フィボナッチ多項式の積分解

関係式
$$F_{m+n}(X)=F_{m+1}(X)F_n(X)+F_m(X)F_{n-1}(X).$$
より
$$\begin{array}{rl}{F}_{(n+2)(2k+1)}(X)& =F_{n(2k+1)+2(2k+1)}(X)\\ & =F_{n(2k+1)+1}(X)F_{2(2k+1)}(X)+F_{n(2k+1)}(X)F_{2(2k+1)-1}(X)\end{array}$$
かつ
$$\begin{array}{rl}{F}_{(n+1)(2k+1)}(X)& =F_{n(2k+1)+(2k+1)}(X)\\ & =F_{n(2k+1)+1}(X)F_{2k+1}(X)+F_{n(2k+1)}(X)F_{(2k+1)-1}(X)\end{array}$$
より,
$$\begin{array}{rl}& F_{(n+2)(2k+1)}(X)F_{2k+1}(X)-F_{(n+1)(2k+1)}(X)F_{2(2k+1)}(X)\\ =& F_{n(2k+1)}(X)F_{2(2k+1)-1}(X)F_{2k+1}(X)-F_{n(2k+1)}(X)F_{(2k+1)-1}(X)F_{2(2k+1)}(X)\\ =& F_{n(2k+1)}(X)(F_{2(2k+1)-1}(X)F_{2k+1}(X)-F_{(2k+1)-1}(X)F_{2(2k+1)}(X))\\ =& F_{n(2k+1)}(X)F_{2k+1}(X)\end{array}$$
となる.
よって,
$$F_{(n+2)(2k+1)}(X)F_{2k+1}(X)=F_{(n+1)(2k+1)}(X)F_{2(2k+1)}(X)+F_{n(2k+1)}(X)F_{2k+1}(X)$$
となる.
ここで, $F_{2(2k+1)}(X)=F_{2k+1}(X)L_{2k+1}(X)$より
$$F_{(n+2)(2k+1)}(X)=L_{2k+1}(X)F_{(n+1)(2k+1)}(X)+F_{n(2k+1)}(X).$$
ここで
$$F_{0(2k+1)}(X)=F_0(X)=0,F_{1(2k+1)}(X)=F_{2k+1}(X),F_{2(2k+1)}(X)=F_{2k+1}(X)L_{2k+1}(X)$$
より,
$$F_{n(2k+1)}(X)=F_{2k+1}(X)F_n(L_{2k+1}(X))$$
となる.