User blog comment:QuasarBooster/Fibonacci/Lucas Sequence Extentions?/@comment-2033667-20150730072029

We can work from Binet's formula to create a closed-form expression for your new sequence. Let $$\psi = \varphi^{-1}$$. Then we have $$\phi^n = \phi^{n-1} + \phi^{n-2}$$ and $$\psi^n = \psi^{n-1} + \psi^{n-2}$$. If we have a sequence of the form $$U_n = x\phi^n + y\psi^n$$, then a little algebra shows that $$U_n$$ satisfies the Fibonacci recurrence relation $$U_{n+2} = U_{n} + U_{n+1}$$. So if we find x and y such that $$U_0 = a$$ and $$U_1 = 1$$, we have $$S_a(n) = U_n$$, which gives us the system of equations $$x + y = a$$ and $$x\phi + b\psi = 1$$. If my midnight algebra brain did things right, this gives us $$x = (1 - a\psi)/(\phi - \psi)$$ and $$y = (1 - a\phi)/(\psi - phi)$$, so $$S_a(n) = \frac{1 - a\psi}{\phi - \psi}\phi^n + \frac{1 - a\phi}{\psi - phi}\psi^n = \frac{\phi^n - \psi^n + a\psi^{n-1} - a\phi^{n-1}}{\phi - psi}$$.