OFFSET
0,1
COMMENTS
Each number in the sequence is the partial sum of A033505 (n starts at 0, each number add one if n is even). We can also find the recursion relation a(n) = 2*a(n-1) + 4*a(n-2) - a(n-4) for the sequence, which can be proved by induction.
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,4,0,-1).
FORMULA
a(n) = 2*a(n-1) + 4*a(n-2) - a(n-4) for n>=4.
G.f.: (2 - x^2) / ((1 + x)*(1 - 3*x - x^2 + x^3)). - Colin Barker, Jul 12 2018
EXAMPLE
For n=4, a(4) = 150 = 2*a(3) + 4*a(2) - a(0).
MATHEMATICA
CoefficientList[ Series[(-x^2 + 1)/(x^4 - 4x^2 - 2x + 1), {x, 0, 27}], x] (* or *) LinearRecurrence[{2, 4, 0, -1}, {2, 4, 15, 46}, 27] (* Robert G. Wilson v, Jul 15 2018 *)
PROG
(PARI) Vec((2 - x^2) / ((1 + x)*(1 - 3*x - x^2 + x^3)) + O(x^30)) \\ Colin Barker, Jul 12 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Zijing Wu, Jul 11 2018
EXTENSIONS
More terms from Colin Barker, Jul 12 2018
STATUS
approved