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
KEYWORD
nonn,easy
AUTHOR
STATUS
approved