OFFSET
0,4
COMMENTS
a(n)=a(n - 1) - sqrt(2*a(n - 1)^2 + (-1)^n) = a(n-1)*(1-sqrt(2-(-1)^n/a(n-1)^2)) for n>0.
So in the limit of n->infinity, a(n+1)/a(n)= 1-Sqrt(2).
This matches the following formula in A000045:
Fibonacci(n) = Fibonacci(n - 1)*(1 + sqrt( 5+4*(-1)^(n-1)/Fibonacci(n-1)^2) )/2.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1).
FORMULA
a(0)=1, a(1)=0, a(2)=-1, a(n)=2*a(n-1)+a(n-2) [From Harvey P. Dale, Dec 24 2011]
G.f.: 1 / (1 + x^2 / (1 - 2*x / (1 - x / (1 + x)))). - Michael Somos, Jan 03 2013
G.f.: 1 - Q(0)*x^2/2 , where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013
EXAMPLE
1 - x^2 - 2*x^3 - 5*x^4 - 12*x^5 - 29*x^6 - 70*x^7 - 169*x^8 - 408*x^9 - 985*x^10 + ...
MATHEMATICA
a[0] = 1; a[n_] := a[n] = a[n - 1] - Sqrt[2*a[n - 1]^2 + (-1)^n]; Table[a[n], {n, 0, 30}]
Join[{1}, LinearRecurrence[{2, 1}, {0, -1}, 30]] (* or *) Join[{1}, Rest[ CoefficientList[Series[1 + (1 - 2 x)/(-1 + 2 x + x^2), {x, 0, 30}] , x]]] (* Harvey P. Dale, Dec 24 2011 *)
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Roger L. Bagula, Apr 30 2010
STATUS
approved