complement of si

[henryzz]

Originally Posted by henryzz

It appears that the even iterations can be worked out using:
\(s_i=4s_{i-2}-s_{i-4}; s_0=2, s_2=4\)

Looks like I am reinventing the wheel. https://oeis.org/A003500 is already connected to https://oeis.org/A003010 Not surprising really.
The standard way of referring to this Lucas sequence of the second kind is \(V_n(4,1)\)
It should be possible to work out SNFS polynomials for this sequence. I am unsure whether this will just boil down to x^2-2 and x^4-4x^2+2 for n a power of 2.

Indeed SNFS polynomials are possible.
If n=4k+r then
r=+-1: f(x)=x^4-2x^3+6x^2-14x+13, g(x)=V_k*x+V_(k+r)
r=+-2: f(x)=7x^4-4x^3+42x^2-388x+1351, g(x)=V_k*x+V_(k+r)

If n=5k+r then
r=+-1: f(x)=x^5-10x^3+40x^2-75x+56, g(x)=V_k*x+V_(k+r)
r=+-2: f(x)=x^5-10x^3+140x^2-975x+2716, g(x)=V_k*x+V_(k+r)

If n=6k+r then
r=+-1: f(x)=x^6-3x^5+15x^4-70x^3+195x^2-291x+181, g(x)=V_k*x+V_(k+r)
r=+-2: f(x)=7x^6-6x^5+105x^4-1940x^3+20265x^2-112902x+262087, g(x)=V_k*x+V_(k+r)
r=+-3: f(x)=13x^6-3x^5+195x^4-13510x^3+526695x^2-10951203x+94875313, g(x)=V_k*x+V_(k+r)

Some of these polynomials may benefit from rotation especially the deg 6 polys.