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A077916
Expansion of (1-x)^(-1)/(1 + 2*x - 2*x^2 - x^3).
11
1, -1, 5, -10, 30, -74, 199, -515, 1355, -3540, 9276, -24276, 63565, -166405, 435665, -1140574, 2986074, -7817630, 20466835, -53582855, 140281751, -367262376, 961505400, -2517253800, 6590256025, -17253514249, 45170286749, -118257345970, 309601751190, -810547907570
OFFSET
0,3
FORMULA
a(n-1) = Sum_{i=1..n} (-1)^(i+1)*Fibonacci(i)*Fibonacci(i+1), n >= 1. - Alexander Adamchuk, Jun 16 2006
From R. J. Mathar, Mar 14 2011: (Start)
G.f.: 1/((1-x)^2*(1+3*x+x^2)).
a(n) = ((-1)^n*A001906(n+2)+n+2)/5. (End)
O.g.f.: exp( Sum_{n >= 1} Lucas(n)^2*(-x)^n/n ) = 1 - x + 5*x^2 - 10*x^3 + .... Cf. A203803. See also A207969 and A207970. - Peter Bala, Apr 03 2014
From Vladimir Reshetnikov, Oct 28 2015: (Start)
Recurrence (5-term): a(0) = 1, a(1) = -1, a(2) = 5, a(3) = -10, a(n) = -a(n-1) + 4*a(n-2) - a(n-3) - a(n-4).
Recurrence (4-term): a(0) = 1, a(1) = -1, a(2) = 5, n*a(n) = (1-2*n)*a(n-1) + (3*n+3)*a(n-2) + (n+1)*a(n-3).
(End)
a(n) = (F(2n+2)+n+1)/5 if n is odd and a(n)= -(F(2n+2)-n-1)/5 if n is even, where F(n) = Fibonacci numbers (A000045). - Rigoberto Florez, May 09 2019
MATHEMATICA
a[0] = 1; a[1] = -1; a[2] = 5; a[3] = -10; a[n_] := a[n] = -a[n-1] + 4 a[n-2] - a[n-3] - a[n-4]; Table[a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
CoefficientList[Series[(1 - x)^(-1)/(1 + 2*x - 2*x^2 - x^3), {x, 0, 50}], x] (* G. C. Greubel, Dec 25 2017 *)
Table[If[OddQ[n], (Fibonacci[2n+2]+n+1)/5, -(Fibonacci[2n+2]-n-1)/5], {n, 1, 20}] (* Rigoberto Florez, May 09 2019 *)
PROG
(PARI) Vec((1-x)^(-1)/(1+2*x-2*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012
(PARI) Vec(1/((1-x)^2*(1+3*x+x^2)) + O(x^100)) \\ Altug Alkan, Oct 28 2015
CROSSREFS
Cf. A002571.
Bisections are A103433 and A103434.
Sequence in context: A156234 A048010 A002571 * A358728 A373568 A189315
KEYWORD
sign,easy
AUTHOR
N. J. A. Sloane, Nov 17 2002
STATUS
approved