Recurrence - using power series

Question

Could you help me in solving this recursion( a closed form ) using power series

$\mu(n)=\mu(n−1)k_0+(n−1)\mu(n−2) k_1 \tag 1$,

where $k_0,k_1$ are constants $\mu(0)=3,\mu(1)=5$

HINT: We can think about a substitution such that it will be converted to a solvable non varying coefficient case.. Just thinking..

NB :: Avoid any methods where $k_i$s come as denominator in any steps. We are looking for a higher dimension like matrices, compatible solution. So a method based on series will be ideal for upward compatibility(Avoid ODE methods,Issue is upward compatibility with matrices. All matrices are not commutative. So it will be an issue when solving ODE.). Keep the positions of $k_i$ as it is in the question(by seeing the upward compatibility for matrices). A solution which satisfies the constraints in the question for non matrices can be easily transformed to matrices.. Thanks

Answer 1

Hint: If $g(x) = \sum_{n=0}^\infty \mu(n) x^n$, the recursion corresponds to a differential equation for $g(x)$. Solve that differential equation with initial conditions $g(0) = 3$, $g'(0) = 5$.

EDIT: Actually that won't work in this case, because the differential equation has a nasty singularity at $0$.

Note that with $\mu(n) = k_0^{n} u(n)$ and $k_1 = c k_0^2 $ the recursion becomes

$$u(n) = u(n-1) + c (n-1) u(n-2),\ n \ge 2$$

Rather than an ordinary generating function, consider an exponential generating function

$$G(z) = \sum_{n=0}^\infty u(n) z^n/n!$$

Using Maple's gfun package, I find that two linearly independent egf's satifying this recurrence are

$$G_1(z) = \exp((cz+1)^2/(2c)),\ G_2(z) = \text{erf}((cz+1)/\sqrt{2c}) \exp((cz+1)^2/(2c))$$

The appropriate linear combination of these will satisfy your initial conditions.