The Lucas polynomials $L_n(x,s)=\sum_{j=0}^{\left\lfloor \frac{n}{2} \right\rfloor} \frac{n}{n-j}\binom{n-j}{j}s^jx^{n-2j}$ satisfy the recursion $L_n(x,s)=xL_{n-1}(x,s)+sL_{n-2}(x,s)$ with initial values $L_0(x,s)=2$ and $L_1(x,s)=x.$

Binet’s formula gives $L_n(x,s)= {\left( {\frac{{x + \sqrt {{x^2} + 4s} }}{2}} \right)^n} + {\left( {\frac{{x - \sqrt {{x^2} + 4s} }}{2}} \right)^n}.$

For $F_n(x)=L_{2n}(2,x)$ this reduces to your formulas.

The Lucas polynomials $L_n(x,s)=\sum_{j=0}^{\left\lfloor \frac{n}{2} \right\rfloor} \frac{n}{n-j}\binom{n-j}{j}s^jx^{n-2j}$ satisfy the recursion $L_n(x,s)=xL_{n-1}(x,s)+sL_{n-2}(x,s)$ with initial values $L_0(x,s)=2$ and $L_1(x,s)=x.$

Binet’s formula gives $L_n(x,s)= {\left( {\frac{{x + \sqrt {{x^2} + 4s} }}{2}} \right)^n} + {\left( {\frac{{x - \sqrt {{x^2} + 4s} }}{2}} \right)^n}.$

For $2F_n(x)=L_{2n}(2,x)$ this reduces to your formulas.