OFFSET
1,2
COMMENTS
All terms are Fibonacci numbers A000045: a(2n-1) = Fibonacci(n), a(2n) = Fibonacci(n+2), a(2n-1) = a(2n+2). - Alexander Adamchuk, Oct 08 2006
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).
FORMULA
From Alexander Adamchuk, Oct 08 2006: (Start)
a(n) = Fibonacci(A028242(n+2)).
a(n) = Fibonacci(A030451(n+1)).
a(n) = Fibonacci(3/4 -(-1)^(n+1)*3/4 +(n+1)/2). (End)
G.f. -x*(1+2*x+x^3) / ( -1+x^2+x^4 ). - R. J. Mathar, Mar 08 2011
a(n) = a(n-2) + a(n-4). - Muniru A Asiru, Oct 12 2018
MAPLE
seq(coeff(series(-x*(1+2*x+x^3)/(x^4+x^2-1), x, n+1), x, n), n = 1 .. 50); # Muniru A Asiru, Oct 12 2018
MATHEMATICA
p[0, x] = 1; p[1, x] = x + 1; p[k_, x_] := p[k, x] = x*p[k - 1, x] + (-1)^(n + 1)p[k - 2, x]; Table[Sum[CoefficientList[p[n, x], x][[m]], {m, 1, n + 1}], {n, 0, 20}]
Rest[Flatten[Reverse/@Partition[Fibonacci[Range[30]], 2, 1]]] (* Harvey P. Dale, Mar 19 2013 *)
PROG
(PARI) vector(50, n, fibonacci(3/4 -(-1)^(n+1)*3/4 +(n+1)/2)) \\ G. C. Greubel, Oct 12 2018
(Magma) m:=50; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 + 2*x + x^3)/(1 - x^2 - x^4))); // G. C. Greubel, Oct 12 2018
(GAP) a:=[1, 2, 1, 3];; for n in [5..50] do a[n]:=a[n-2]+a[n-4]; od; a; # Muniru A Asiru, Oct 12 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Roger L. Bagula, Oct 06 2006
EXTENSIONS
More terms from Alexander Adamchuk, Oct 08 2006
STATUS
approved