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For each (commutative unitary) ring $R$, let $\mathfrak{R}(R)$ be the set of all linear recurrences over $R$, that is, the set of all sequences $(a(n))_{n \geq 0}$ in $R$ such that $$a(n+k) = r_1 a(n+k-1) + \cdots + r_k a(n) \quad \forall n \geq 0 ,$$ for some $r_1, \ldots, r_k \in R$.

It is known that $\mathfrak{R}(R)$ equipped with pointwise addition and pointwise multiplication of sequences is a ring (see [1]).

My question is: How much is known about the factorization of elements of $\mathfrak{R}(R)$ ?

Well, to ask that for a generic ring is probably too much. It seems reasonable to start from the case $R = D$ integral domain of characteristic $0$. In fact, the case $R = \mathbb{Z}$ is already enough interesting in my opinion. Thus, if you known only partial or special results, do not hesitate to let me know.

Let $K$ be the field of fractions of $D$ and let $L$ be the algebraic closure of $K$.

I try to summarize what I found in the literature.

Generalized power sums

It is well-known that any $(a(n))_{n \geq 0} \in \mathfrak{R}(D)$ satisfies $$(\star) \quad a(n) = \sum_{i=1}^m A_i(n) \alpha_i^n \quad \forall n \geq 0, $$ where $\alpha_1, \ldots, \alpha_m \in L$ are all the distinct roots of the polynomial $$x^k - r_1 x^{k-1} - \cdots - r_k$$ and $A_1, \ldots, A_n \in L[X]$ are such that $\deg(A_i) + 1$ is the multiplicity of the root $\alpha_i$. Functions of the form $(\star)$ are called generalized power sums. Actually, one can prove (in several ways) that the generalized power sums are exactly the elements of $\mathfrak{R}(L)$.

It seems that Bézivin developed [2] a theory of factorization for generalized power sums. Unfortunately, his paper is in French and I can't understand it.

Exponential polynomials

When $R$ is a subring of $\mathbb{C}$ it seems that the theory of exponential polynomials could be of help. Broadly speaking, an exponential polynomials is a formal expression of the form $$\sum_{i=1}^m A_i(Z) e^{\omega_i Z} ,$$ where $A_1, \ldots, A_m \in \mathbb{C}[Z]$ and $\omega_1, \ldots, \omega_k \in \mathbb{C}^*$ are pairwise distinct. The set of exponential polynomials is a ring $\mathbb{C}_Z$ in a natural way. The units of $\mathbb{C}_Z$ are $A e^{\omega Z}$, with $A,\omega \in \mathbb{C}^*$. The ring $\mathbb{C}_Z$ is not a UFD since any so called simple exponential polynomials $1 - Ae^{\omega Z}$ has the factor $1 - A^{1/n}e^{(\omega/n) Z}$, for any positive integer $n$. However, it holds [3] a kind of unique factorization theorem for $\mathbb{C}_Z$.

Theorem (Ritt). An exponential polynomial in $Z$ factors, uniquely up to units and associates, as a product of a polynomial in $Z$, a finite product of simple polynomials $1 - A_i e^{\omega_i Z}$ with the $\omega_i$ not rational multiples one of the other, and a finite product of exponential polynomials each irreducible in the ring of exponential polynomials.

With the notation of $(\star)$, for any $(a(n))_{n \geq 0}$ there exists an exponential polynomial $$\tilde{a}(Z) = \sum_{i=1}^m A_i(Z) e^{\log(\alpha_i) Z} ,$$ where $\log(\alpha_i)$ is some choice of a complex logarithm for $\alpha_i$, such that $a(n) = \tilde{a}(n)$ for each integer $n \geq 0$. Hence, one can hope to use Ritt's theorem to prove a factorization theorem in $\mathfrak{R}(R)$. Unfortunately, there is no global lift $\mathfrak{R}(R) \to \mathbb{C}_Z$ such that if $a(n) = b(n)c(n)$ in $\mathfrak{R}(R)$ then $\tilde{a}(Z) = \tilde{b}(n)\tilde{c}(n)$ in $\mathbb{C}_Z$. In fact, let $a(n) = 1^n$ and $b(n) = (-1)^n$, so that $\tilde{a}(Z) = e^{\log(1)Z}$ and $\tilde{b}(Z) = e^{\log(-1)Z}$. Now for any positive integer $k$ it holds $a(n) = (b(n))^{2k}$ in $\mathfrak{R}(R)$. However, for any choice of $\log(1)$ and $\log(-1)$ there exists a positive integer $k$ such that $\tilde{a}(Z) \neq (\tilde{b}(Z))^{2k}$ in $\mathbb{C}_Z$. Notably, when only some particular factorizations of $a(n)$ are considered, a lift may exists (see Sec. 5.1 of [4]).

Thank you in advance for any suggestion/reference.

[1] U. Cerruti and F. Vaccarino, R-algebras of linear recurrent sequences, J. Algebra 175 (1995), no. 1, 332--338.

[2] J.-P. Bézivin, Factorisation de suites récurrentes linéaires, Groupe d'étude d'Analyse ultramétrique, tome 7-8 (1979-1981), no. 33, 1--9.

[3] G. R. Everest and A. J. van der Poorten, Factorisation in the ring of exponential polynomials, Proc. Amer. Math. Soc. 125 (1997), no. 5, 1293--1298.

[4] J.-P. Bézivin, A. Pethö, and A. J. van der Porten, A full characterization of divisibility sequences, Am. J. Math. 112 (1990), no. 6, 985--1001.

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  • $\begingroup$ If $R$ is a domain of characteristic zero (do you actually need the positive characteristic case?), there is a simple description of the units (they are piecewise monomials over suitable arithmetic progressions), and, restricting to linear sequences with only finitely many zeros (clearly this restriction is necessary), Bézivin proved that there is up to the units a unique factorization into irreducibles. What more could you ask for? I suppose similar results will be true in characteristic $p$. $\endgroup$ Nov 18, 2015 at 17:53
  • $\begingroup$ @VesselinDimitrov Thank you. As I said I do not understand Bézivin paper because it is in French. Do you know some English paper about Bézivin's result? Anyway unique factorization sounds strange to me: for any $k \geq 1$ you can write the sequence of Fibonacci numbers $(F_n)_{n \geq 0}$ as the product of the periodic (hence linear recurrent) sequence $1,1,\ldots,1(k\text{ times 1}),F_k, 1,1,\ldots,1(k\text{ times 1}),F_k,\ldots$ and another linear recurrence (thanks to Hadamard quotient theorem). So I guess that these factorization are in some way the same modulo units (?). $\endgroup$
    – user40023
    Nov 18, 2015 at 21:19
  • $\begingroup$ The periodic sequences are units. No, I don't know if Bézivin's proof has been exposed in English. $\endgroup$ Nov 18, 2015 at 21:40
  • $\begingroup$ @VesselinDimitrov OK, however I can make the same example with the periodic sequence replaced with the (nonperiodic) $1,1,\ldots,1(k\text{ times } 1), F_k, 1,1,\ldots,1(k\text{ times } 1), F_{2k}, \ldots$, which is a linear recurrence since it is the interlacing of linear recurrence. $\endgroup$
    – user40023
    Nov 18, 2015 at 22:20
  • $\begingroup$ Right. That's a good point. I don't know, I would have to look into Bézivin's proof to see what is really proved. Perhaps we should restrict to non-degenerate sequences. (?) $\endgroup$ Nov 18, 2015 at 22:36

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