OFFSET
1,2
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,4,-4,1,-1).
FORMULA
a(2n) = (1/2)*(Fibonacci(3n+2)-1).
a(2n+1) = (1/2)*(Fibonacci(3n+1)+Fibonacci(3n+3)-1).
a(n) = a(n-1)+4*a(n-2)-4*a(n-3)+a(n-4)-a(n-5) for n>4. - Colin Barker, Oct 26 2014
G.f.: x*(x^3-x^2+x+1) / ((x-1)*(x^4+4*x^2-1)). - Colin Barker, Oct 26 2014
From Vladimir Reshetnikov, Oct 30 2015: (Start)
a(n) = ((sin(Pi*n/2)*sqrt(5/phi) - cos(Pi*n/2)/phi^2)/phi^(3*n/2) + (sqrt(5*phi)*sin(Pi*n/2)^2 + phi^2*cos(Pi*n/2)^2)*phi^(3*n/2))/(2*sqrt(5)) - 1/2, where phi=(1+sqrt(5))/2.
E.g.f.: phi^2*cosh(phi^(3/2)*x)/(2*sqrt(5)) + sqrt(phi)*sinh(phi^(3/2)*x)/2 - cos(x/phi^(3/2))/(2*sqrt(5)*phi^2) + sin(x/phi^(3/2))/(2*sqrt(phi)) - exp(x)/2.
(End)
MATHEMATICA
s=0; lst={}; Do[f=Fibonacci[n]; If[OddQ[f], AppendTo[lst, s+=f]], {n, 0, 5!}]; lst
CoefficientList[Series[(x^3 - x^2 + x + 1)/((x - 1) (x^4 + 4 x^2 - 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 27 2014 *)
Accumulate[Select[Fibonacci[Range[50]], OddQ]] (* or *) LinearRecurrence[{1, 4, -4, 1, -1}, {1, 2, 5, 10, 23}, 40] (* Harvey P. Dale, Sep 05 2023 *)
PROG
(PARI) Vec(x*(x^3-x^2+x+1)/((x-1)*(x^4+4*x^2-1)) + O(x^100)) \\ Colin Barker, Oct 26 2014
(Magma) &cat[[(1/2)*(Fibonacci(3*n+2)-1), (1/2)*(Fibonacci(3*n+1)+Fibonacci(3*n+3)-1)]: n in [1..30]]; // Vincenzo Librandi, Oct 27 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, Nov 28 2010
STATUS
approved