What is closed-form expression for $F(n)$ when $F(n)=F(n-1)+F(n-2)$ and $F(0)=a$,$F(1)=b$ and $a,b>0$? What is closed-form expression for $F(n)$ when $F(n)=F(n-1)+F(n-2)$ and $F(0)=a$, $F(1)=b$ and  $a,b>0$ ? It seems to be simple generalization of Fibonacci sequence but I can't find closed form for it nor by myself neither using google.
 A: All solutions to the recurrence $F(n)=F(n-1)+F(n-2)$ are of the form
$$F(n)=A\varphi^n+B\widehat\varphi^n\tag{1}$$
for some constants $A$ and $B$, where $$\varphi=\frac{1+\sqrt5}2\quad\text{and}\quad\widehat\varphi=\frac{1-\sqrt5}2\;.$$
To find $A$ and $B$, just substitute your initial values into $(1)$ to get
$$\left\{\begin{align*}
&a=F(0)=A+B\\
&b=F(1)=\varphi A+\widehat\varphi B\;,
\end{align*}\right.$$
and solve the resulting system for $A$ and $B$.
A: Note that
$$
  \binom{F(n)}{F(n+1)}
=
  \begin{bmatrix}
    0 & 1\\[6pt]
    1 & 1
  \end{bmatrix}
  \cdot
  \binom{F(n-1)}{F(n)}
=
  \begin{bmatrix}
    0 & 1\\[6pt]
    1 & 1
  \end{bmatrix}^n
  \binom{a}{b}
$$
so the problem essentially reduces to diagonalize the matrix $M=\begin{bmatrix}0&1\\1&1\end{bmatrix}$.
Denoting $\phi_1=\frac{1-\sqrt 5}{2}$ and $\phi_2=\frac{1+\sqrt 5}{2}$ (the roots of $x^2-x-1=0$) you have
$$
  \begin{bmatrix}
    0 & 1\\
    1 & 1
  \end{bmatrix}
=
  T
  \begin{bmatrix}
    \phi_1 & 0\\
    0 & \phi_2
  \end{bmatrix}
  T^{-1}
$$
where,
$$
  T
=
  \begin{bmatrix}
    -\phi_2 & -\phi_1\\
    1 & 1
  \end{bmatrix}
\quad\text{and}\quad
  T^{-1}
=
  \frac{1}{\sqrt 5}
  \begin{bmatrix}
    -1 & -\phi_1\\
    1 & \phi_2
  \end{bmatrix}
$$
so that
$$
  \binom{F(n)}{F(n+1)}
=
  T
  \begin{bmatrix}
    \phi_1^n & 0\\[6pt]
    0 & \phi_2^n
  \end{bmatrix}
  T^{-1}
  \binom{a}{b}
$$
The first row tells you that
$$
  F(n)
=
  \phi_1\phi_2
  \frac
  {
    (\phi_1^{n-1}-\phi_2^{n-1})a
    +
    (\phi_1^n-\phi_2^n)b
  }
  {
    \sqrt 5
  }
$$
A: The analysis is essentially the same as the one that gives the "Binet" formula for the Fibonacci sequence. Let $\alpha$, $\beta$ be the two roots of the equation $x^2-x-1=0$. Then
$$F(n)=A\alpha^n+B\beta^n,$$ 
where $A$ and $B$ are chosen so that the initial conditions are satisfied.
So set $A+B=a$ and $A\alpha +B\beta=b$, and solve this system of two linear equations for $A$ and $B$. For example, with $a=2$ and $b=1$, we get the Lucas numbers,
A: You can form the generating function for this particular sequence; it is the same as that for the Fib Seq, but the different initial conditions lead to a different function:
$$g(z) = \frac{a+(b-a)z}{1-z-z^2}$$
To get the numbers in the sequence, Taylor expand this function.
A: $F(n)=aFib(n-1)+bFib(n)$ (This is assuming Fib(0)=0Fib(1)=1)
You get this by assuming   $F(n)=ah_n+b g_n$ and realizing that h and g satisfy the Fibonacci recurrence and the initial conditions are shifted ahead.
