A022121 Fibonacci sequence beginning 3, 8.
3, 8, 11, 19, 30, 49, 79, 128, 207, 335, 542, 877, 1419, 2296, 3715, 6011, 9726, 15737, 25463, 41200, 66663, 107863, 174526, 282389, 456915, 739304, 1196219, 1935523, 3131742, 5067265, 8199007, 13266272, 21465279, 34731551, 56196830, 90928381, 147125211
Offset: 0
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).
Programs
-
Magma
a0:=3; a1:=8; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Vincenzo Librandi, Jan 25 2017
-
Mathematica
LinearRecurrence[{1, 1}, {3, 8}, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 17 2008 *) Table[2 LucasL[n+1] + Fibonacci[n+2], {n,0,40}] (* Rigoberto Florez, Jan 29 2020 *) -
PARI
apply( {A022121(n)=[3,8]*([0,1;1,1]^n)[,1]}, [0..30]) \\ M. F. Hasler, Feb 28 2020
Formula
G.f.: (3+5*x)/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = F(n+2) + 2*L(n+1), where F(n) and L(n) are the n-th Fibonacci and Lucas number, respectively. - Rigoberto Florez, Jan 29 2020
E.g.f.: exp(-2*x/(1+sqrt(5)))*(-2*(5+6*sqrt(5))+(55+27*sqrt(5))*exp(sqrt(5)*x))/(5*(3+sqrt(5))). - Stefano Spezia, Jan 29 2020 after Rigoberto Florez
a(n) = Lucas(n+3) + Fibonacci(n-2). - Greg Dresden and Steve Warren, Feb 28 2022
Comments