Hi everyone!

Let $$$R$$$ be a ring, $$$d_0, d_1, d_2, \dots \in R$$$ and $$$e_0, e_1, e_2, \dots \in R$$$ be linear recurrence sequences, such that

In some applications, the following two sequences arise:

Today I'd like to write about the framework that allows to prove that both the sequences defined above are also linear recurrences. It would also allow to compute their characteristic polynomials in $$$O(kl \log kl)$$$, which is optimal as their degrees are $$$O(kl)$$$ in both cases.

### Umbral calculus

Generally, a linear recurrence $$$f_k$$$ can be described and analyzed with the help of the linear functional $$$T : R[f] \mapsto R$$$ such that

For such functional, $$$T(P(f)) = 0$$$ when $$$P(f)$$$ is a multiple of the characteristic polynomial of $$$f_k$$$. The existence of the characteristic polynomial is the criterion of $$$f_k$$$ being a linear recurrence. So, we need to prove that there is such a polynomial for $$$f_k$$$ defined above.

#### Joint umbral calculus

To analyze joint properties of $$$d_k$$$ and $$$e_k$$$, we define a linear functional $$$T: R[d, e] \to R$$$ such that

Similarly to the case of a single recurrence, $$$T(f(d, e))=0$$$ whenever $$$f(d, e)$$$ is a linear combination of $$$a(d)$$$ and $$$b(e)$$$, where

are the characteristic polynomials of $$$d_i$$$ and $$$e_j$$$. In other words, $$$T(f(d, e))=0$$$ whenever $$$f(d, e)$$$ lies in the ideal $$$\langle a(d), b(e) \rangle$$$.

### Composed sum

For the binomial convolution let $$$f=d+e$$$, then

To show that $$$f_k$$$ is a linear recurrence obeying to the rule

it is sufficient to show that there is a characteristics polynomial $$$c(f)$$$ such that $$$c(f) \in \langle a(d), b(e) \rangle$$$.

Assume that $$$R$$$ is an integral domain. Then the polynomial exists and can be defined explicitly as

where

The fact that $$$c(d+e) \in \langle a(d), b(e) \rangle$$$ is proven as follows:

In the sum above, there are $$$2^{kl}$$$ summands, each of them is divisible by either $$$a(d)$$$ or $$$b(e)$$$, so $$$c(d+e) \in \langle a(d), b(e)\rangle$$$.

The polynomial $$$c(f)$$$ defined above is called the composed sum of $$$a(d)$$$ and $$$b(e)$$$.

### Composed product

Now the question is, how to prove that the Hadamard product $$$f_k = d_k e_k$$$ is a linear recurrence?

Using similar logic as above, one would define $$$f = de$$$ and then look for $$$c(f) \in \langle a(d), b(e) \rangle$$$. Let

This one is a bit trickier to prove. Let's start with $$$k=l=1$$$:

Rewriting it in the same way for arbitrary $$$k$$$ and $$$l$$$, we get

Then the same logic applies as to $$$c(d+e)$$$ in the binomial convolution case.

The polynomial $$$c(f) = c(de)$$$ defined above is called the composed product of $$$a(d)$$$ and $$$b(e)$$$.

### Computing composed products and sums

Let $$$s_i$$$ be the sum of $$$i$$$-th powers of all $$$\lambda$$$ and $$$t_j$$$ be the sum of $$$j$$$-th powers of all $$$\mu$$$, that is

The roots of the composed sum are $$$\lambda_i + \mu_j$$$ for all $$$i$$$ and $$$j$$$ and of the composed product are $$$\lambda_i \mu_j$$$, from which we can see that

So, if we're able to transform from $$$a(d)$$$ to $$$s_i$$$, then from $$$b(e)$$$ to $$$t_j$$$, compute the transforms above on them and then recover the characteristics polynomials from the result, it would solve the problem.

Next thing we should note is that the generating function of $$$s_i$$$ is

It can be further expanded as

where

is the reversed characteristic polynomial of $$$d_k$$$. Its log-derivative is indeed

Finally, to inverse this transform, we could make use of the fact that $$$\frac{A'}{A} = (\log A)'$$$, hence for

it holds that

The resulting $$$f(x)$$$ has degree $$$kl$$$, so only $$$kl$$$ terms of $$$\frac{A'}{A}$$$ and $$$\exp$$$ are needed and they may be computed in $$$O(kl \log kl)$$$.