A constant-recursive sequence is any sequence of integers, rational numbers, algebraic numbers, real numbers, or complex numbers ${\displaystyle s_{0},s_{1},s_{2},s_{3},\ldots }$ (written as ${\displaystyle (s_{n})_{n=0}^{\infty }}$ as a shorthand) satisfying a formula of the form

$${\displaystyle s_{n}=c_{1}s_{n-1}+c_{2}s_{n-2}+\dots +c_{d}s_{n-d},}$$

for all ${\displaystyle n\geq d,}$ for some fixed coefficients ${\displaystyle c_{1},c_{2},\dots ,c_{d}}$ ranging over the same domain as the sequence (integers, rational numbers, algebraic numbers, real numbers, or complex numbers). The equation is called a linear recurrence with constant coefficients of order d. The order of the sequence is the smallest positive integer ${\displaystyle d}$ such that the sequence satisfies a recurrence of order d, or ${\displaystyle d=0}$ for the everywhere-zero sequence.^{[citation needed]}

The definition above allows eventually-periodic sequences such as ${\displaystyle 1,0,0,0,\ldots }$ and ${\displaystyle 0,1,0,0,\ldots }$. Some authors require that ${\displaystyle c_{d}\neq 0}$, which excludes such sequences.^[3][4][5]

Selected examples of integer constant-recursive sequences^[6]

Name	Order (${\displaystyle d}$)	First few values	Recurrence (for ${\displaystyle n\geq d}$)	Generating function	OEIS
Zero sequence	0	0, 0, 0, 0, 0, 0, ...	${\displaystyle s_{n}=0}$	${\displaystyle {\frac {0}{1}}}$	A000004
One sequence	1	1, 1, 1, 1, 1, 1, ...	${\displaystyle s_{n}=s_{n-1}}$	${\displaystyle {\frac {1}{1-x}}}$	A000012
Characteristic function of ${\displaystyle \{0\}}$	1	1, 0, 0, 0, 0, 0, ...	${\displaystyle s_{n}=0}$	${\displaystyle {\frac {1}{1}}}$	A000007
Powers of two	1	1, 2, 4, 8, 16, 32, ...	${\displaystyle s_{n}=2s_{n-1}}$	${\displaystyle {\frac {1}{1-2x}}}$	A000079
Powers of −1	1	1, −1, 1, −1, 1, −1, ...	${\displaystyle s_{n}=-s_{n-1}}$	${\displaystyle {\frac {1}{1+x}}}$	A033999
Characteristic function of ${\displaystyle \{1\}}$	2	0, 1, 0, 0, 0, 0, ...	${\displaystyle s_{n}=0}$	${\displaystyle {\frac {x}{1}}}$	A063524
Decimal expansion of 1/6	2	1, 6, 6, 6, 6, 6, ...	${\displaystyle s_{n}=s_{n-1}}$	${\displaystyle {\frac {1+5x}{1-x}}}$	A020793
Decimal expansion of 1/11	2	0, 9, 0, 9, 0, 9, ...	${\displaystyle s_{n}=s_{n-2}}$	${\displaystyle {\frac {9x}{1-x^{2}}}}$	A010680
Nonnegative integers	2	0, 1, 2, 3, 4, 5, ...	${\displaystyle s_{n}=2s_{n-1}-s_{n-2}}$	${\displaystyle {\frac {x}{(1-x)^{2}}}}$	A001477
Odd positive integers	2	1, 3, 5, 7, 9, 11, ...	${\displaystyle s_{n}=2s_{n-1}-s_{n-2}}$	${\displaystyle {\frac {1+x}{(1-x)^{2}}}}$	A005408
Fibonacci numbers	2	0, 1, 1, 2, 3, 5, 8, 13, ...	${\displaystyle s_{n}=s_{n-1}+s_{n-2}}$	${\displaystyle {\frac {x}{1-x-x^{2}}}}$	A000045
Lucas numbers	2	2, 1, 3, 4, 7, 11, 18, 29, ...	${\displaystyle s_{n}=s_{n-1}+s_{n-2}}$	${\displaystyle {\frac {2-x}{1-x-x^{2}}}}$	A000032
Pell numbers	2	0, 1, 2, 5, 12, 29, 70, ...	${\displaystyle s_{n}=2s_{n-1}+s_{n-2}}$	${\displaystyle {\frac {x}{1-2x-x^{2}}}}$	A000129
Powers of two interleaved with 0s	2	1, 0, 2, 0, 4, 0, 8, 0, ...	${\displaystyle s_{n}=2s_{n-2}}$	${\displaystyle {\frac {1}{1-2x^{2}}}}$	A077957
Inverse of 6th cyclotomic polynomial	2	1, 1, 0, −1, −1, 0, 1, 1, ...	${\displaystyle s_{n}=s_{n-1}-s_{n-2}}$	${\displaystyle {\frac {1}{1-x+x^{2}}}}$	A010892
Triangular numbers	3	0, 1, 3, 6, 10, 15, 21, ...	${\displaystyle s_{n}=3s_{n-1}-3s_{n-2}+s_{n-3}}$	${\displaystyle {\frac {x}{(1-x)^{3}}}}$	A000217

The sequence 0, 1, 1, 2, 3, 5, 8, 13, ... of Fibonacci numbers is constant-recursive of order 2 because it satisfies the recurrence ${\displaystyle F_{n}=F_{n-1}+F_{n-2}}$ with ${\displaystyle F_{0}=0,F_{1}=1}$. For example, ${\displaystyle F_{2}=F_{1}+F_{0}=1+0=1}$ and ${\displaystyle F_{6}=F_{5}+F_{4}=5+3=8}$. The sequence 2, 1, 3, 4, 7, 11, ... of Lucas numbers satisfies the same recurrence as the Fibonacci sequence but with initial conditions ${\displaystyle L_{0}=2}$ and ${\displaystyle L_{1}=1}$. More generally, every Lucas sequence is constant-recursive of order 2.^[2]

For any ${\displaystyle a}$ and any ${\displaystyle r\neq 0}$, the arithmetic progression ${\displaystyle a,a+r,a+2r,\ldots }$ is constant-recursive of order 2, because it satisfies ${\displaystyle s_{n}=2s_{n-1}-s_{n-2}}$. Generalizing this, see polynomial sequences below.^{[citation needed]}

For any ${\displaystyle a\neq 0}$ and ${\displaystyle r}$, the geometric progression ${\displaystyle a,ar,ar^{2},\ldots }$ is constant-recursive of order 1, because it satisfies ${\displaystyle s_{n}=rs_{n-1}}$. This includes, for example, the sequence 1, 2, 4, 8, 16, ... as well as the rational number sequence ${\textstyle 1,{\frac {1}{2}},{\frac {1}{4}},{\frac {1}{8}},{\frac {1}{16}},...}$.^{[citation needed]}

A sequence that is eventually periodic with period length ${\displaystyle \ell }$ is constant-recursive, since it satisfies ${\displaystyle s_{n}=s_{n-\ell }}$ for all ${\displaystyle n\geq d}$, where the order ${\displaystyle d}$ is the length of the initial segment including the first repeating block. Examples of such sequences are 1, 0, 0, 0, ... (order 1) and 1, 6, 6, 6, ... (order 2).^{[citation needed]}

A sequence defined by a polynomial ${\displaystyle s_{n}=a_{0}+a_{1}n+a_{2}n^{2}+\cdots +a_{d}n^{d}}$ is constant-recursive. The sequence satisfies a recurrence of order ${\displaystyle d+1}$ (where ${\displaystyle d}$ is the degree of the polynomial), with coefficients given by the corresponding element of the binomial transform.^[7][8] The first few such equations are

${\displaystyle s_{n}=1\cdot s_{n-1}}$ for a degree 0 (that is, constant) polynomial,

${\displaystyle s_{n}=2\cdot s_{n-1}-1\cdot s_{n-2}}$ for a degree 1 or less polynomial,

${\displaystyle s_{n}=3\cdot s_{n-1}-3\cdot s_{n-2}+1\cdot s_{n-3}}$ for a degree 2 or less polynomial, and

${\displaystyle s_{n}=4\cdot s_{n-1}-6\cdot s_{n-2}+4\cdot s_{n-3}-1\cdot s_{n-4}}$ for a degree 3 or less polynomial.

A sequence obeying the order-d equation also obeys all higher order equations. These identities may be proved in a number of ways, including via the theory of finite differences.^[9] Any sequence of ${\displaystyle d+1}$ integer, real, or complex values can be used as initial conditions for a constant-recursive sequence of order ${\displaystyle d+1}$. If the initial conditions lie on a polynomial of degree ${\displaystyle d-1}$ or less, then the constant-recursive sequence also obeys a lower order equation.

Let ${\displaystyle L}$ be a regular language, and let ${\displaystyle s_{n}}$ be the number of words of length ${\displaystyle n}$ in ${\displaystyle L}$. Then ${\displaystyle (s_{n})_{n=0}^{\infty }}$ is constant-recursive.^[10] For example, ${\displaystyle s_{n}=2^{n}}$ for the language of all binary strings, ${\displaystyle s_{n}=1}$ for the language of all unary strings, and ${\displaystyle s_{n}=F_{n+2}}$ for the language of all binary strings that do not have two consecutive ones. More generally, any function accepted by a weighted automaton over the unary alphabet ${\displaystyle \Sigma =\{a\}}$ over the semiring ${\displaystyle (\mathbb {R} ,+,\times )}$ (which is in fact a ring, and even a field) is constant-recursive.^{[citation needed]}

The sequences of Jacobsthal numbers, Padovan numbers, Pell numbers, and Perrin numbers^[2] are constant-recursive.

The factorial sequence ${\displaystyle 1,1,2,6,24,120,720,\ldots }$ is not constant-recursive. More generally, every constant-recursive function is asymptotically bounded by an exponential function (see #Closed-form characterization) and the factorial sequence grows faster than this.

The Catalan sequence ${\displaystyle 1,1,2,5,14,42,132,\ldots }$ is not constant-recursive. This is because the generating function of the Catalan numbers is not a rational function (see #Equivalent definitions).

${\displaystyle F_{n}={\begin{bmatrix}0&1\end{bmatrix}}{\begin{bmatrix}1&1\\1&0\end{bmatrix}}^{n}{\begin{bmatrix}1\\0\end{bmatrix}}.}$

Definition of the Fibonacci sequence using matrices.

A sequence ${\displaystyle (s_{n})_{n=0}^{\infty }}$ is constant-recursive of order less than or equal to ${\displaystyle d}$ if and only if it can be written as

${\displaystyle s_{n}=uA^{n}v}$

where ${\displaystyle u}$ is a ${\displaystyle 1\times d}$ vector, ${\displaystyle A}$ is a ${\displaystyle d\times d}$ matrix, and ${\displaystyle v}$ is a ${\displaystyle d\times 1}$ vector, where the elements come from the same domain (integers, rational numbers, algebraic numbers, real numbers, or complex numbers) as the original sequence. Specifically, ${\displaystyle v}$ can be taken to be the first ${\displaystyle d}$ values of the sequence, ${\displaystyle A}$ the linear transformation that computes ${\displaystyle s_{n+1},s_{n+2},\ldots ,s_{n+d}}$ from ${\displaystyle s_{n},s_{n+1},\ldots ,s_{n+d-1}}$, and ${\displaystyle u}$ the vector ${\displaystyle [0,0,\ldots ,0,1]}$.^[11]

Non-homogeneous	Homogeneous
${\displaystyle s_{n}=1+s_{n-1}}$	${\displaystyle s_{n}=2s_{n-1}-s_{n-2}}$
${\displaystyle s_{0}=0}$	${\displaystyle s_{0}=0;s_{1}=1}$

Definition of the sequence of natural numbers ${\displaystyle s_{n}=n}$, using a non-homogeneous recurrence and the equivalent homogeneous version.

A non-homogeneous linear recurrence is an equation of the form

${\displaystyle s_{n}=c_{1}s_{n-1}+c_{2}s_{n-2}+\dots +c_{d}s_{n-d}+c}$

where ${\displaystyle c}$ is an additional constant. Any sequence satisfying a non-homogeneous linear recurrence is constant-recursive. This is because subtracting the equation for ${\displaystyle s_{n-1}}$ from the equation for ${\displaystyle s_{n}}$ yields a homogeneous recurrence for ${\displaystyle s_{n}-s_{n-1}}$, from which we can solve for ${\displaystyle s_{n}}$ to obtain^{[citation needed]}

${\displaystyle {\begin{aligned}s_{n}=&(c_{1}+1)s_{n-1}\\&+(c_{2}-c_{1})s_{n-2}+\dots +(c_{d}-c_{d-1})s_{n-d}\\&-c_{d}s_{n-d-1}.\end{aligned}}}$

${\displaystyle \sum _{n=0}^{\infty }F_{n}x^{n}={\frac {x}{1-x-x^{2}}}.}$

Definition of the Fibonacci sequence using a generating function.

A sequence is constant-recursive precisely when its generating function

${\displaystyle \sum _{n=0}^{\infty }s_{n}x^{n}=s_{0}+s_{1}x^{1}+s_{2}x^{2}+s_{3}x^{3}+\cdots }$

is a rational function ${\displaystyle p(x)\,/\,q(x)}$, where ${\displaystyle p}$ and ${\displaystyle q}$ are polynomials and ${\displaystyle q(0)=1}$.^[3] Moreover, the order of the sequence is the minimum ${\displaystyle d}$ such that it has such a form with ${\displaystyle {\text{deg }}q(x)\leq d}$ and ${\displaystyle {\text{deg }}p(x)<d}$.^[12]

The denominator is the polynomial obtained from the auxiliary polynomial by reversing the order of the coefficients, and the numerator is determined by the initial values of the sequence:^[13][14]

${\displaystyle \sum _{n=0}^{\infty }s_{n}x^{n}={\frac {b_{0}+b_{1}x^{1}+b_{2}x^{2}+\dots +b_{d-1}x^{d-1}}{1-c_{1}x^{1}-c_{2}x^{2}-\dots -c_{d}x^{d}}},}$

where

${\displaystyle b_{n}=s_{n}-c_{1}s_{n-1}-c_{2}s_{n-2}-\dots -c_{d}s_{n-d}.}$^[15]

It follows from the above that the denominator ${\displaystyle q(x)}$ must be a polynomial not divisible by ${\displaystyle x}$ (and in particular nonzero).

${\displaystyle \{(an+b)_{n=0}^{\infty }:a,b\in \mathbb {R} \}}$

2-dimensional vector space of sequences generated by the sequence ${\displaystyle s_{n}=n}$.

A sequence ${\displaystyle (s_{n})_{n=0}^{\infty }}$ is constant-recursive if and only if the set of sequences

${\displaystyle \left\{(s_{n+r})_{n=0}^{\infty }:r\geq 0\right\}}$

is contained in a sequence space (vector space of sequences) whose dimension is finite. That is, ${\displaystyle (s_{n})_{n=0}^{\infty }}$ is contained in a finite-dimensional subspace of ${\displaystyle \mathbb {C} ^{\mathbb {N} }}$ closed under the left-shift operator.^[16][17]

This characterization is because the order-${\displaystyle d}$ linear recurrence relation can be understood as a proof of linear dependence between the sequences ${\displaystyle (s_{n+r})_{n=0}^{\infty }}$ for ${\displaystyle r=0,\ldots ,d}$. An extension of this argument shows that the order of the sequence is equal to the dimension of the sequence space generated by ${\displaystyle (s_{n+r})_{n=0}^{\infty }}$ for all ${\displaystyle r}$.^[18][17]

${\displaystyle F_{n}={\frac {1}{\sqrt {5}}}(1.618\ldots )^{n}-{\frac {1}{\sqrt {5}}}(-0.618\ldots )^{n}}$

Closed-form characterization of the Fibonacci sequence (Binet's formula)

Constant-recursive sequences admit the following unique closed form characterization using exponential polynomials: every constant-recursive sequence can be written in the form

${\displaystyle s_{n}=z_{n}+k_{1}(n)r_{1}^{n}+k_{2}(n)r_{2}^{n}+\cdots +k_{e}(n)r_{e}^{n},}$

for all ${\displaystyle n\geq 0}$, where

• The term ${\displaystyle z_{n}}$ is a sequence which is zero for all ${\displaystyle n\geq d}$ (where ${\displaystyle d}$ is the order of the sequence);
• The terms ${\displaystyle k_{1}(n),k_{2}(n),\ldots ,k_{e}(n)}$ are complex polynomials; and
• The terms ${\displaystyle r_{1},r_{2},\ldots ,r_{k}}$ are distinct complex constants.^[19][3]

This characterization is exact: every sequence of complex numbers that can be written in the above form is constant-recursive.^[20]

For example, the Fibonacci number ${\displaystyle F_{n}}$ is written in this form using Binet's formula:^[21]

${\displaystyle F_{n}={\frac {1}{\sqrt {5}}}\varphi ^{n}-{\frac {1}{\sqrt {5}}}\psi ^{n},}$

where ${\displaystyle \varphi =(1+{\sqrt {5}})\,/\,2\approx 1.61803\ldots }$ is the golden ratio and ${\displaystyle \psi =-1\,/\,\varphi }$. These are the roots of the equation ${\displaystyle x^{2}-x-1=0}$. In this case, ${\displaystyle e=2}$, ${\displaystyle z_{n}=0}$ for all ${\displaystyle n}$, ${\displaystyle k_{1}(n)=k_{2}(n)=1\,/\,{\sqrt {5}}}$ are both constant polynomials, ${\displaystyle r_{1}=\varphi }$, and ${\displaystyle r_{2}=\psi }$.

The term ${\displaystyle z_{n}}$ is only needed when ${\displaystyle c_{d}\neq 0}$; if ${\displaystyle c_{d}=0}$ then it corrects for the fact that some initial values may be exceptions to the general recurrence. In particular, ${\displaystyle z_{n}=0}$ for all ${\displaystyle n\geq d}$.^{[citation needed]}

The complex numbers ${\displaystyle r_{1},\ldots ,r_{n}}$ are the roots of the characteristic polynomial of the recurrence:

${\displaystyle x^{d}-c_{1}x^{d-1}-\dots -c_{d-1}x-c_{d}}$

whose coefficients are the same as those of the recurrence.^[22] We call ${\displaystyle r_{1},\ldots ,r_{n}}$ the characteristic roots of the recurrence. If the sequence consists of integers or rational numbers, the roots will be algebraic numbers. If the ${\displaystyle d}$ roots ${\displaystyle r_{1},r_{2},\dots ,r_{d}}$ are all distinct, then the polynomials ${\displaystyle k_{i}(n)}$ are all constants, which can be determined from the initial values of the sequence. If the roots of the characteristic polynomial are not distinct, and ${\displaystyle r_{i}}$ is a root of multiplicity ${\displaystyle m}$, then ${\displaystyle k_{i}(n)}$ in the formula has degree ${\displaystyle m-1}$. For instance, if the characteristic polynomial factors as ${\displaystyle (x-r)^{3}}$, with the same root r occurring three times, then the ${\displaystyle n}$th term is of the form ${\displaystyle s_{n}=(a+bn+cn^{2})r^{n}.}$^[23][24]

The sum of two constant-recursive sequences is also constant-recursive.^[25][26] For example, the sum of ${\displaystyle s_{n}=2^{n}}$ and ${\displaystyle t_{n}=n}$ is ${\displaystyle u_{n}=2^{n}+n}$ (${\displaystyle 1,3,6,11,20,\ldots }$), which satisfies the recurrence ${\displaystyle u_{n}=4u_{n-1}-5u_{n-2}+2u_{n-3}}$. The new recurrence can be found by adding the generating functions for each sequence.

Similarly, the product of two constant-recursive sequences is constant-recursive.^[25] For example, the product of ${\displaystyle s_{n}=2^{n}}$ and ${\displaystyle t_{n}=n}$ is ${\displaystyle u_{n}=n\cdot 2^{n}}$ (${\displaystyle 0,2,8,24,64,\ldots }$), which satisfies the recurrence ${\displaystyle u_{n}=4u_{n-1}-4u_{n-2}}$.

The left-shift sequence ${\displaystyle u_{n}=s_{n+1}}$ and the right-shift sequence ${\displaystyle u_{n}=s_{n-1}}$ (with ${\displaystyle u_{0}=0}$) are constant-recursive because they satisfy the same recurrence relation. For example, because ${\displaystyle s_{n}=2^{n}}$ is constant-recursive, so is ${\displaystyle u_{n}=2^{n+1}}$.

In general, constant-recursive sequences are closed under the following operations, where ${\displaystyle s=(s_{n})_{n\in \mathbb {N} },t=(t_{n})_{n\in \mathbb {N} }}$ denote constant-recursive sequences, ${\displaystyle f(x),g(x)}$ are their generating functions, and ${\displaystyle d,e}$ are their orders, respectively.^[27]

Operations on constant-recursive sequences

Operation	Definition	Requirement	Generating function equivalent	Order
Term-wise sum ${\displaystyle s+t}$	${\displaystyle (s+t)_{n}=s_{n}+t_{n}}$	—	${\displaystyle f(x)+g(x)}$	${\displaystyle \leq d+e}$[25]
Term-wise product ${\displaystyle s\cdot t}$	${\displaystyle (s\cdot t)_{n}=s_{n}\cdot t_{n}}$	—	${\displaystyle {\frac {1}{2\pi i}}\int _{\gamma }{\frac {f(\zeta )}{\zeta }}g\left({\frac {x}{\zeta }}\right)\;\mathrm {d} \zeta }$[28][29]	${\displaystyle \leq d\cdot e}$[11][25]
Cauchy product ${\displaystyle s*t}$	${\displaystyle (s*t)_{n}=\sum _{i=0}^{n}s_{i}t_{n-i}}$	—	${\displaystyle f(x)g(x)}$	${\displaystyle \leq d+e}$[27]
Left shift ${\displaystyle Ls}$	${\displaystyle (Ls)_{n}=s_{n+1}}$	—	${\displaystyle {\frac {f(x)-s_{0}}{x}}}$	${\displaystyle \leq d}$[27]
Right shift ${\displaystyle Rs}$	${\displaystyle (Rs)_{n}={\begin{cases}s_{n-1}&n\geq 1\\0&n=0\end{cases}}}$	—	${\displaystyle xf(x)}$	${\displaystyle \leq d+1}$[27]
Cauchy inverse ${\displaystyle s^{(-1)}}$	${\displaystyle (s^{(-1)})_{n}=\sum _{{i_{1}+\dots +i_{k}=n} \atop {i_{1},\ldots ,i_{k}\neq 0}}(-1)^{k}s_{i_{1}}s_{i_{2}}\cdots s_{i_{k}}}$	${\displaystyle s_{0}=1}$	${\displaystyle {\frac {1}{f(x)}}}$	${\displaystyle \leq d+1}$[27]
Kleene star ${\displaystyle s^{(*)}}$	${\displaystyle (s^{(*)})_{n}=\sum _{{i_{1}+\dots +i_{k}=n} \atop {i_{1},\ldots ,i_{k}\neq 0}}s_{i_{1}}s_{i_{2}}\cdots s_{i_{k}}}$	${\displaystyle s_{0}=0}$	${\displaystyle {\frac {1}{1-f(x)}}}$	${\displaystyle \leq d+1}$[27]

The closure under term-wise addition and multiplication follows from the closed-form characterization in terms of exponential polynomials. The closure under Cauchy product follows from the generating function characterization.^[27] The requirement ${\displaystyle s_{0}=1}$ for Cauchy inverse is necessary for the case of integer sequences, but can be replaced by ${\displaystyle s_{0}\neq 0}$ if the sequence is over any field (rational, algebraic, real, or complex numbers).^[27]

Despite satisfying a simple local formula, a constant-recursive sequence can exhibit complicated global behavior. Define a zero of a constant-recursive sequence to be a nonnegative integer ${\displaystyle n}$ such that ${\displaystyle s_{n}=0}$. The Skolem–Mahler–Lech theorem states that the zeros of the sequence are eventually repeating: there exists constants ${\displaystyle M}$ and ${\displaystyle N}$ such that for all ${\displaystyle n>M}$, ${\displaystyle s_{n}=0}$ if and only if ${\displaystyle s_{n+N}=0}$. This result holds for a constant-recursive sequence over the complex numbers, or more generally, over any field of characteristic zero.^[30]

The pattern of zeros in a constant-recursive sequence can also be investigated from the perspective of computability theory. To do so, the description of the sequence ${\displaystyle s_{n}}$ must be given a finite description; this can be done if the sequence is over the integers, rational numbers, or algebraic numbers.^[11] Given such an encoding for sequences ${\displaystyle s_{n}}$, the following problems can be studied:

Notable decision problems

Problem	Description	Status[11][31]
Existence of a zero (Skolem problem)	On input ${\displaystyle (s_{n})_{n=0}^{\infty }}$, is ${\displaystyle s_{n}=0}$ for some ${\displaystyle n}$?	Open
Infinitely many zeros	On input ${\displaystyle (s_{n})_{n=0}^{\infty }}$, is ${\displaystyle s_{n}=0}$ for infinitely many ${\displaystyle n}$?	Decidable
Eventually all zero	On input ${\displaystyle (s_{n})_{n=0}^{\infty }}$, is ${\displaystyle s_{n}=0}$ for all sufficiently large ${\displaystyle n}$?	Decidable
Positivity	On input ${\displaystyle (s_{n})_{n=0}^{\infty }}$, is ${\displaystyle s_{n}>0}$ for all ${\displaystyle n}$?	Open
Eventual positivity	On input ${\displaystyle (s_{n})_{n=0}^{\infty }}$, is ${\displaystyle s_{n}>0}$ for all sufficiently large ${\displaystyle n}$?	Open

Because the square of a constant-recursive sequence ${\displaystyle s_{n}^{2}}$ is still constant-recursive (see closure properties), the existence-of-a-zero problem in the table above reduces to positivity, and infinitely-many-zeros reduces to eventual positivity. Other problems also reduce to those in the above table: for example, whether ${\displaystyle s_{n}=c}$ for some ${\displaystyle n}$ reduces to existence-of-a-zero for the sequence ${\displaystyle s_{n}-c}$. As a second example, for sequences in the real numbers, weak positivity (is ${\displaystyle s_{n}\geq 0}$ for all ${\displaystyle n}$?) reduces to positivity of the sequence ${\displaystyle -s_{n}}$ (because the answer must be negated, this is a Turing reduction).

The Skolem-Mahler-Lech theorem would provide answers to some of these questions, except that its proof is non-constructive. It states that for all ${\displaystyle n>M}$, the zeros are repeating; however, the value of ${\displaystyle M}$ is not known to be computable, so this does not lead to a solution to the existence-of-a-zero problem.^[11] On the other hand, the exact pattern which repeats after ${\displaystyle n>M}$ is computable.^[11][32] This is why the infinitely-many-zeros problem is decidable: just determine if the infinitely-repeating pattern is empty.

Decidability results are known when the order of a sequence is restricted to be small. For example, the Skolem problem is decidable for algebraic sequences of order up to 4.^[33][34][35] It is also known to be decidable for reversible integer sequences up to order 7, that is, sequences that may be continued backwards in the integers.^[31]

Decidability results are also known under the assumption of certain unproven conjectures in number theory. For example, decidability is known for rational sequences of order up to 5 subject to a conjecture known as Skolem's conjecture or the exponential local-global principle. Decidability is also known for all simple rational sequences (those with simple characteristic polynomial) subject to the Skolem conjecture and the weak p-adic Schanuel conjecture.^[36]

Let ${\displaystyle r_{1},\ldots ,r_{n}}$ be the characteristic roots of a constant recursive sequence ${\displaystyle s}$. We say that the sequence is degenerate if the ratio ${\displaystyle r_{i}/r_{j}}$ is a root of unity, for any ${\displaystyle i\neq j}$. It is often easier to study non-degenerate sequences, and one can reduce to this using the following theorem: if ${\displaystyle s}$ has order ${\displaystyle d}$ and is contained in a number field ${\displaystyle K}$ of degree ${\displaystyle k}$ over ${\displaystyle \mathbb {Q} }$, then there is a constant ${\displaystyle M(k,d)\leq {\begin{cases}\exp(2d(3\log d)^{1/2})&{\text{if }}k=1,\\2^{kd+1}&{\text{if }}k\geq 2\end{cases}}}$

such that for some ${\displaystyle M\leq M(k,d)}$ each subsequence ${\displaystyle s_{Mn+\ell }}$ is either identically zero or non-degenerate.^[37]

A D-finite or holonomic sequence is a natural generalization where the coefficients of the recurrence are allowed to be polynomial functions of ${\displaystyle n}$ rather than constants.^[38]

A ${\displaystyle k}$-regular sequence satisfies a linear recurrences with constant coefficients, but the recurrences take a different form. Rather than ${\displaystyle s_{n}}$ being a linear combination of ${\displaystyle s_{m}}$ for some integers ${\displaystyle m}$ that are close to ${\displaystyle n}$, each term ${\displaystyle s_{n}}$ in a ${\displaystyle k}$-regular sequence is a linear combination of ${\displaystyle s_{m}}$ for some integers ${\displaystyle m}$ whose base-${\displaystyle k}$ representations are close to that of ${\displaystyle n}$.^[39] Constant-recursive sequences can be thought of as ${\displaystyle 1}$-regular sequences, where the base-1 representation of ${\displaystyle n}$ consists of ${\displaystyle n}$ copies of the digit ${\displaystyle 1}$.^{[citation needed]}