User blog comment:Mh314159/A new approach/@comment-35470197-20190702151646

Let \(M\) denote the maximum of the left had side. The we have \begin{eqnarray*} f_a(n) & \sim & F_{2a}(n) \\ [a] & \sim & F_{2a}(a) = F_{2 \omega}(a) \\ [a,1] & \leq & F_{2 \omega + 1}(a) \\ [a,2] & \leq & F_{2 \omega + 2}^2(a) \\ [a,b] & \leq & F_{2 \omega + b}^2(a) \\ [a,b,1] & \leq & F_{2 \omega + \omega + 1}^2(M) \\ [a,b,2] & \leq & F_{2 \omega + \omega + 2}^2(M) \\ [a,b,c] & \leq & F_{2 \omega + \omega + c}^2(M) \\ [n,\ldots,n] & \leq & F_{\omega^2}^2(n) \\ [x][1] & \leq & F_{\omega^2}^4(M) \\ [x][2] & \leq & F_{\omega^2}^6(M) \\ [x][n] & \leq & F_{\omega^2}^{2n}(M) \\ [x][y] & \leq & F_{\omega^2 + 1}^2(F_{\omega^2}(M)) \\ [x][y][1] & \leq & F_{\omega^2 + 1}^2(F_{\omega^2}^4(M)). \end{eqnarray*} This is the limit of the current version, because [x][1][2] has an infinite loop.