A022128 A022129

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved