A176981 Expansion of 2+(1-2*x)/(-1+2*x+x^2).
1, 0, -1, -2, -5, -12, -29, -70, -169, -408, -985, -2378, -5741, -13860, -33461, -80782, -195025, -470832, -1136689, -2744210, -6625109, -15994428, -38613965, -93222358, -225058681, -543339720, -1311738121, -3166815962, -7645370045, -18457556052, -44560482149
Offset: 0
Examples
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 + ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (2,1).
Crossrefs
Cf. A000129.
Programs
-
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 *) FullSimplify[Join[{1}, Table[((1 - Sqrt[2])^(n-1) - (1 + Sqrt[2])^(n-1)) / 2^(3/2), {n, 1, 30}]]] (* Vaclav Kotesovec, Sep 01 2025 *)
Formula
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 lim_{n->infinity} a(n+1)/a(n) = 1+sqrt(2).
Matches the A000045 formula: Fibonacci(n) = Fibonacci(n-1)*(1 + sqrt(5+4*(-1)^(n-1)/Fibonacci(n-1)^2))/2.
a(0)=1, a(1)=0, a(2)=-1, a(n) = 2*a(n-1)+a(n-2). - 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
For n>0, a(n) = ((1 - sqrt(2))^(n-1) - (1 + sqrt(2))^(n-1)) / 2^(3/2). - Vaclav Kotesovec, Sep 01 2025
Extensions
Name corrected by Jason Yuen, Sep 01 2025
Comments