THC Science

In mathematics, a constant-recursive sequence or C-finite sequence is a sequence satisfying a linear recurrence with constant coefficients.

Definition

An order-d homogeneous linear recurrence with constant coefficients is an equation of the form

s(n)=c_{1}s(n-1)+c_{2}s(n-2)+\cdots +c_{d}s(n-d),

where the d coefficients $c_{1},c_{2},\dots ,c_{d}$ are constants.

A sequence $s(0),s(1),s(2),\dots$ is a constant-recursive sequence if there is an integer N such that the sequence satisfies an equation of this form for all $n\geq N$ .

Examples

Fibonacci sequence

The sequence of Fibonacci numbers satisfies the recurrence

F_{n}=F_{n-1}+F_{n-2}

with initial conditions

F_{0}=0

F_{1}=1.

Explicitly, the recurrence yields the values

F_{2}=F_{1}+F_{0}=1

F_{3}=F_{2}+F_{1}=2

F_{4}=F_{3}+F_{2}=3

F_{5}=F_{4}+F_{3}=5

etc. The sequence of Fibonacci numbers is

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... .

Lucas sequences

The sequence 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, ... of Lucas numbers satisfies the same recurrence as the Fibonacci sequence but with initial conditions

L_{0}=2

L_{1}=1.

More generally, every Lucas sequence is a constant-recursive sequence.

Eventually periodic sequences

A sequence that is eventually periodic with period length $\ell$ is constant-recursive, since it satisfies the recurrence $s(n+\ell )=s(n)$ for all $n\geq N$ for some N.

Characterization in terms of exponential polynomials

The characteristic polynomial (or "auxiliary polynomial") of the recurrence is the polynomial

t^{d}-c_{1}t^{d-1}-c_{2}t^{d-2}-\cdots -c_{d}

whose coefficients are the same as those of the recurrence. The nth term $s(n)$ of the sequence can be written in terms of the roots of the characteristic polynomial. If the d roots $r_{1},r_{2},\dots ,r_{d}$ are all distinct, then, for sufficiently large n, the nth term of the sequence is

s(n)=k_{1}r_{1}^{n}+k_{2}r_{2}^{n}+\cdots +k_{d}r_{d}^{n}

where the coefficients k_i are constants that can be determined by the initial conditions.

For the Fibonacci sequence, the characteristic polynomial is $t^{2}-t-1$ , whose roots $\phi ={\frac {1+{\sqrt {5}}}{2}}$ and ${\overline {\phi }}={\frac {1-{\sqrt {5}}}{2}}$ appear in Binet's formula

F_{n}={\frac {\phi ^{n}-{\overline {\phi }}^{n}}{\sqrt {5}}}.

More generally, if a root r of the characteristic polynomial has multiplicity m, then the term $r^{n}$ is multiplied by a polynomial in n of degree m. That is, let $r_{1},\dots ,r_{e}$ be the distinct roots of the characteristic polynomial. Then

s(n)=k_{1}(n)r_{1}^{n}+k_{2}(n)r_{2}^{n}+\cdots +k_{e}(n)r_{e}^{n}

where $k_{i}(n)$ is a polynomial of degree $m_{i}$ . For instance, if the characteristic polynomial factors as $(x-r)^{3}$ , with the same root r occurring three times, then the solution takes the form

s(n)=k_{1}r^{n}+k_{2}nr^{n}+k_{3}n^{2}r^{n}.

^[1]

Conversely, if there are polynomials $k_{i}(n)$ such that

s(n)=k_{1}(n)r_{1}^{n}+k_{2}(n)r_{2}^{n}+\cdots +k_{e}(n)r_{e}^{n}

for sufficiently large n, then $s(n)_{n\geq 0}$ is constant-recursive.

Characterization in terms of rational generating functions

Theorem: Linear recursive sequences are precisely the sequences whose generating function is a rational function. 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.^[2]

The simplest cases are periodic sequences, $a_{n}=a_{n-d},n\geq d$ , which have sequence $a_{0},a_{1},\dots ,a_{d-1},a_{0},\dots$ and generating function a sum of geometric series:

{\begin{aligned}{\frac {a_{0}+a_{1}x^{1}+\cdots +a_{d-1}x^{d-1}}{1-x^{d}}}=&\left(a_{0}+a_{1}x^{1}+\cdots +a_{d-1}x^{d-1}\right)+\left(a_{0}+a_{1}x^{1}+\cdots +a_{d-1}x^{d-1}\right)x^{d}+\\&\left(a_{0}+a_{1}x^{1}+\cdots +a_{d-1}x^{d-1}\right)x^{2d}+\cdots .\end{aligned}}

More generally, given the recurrence relation:

a_{n}=c_{1}a_{n-1}+c_{2}a_{n-2}+\cdots +c_{d}a_{n-d}

with generating function

a_{0}+a_{1}x^{1}+a_{2}x^{2}+\cdots ,

the series is annihilated at a_d and above by the polynomial:

1-c_{1}x^{1}-c_{2}x^{2}-\cdots -c_{d}x^{d}.

That is, multiplying the generating function by the polynomial yields

b_{n}=a_{n}-c_{1}a_{n-1}-c_{2}a_{n-2}-\cdots -c_{d}a_{n-d}

as the coefficient on $x^{n}$ , which vanishes (by the recurrence relation) for n ≥ d. Thus

\left(a_{0}+a_{1}x^{1}+a_{2}x^{2}+\cdots \right)\left(1-c_{1}x^{1}-c_{2}x^{2}-\cdots -c_{d}x^{d}\right)=\left(b_{0}+b_{1}x^{1}+b_{2}x^{2}+\cdots +b_{d-1}x^{d-1}\right)

so dividing yields

a_{0}+a_{1}x^{1}+a_{2}x^{2}+\cdots ={\frac {b_{0}+b_{1}x^{1}+b_{2}x^{2}+\cdots +b_{d-1}x^{d-1}}{1-c_{1}x^{1}-c_{2}x^{2}-\cdots -c_{d}x^{d}}},

expressing the generating function as a rational function.

The denominator is $x^{d}p\left(x^{-1}\right),$ a transform of the auxiliary polynomial (equivalently, reversing the order of coefficients); one could also use any multiple of this, but this normalization is chosen both because of the simple relation to the auxiliary polynomial, and so that $b_{0}=a_{0}$ .

The generating function of the Fibonacci sequence is

{\frac {t}{1-t-t^{2}}}.

The generating function of the Catalan numbers is not a rational function, so the theorem tells us that the Catalan numbers do not satisfy a linear recurrence with constant coefficients.

Solving homogeneous linear recurrence relations with constant coefficients

For order 1, the recurrence

a_{n}=ra_{n-1}

has the solution a_n = rⁿ with a₀ = 1 and the most general solution is a_n = krⁿ with a₀ = k. The characteristic polynomial equated to zero (the characteristic equation) is simply t − r = 0.

Solutions to such recurrence relations of higher order are found by systematic means, often using the fact that a_n = rⁿ is a solution for the recurrence exactly when t = r is a root of the characteristic polynomial. This can be approached directly or using generating functions (formal power series) or matrices.

Consider, for example, a recurrence relation of the form

a_{n}=Aa_{n-1}+Ba_{n-2}.

When does it have a solution of the same general form as a_n = rⁿ? Substituting this guess (ansatz) in the recurrence relation, we find that

r^{n}=Ar^{n-1}+Br^{n-2}

must be true for all n > 1.

Dividing through by rⁿ⁻², we get that all these equations reduce to the same thing:

r^{2}=Ar+B,

r^{2}-Ar-B=0,

which is the characteristic equation of the recurrence relation. Solve for r to obtain the two roots λ₁, λ₂: these roots are known as the characteristic roots or eigenvalues of the characteristic equation. Different solutions are obtained depending on the nature of the roots: If these roots are distinct, we have the general solution

a_{n}=C\lambda _{1}^{n}+D\lambda _{2}^{n}

while if they are identical (when A² + 4B = 0), we have

a_{n}=C\lambda ^{n}+Dn\lambda ^{n}

This is the most general solution; the two constants C and D can be chosen based on two given initial conditions a₀ and a₁ to produce a specific solution.

In the case of complex eigenvalues (which also gives rise to complex values for the solution parameters C and D), the use of complex numbers can be eliminated by rewriting the solution in trigonometric form. In this case we can write the eigenvalues as $\lambda _{1},\lambda _{2}=\alpha \pm \beta i.$ Then it can be shown that

a_{n}=C\lambda _{1}^{n}+D\lambda _{2}^{n}

can be rewritten as^[3]^{: 576–585}

a_{n}=2M^{n}\left(E\cos(\theta n)+F\sin(\theta n)\right)=2GM^{n}\cos(\theta n-\delta ),

where

{\begin{array}{lcl}M={\sqrt {\alpha ^{2}+\beta ^{2}}}&\cos(\theta )={\tfrac {\alpha }{M}}&\sin(\theta )={\tfrac {\beta }{M}}\\C,D=E\mp Fi&&\\G={\sqrt {E^{2}+F^{2}}}&\cos(\delta )={\tfrac {E}{G}}&\sin(\delta )={\tfrac {F}{G}}\end{array}}

Here E and F (or equivalently, G and δ) are real constants which depend on the initial conditions. Using

\lambda _{1}+\lambda _{2}=2\alpha =A,

\lambda _{1}\cdot \lambda _{2}=\alpha ^{2}+\beta ^{2}=-B,

one may simplify the solution given above as

a_{n}=(-B)^{\frac {n}{2}}\left(E\cos(\theta n)+F\sin(\theta n)\right),

where a₁ and a₂ are the initial conditions and

{\begin{aligned}E&={\frac {-Aa_{1}+a_{2}}{B}}\\F&=-i{\frac {A^{2}a_{1}-Aa_{2}+2a_{1}B}{B{\sqrt {A^{2}+4B}}}}\\\theta &=\arccos \left({\frac {A}{2{\sqrt {-B}}}}\right)\end{aligned}}

In this way there is no need to solve for λ₁ and λ₂.

In all cases—real distinct eigenvalues, real duplicated eigenvalues, and complex conjugate eigenvalues—the equation is stable (that is, the variable a converges to a fixed value [specifically, zero]) if and only if both eigenvalues are smaller than one in absolute value. In this second-order case, this condition on the eigenvalues can be shown^[4] to be equivalent to |A| < 1 − B < 2, which is equivalent to |B| < 1 and |A| < 1 − B.

The equation in the above example was homogeneous, in that there was no constant term. If one starts with the non-homogeneous recurrence

b_{n}=Ab_{n-1}+Bb_{n-2}+K

with constant term K, this can be converted into homogeneous form as follows: The steady state is found by setting b_n = b_n−1 = b_n−2 = b* to obtain

b^{*}={\frac {K}{1-A-B}}.

Then the non-homogeneous recurrence can be rewritten in homogeneous form as

[b_{n}-b^{*}]=A[b_{n-1}-b^{*}]+B[b_{n-2}-b^{*}],

which can be solved as above.

The stability condition stated above in terms of eigenvalues for the second-order case remains valid for the general n^th-order case: the equation is stable if and only if all eigenvalues of the characteristic equation are less than one in absolute value.

Solving via linear algebra

A linearly recursive sequence y of order n

y_{n+k}-c_{n-1}y_{n-1+k}-c_{n-2}y_{n-2+k}+\cdots -c_{0}y_{k}=0

is identical to

y_{n}=c_{n-1}y_{n-1}+c_{n-2}y_{n-2}+\cdots +c_{0}y_{0}.

Expanded with n-1 identities of kind $y_{n-k}=y_{n-k},$ this n-th order equation is translated into a system of n first order linear equations,

{\vec {y}}_{n}={\begin{bmatrix}y_{n}\\y_{n-1}\\\vdots \\\vdots \\y_{1}\end{bmatrix}}={\begin{bmatrix}c_{n-1}&c_{n-2}&\cdots &\cdots &c_{0}\\1&0&\cdots &\cdots &0\\0&\ddots &\ddots &&\vdots \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&0\end{bmatrix}}{\begin{bmatrix}y_{n-1}\\y_{n-2}\\\vdots \\\vdots \\y_{0}\end{bmatrix}}=C\ {\vec {y}}_{n-1}=C^{n}{\vec {y}}_{0}.

Observe that the vector ${\vec {y}}_{n}$ can be computed by n applications of the companion matrix, C, to the initial state vector, $y_{0}$ . Thereby, n-th entry of the sought sequence y, is the top component of ${\vec {y}}_{n},y_{n}={\vec {y}}_{n}[n]$ .

Eigendecomposition, ${\vec {y}}_{n}={\vec {C}}^{n}\,{\vec {y}}_{0}=c_{1}\,\lambda _{1}^{n}\,{\vec {e}}_{1}+c_{2}\,\lambda _{2}^{n}\,{\vec {e}}_{2}+\cdots +c_{n}\,\lambda _{n}^{n}\,{\vec {e}}_{n}$ into eigenvalues, $\lambda _{1},\lambda _{2},\ldots ,\lambda _{n}$ , and eigenvectors, ${\vec {e}}_{1},{\vec {e}}_{2},\ldots ,{\vec {e}}_{n}$ , is used to compute ${\vec {y}}_{n}.$ Thanks to the crucial fact that system C time-shifts every eigenvector, e, by simply scaling its components λ times,

C\,{\vec {e}}_{i}=\lambda _{i}{\vec {e}}_{i}=C{\begin{bmatrix}e_{i,n}\\e_{i,n-1}\\\vdots \\e_{i,1}\end{bmatrix}}={\begin{bmatrix}\lambda _{i}\,e_{i,n}\\\lambda _{i}\,e_{i,n-1}\\\vdots \\\lambda _{i}\,e_{i,1}\end{bmatrix}}

that is, time-shifted version of eigenvector,e, has components λ times larger, the eigenvector components are powers of λ, ${\vec {e}}_{i}={\begin{bmatrix}\lambda _{i}^{n-1}&\cdots &\lambda _{i}^{2}&\lambda _{i}&1\end{bmatrix}}^{T},$ and, thus, recurrent homogeneous linear equation solution is a combination of exponential functions, ${\vec {y}}_{n}=\sum _{1}^{n}{c_{i}\,\lambda _{i}^{n}\,{\vec {e}}_{i}}$ . The components $c_{i}$ can be determined out of initial conditions:

{\vec {y}}_{0}={\begin{bmatrix}y_{0}\\y_{-1}\\\vdots \\y_{-n+1}\end{bmatrix}}=\sum _{i=1}^{n}{c_{i}\,\lambda _{i}^{0}\,{\vec {e}}_{i}}={\begin{bmatrix}{\vec {e}}_{1}&{\vec {e}}_{2}&\cdots &{\vec {e}}_{n}\end{bmatrix}}\,{\begin{bmatrix}c_{1}\\c_{2}\\\cdots \\c_{n}\end{bmatrix}}=E\,{\begin{bmatrix}c_{1}\\c_{2}\\\cdots \\c_{n}\end{bmatrix}}

Solving for coefficients,

{\begin{bmatrix}c_{1}\\c_{2}\\\cdots \\c_{n}\end{bmatrix}}=E^{-1}{\vec {y}}_{0}={\begin{bmatrix}\lambda _{1}^{n-1}&\lambda _{2}^{n-1}&\cdots &\lambda _{n}^{n-1}\\\vdots &\vdots &\ddots &\vdots \\\lambda _{1}&\lambda _{2}&\cdots &\lambda _{n}\\1&1&\cdots &1\end{bmatrix}}^{-1}\,{\begin{bmatrix}y_{0}\\y_{-1}\\\vdots \\y_{-n+1}\end{bmatrix}}.

This also works with arbitrary boundary conditions $\underbrace {y_{a},y_{b},\ldots } _{\text{n}}$ , not necessary the initial ones,

{\begin{bmatrix}y_{a}\\y_{b}\\\vdots \end{bmatrix}}={\begin{bmatrix}{\vec {y}}_{a}[n]\\{\vec {y}}_{b}[n]\\\vdots \end{bmatrix}}={\begin{bmatrix}\sum _{i=1}^{n}{c_{i}\,\lambda _{i}^{a}\,{\vec {e}}_{i}[n]}\\\sum _{i=1}^{n}{c_{i}\,\lambda _{i}^{b}\,{\vec {e}}_{i}[n]}\\\vdots \end{bmatrix}}={\begin{bmatrix}\sum _{i=1}^{n}{c_{i}\,\lambda _{i}^{a}\,\lambda _{i}^{n-1}}\\\sum _{i=1}^{n}{c_{i}\,\lambda _{i}^{b}\,\lambda _{i}^{n-1}}\\\vdots \end{bmatrix}}=

={\begin{bmatrix}\sum {c_{i}\,\lambda _{i}^{a+n-1}}\\\sum {c_{i}\,\lambda _{i}^{b+n-1}}\\\vdots \end{bmatrix}}={\begin{bmatrix}\lambda _{1}^{a+n-1}&\lambda _{2}^{a+n-1}&\cdots &\lambda _{n}^{a+n-1}\\\lambda _{1}^{b+n-1}&\lambda _{2}^{b+n-1}&\cdots &\lambda _{n}^{b+n-1}\\\vdots &\vdots &\ddots &\vdots \end{bmatrix}}\,{\begin{bmatrix}c_{1}\\c_{2}\\\vdots \\c_{n}\end{bmatrix}}.

This description is really no different from general method above, however it is more succinct. It also works nicely for situations like

{\begin{cases}a_{n}=a_{n-1}-b_{n-1}\\b_{n}=2a_{n-1}+b_{n-1}.\end{cases}}

where there are several linked recurrences.^[5]

Solving with z-transforms

Certain difference equations - in particular, linear constant coefficient difference equations - can be solved using z-transforms. The z-transforms are a class of integral transforms that lead to more convenient algebraic manipulations and more straightforward solutions. There are cases in which obtaining a direct solution would be all but impossible, yet solving the problem via a thoughtfully chosen integral transform is straightforward.

Theorem

Given a homogeneous linear recurrence relation with constant coefficients of order d, let p(t) be the characteristic polynomial (also "auxiliary polynomial")

t^{d}-c_{1}t^{d-1}-c_{2}t^{d-2}-\cdots -c_{d}=0\,

such that each c_i corresponds to each c_i in the original recurrence relation (see the general form above). Suppose λ is a root of p(t) having multiplicity r. This is to say that (t−λ)^r divides p(t). The following two properties hold:

Each of the r sequences $\lambda ^{n},n\lambda ^{n},n^{2}\lambda ^{n},\dots ,n^{r-1}\lambda ^{n}$ satisfies the recurrence relation.
Any sequence satisfying the recurrence relation can be written uniquely as a linear combination of solutions constructed in part 1 as λ varies over all distinct roots of p(t).

As a result of this theorem a homogeneous linear recurrence relation with constant coefficients can be solved in the following manner:

Find the characteristic polynomial p(t).
Find the roots of p(t) counting multiplicity.
Write a_n as a linear combination of all the roots (counting multiplicity as shown in the theorem above) with unknown coefficients b_i.

a_{n}=\left(b_{1}\lambda _{1}^{n}+b_{2}n\lambda _{1}^{n}+b_{3}n^{2}\lambda _{1}^{n}+\cdots +b_{r}n^{r-1}\lambda _{1}^{n}\right)+\cdots +\left(b_{d-q+1}\lambda _{*}^{n}+\cdots +b_{d}n^{q-1}\lambda _{*}^{n}\right)

This is the general solution to the original recurrence relation. (q is the multiplicity of λ_*)

4. Equate each

a_{0},a_{1},\dots ,a_{d}

from part 3 (plugging in n = 0, ..., d into the general solution of the recurrence relation) with the known values

a_{0},a_{1},\dots ,a_{d}

from the original recurrence relation. However, the values a_n from the original recurrence relation used do not usually have to be contiguous: excluding exceptional cases, just d of them are needed (i.e., for an original homogeneous linear recurrence relation of order 3 one could use the values a₀, a₁, a₄). This process will produce a linear system of d equations with d unknowns. Solving these equations for the unknown coefficients

b_{1},b_{2},\dots ,b_{d}

of the general solution and plugging these values back into the general solution will produce the particular solution to the original recurrence relation that fits the original recurrence relation's initial conditions (as well as all subsequent values

a_{0},a_{1},a_{2},\dots

of the original recurrence relation).

The method for solving linear differential equations is similar to the method above—the "intelligent guess" (ansatz) for linear differential equations with constant coefficients is e^λx where λ is a complex number that is determined by substituting the guess into the differential equation.

This is not a coincidence. Considering the Taylor series of the solution to a linear differential equation:

\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}

it can be seen that the coefficients of the series are given by the n^th derivative of f(x) evaluated at the point a. The differential equation provides a linear difference equation relating these coefficients.

This equivalence can be used to quickly solve for the recurrence relationship for the coefficients in the power series solution of a linear differential equation.

The rule of thumb (for equations in which the polynomial multiplying the first term is non-zero at zero) is that:

y^{[k]}\to f[n+k]

and more generally

x^{m}*y^{[k]}\to n(n-1)(n-m+1)f[n+k-m]

Example: The recurrence relationship for the Taylor series coefficients of the equation:

(x^{2}+3x-4)y^{[3]}-(3x+1)y^{[2]}+2y=0

is given by

n(n-1)f[n+1]+3nf[n+2]-4f[n+3]-3nf[n+1]-f[n+2]+2f[n]=0

or

-4f[n+3]+2nf[n+2]+n(n-4)f[n+1]+2f[n]=0.

This example shows how problems generally solved using the power series solution method taught in normal differential equation classes can be solved in a much easier way.

Example: The differential equation

ay''+by'+cy=0

has solution

y=e^{ax}.

The conversion of the differential equation to a difference equation of the Taylor coefficients is

af[n+2]+bf[n+1]+cf[n]=0.

It is easy to see that the nth derivative of e^ax evaluated at 0 is aⁿ

Non-homogeneous linear recurrence relations with constant coefficients

If the recurrence is non-homogeneous, a particular solution can be found by the method of undetermined coefficients and the solution is the sum of the solution of the homogeneous and the particular solutions. Another method to solve an non-homogeneous recurrence is the method of symbolic differentiation. For example, consider the following recurrence:

a_{n+1}=a_{n}+1

This is an non-homogeneous recurrence. If we substitute n ↦ n+1, we obtain the recurrence

a_{n+2}=a_{n+1}+1

Subtracting the original recurrence from this equation yields

a_{n+2}-a_{n+1}=a_{n+1}-a_{n}

or equivalently

a_{n+2}=2a_{n+1}-a_{n}

This is a homogeneous recurrence, which can be solved by the methods explained above. In general, if a linear recurrence has the form

a_{n+k}=\lambda _{k-1}a_{n+k-1}+\lambda _{k-2}a_{n+k-2}+\cdots +\lambda _{1}a_{n+1}+\lambda _{0}a_{n}+p(n)

where $\lambda _{0},\lambda _{1},\dots ,\lambda _{k-1}$ are constant coefficients and p(n) is the inhomogeneity, then if p(n) is a polynomial with degree r, then this non-homogeneous recurrence can be reduced to a homogeneous recurrence by applying the method of symbolic differencing r times.

If

P(x)=\sum _{n=0}^{\infty }p_{n}x^{n}

is the generating function of the inhomogeneity, the generating function

A(x)=\sum _{n=0}^{\infty }a(n)x^{n}

of the non-homogeneous recurrence

a_{n}=\sum _{i=1}^{s}c_{i}a_{n-i}+p_{n},\quad n\geq n_{r},

with constant coefficients $c i$ is derived from

\left(1-\sum _{i=1}^{s}c_{i}x^{i}\right)A(x)=P(x)+\sum _{n=0}^{n_{r}-1}[a_{n}-p_{n}]x^{n}-\sum _{i=1}^{s}c_{i}x^{i}\sum _{n=0}^{n_{r}-i-1}a_{n}x^{n}.

If P(x) is a rational generating function, A(x) is also one. The case discussed above, where p_n = K is a constant, emerges as one example of this formula, with P(x) = K/(1−x). Another example, the recurrence $a_{n}=10a_{n-1}+n$ with linear inhomogeneity, arises in the definition of the schizophrenic numbers. The solution of homogeneous recurrences is incorporated as p = P = 0.

Notes

^ Greene, Daniel H.; Knuth, Donald E. (1982), "2.1.1 Constant coefficients – A) Homogeneous equations", Mathematics for the Analysis of Algorithms (2nd ed.), Birkhäuser, p. 17.
^ Martino, Ivan; Martino, Luca (2013-11-14). "On the variety of linear recurrences and numerical semigroups". Semigroup Forum. 88 (3): 569–574. doi:10.1007/s00233-013-9551-2. ISSN 0037-1912.
^ Chiang, Alpha C., Fundamental Methods of Mathematical Economics, third edition, McGraw-Hill, 1984.
^ Papanicolaou, Vassilis, "On the asymptotic stability of a class of linear difference equations," Mathematics Magazine 69(1), February 1996, 34–43.
^ Maurer, Stephen B.; Ralston, Anthony (1998), Discrete Algorithmic Mathematics (2nd ed.), A K Peters, p. 609, ISBN 9781568810911.

References

Brousseau, Alfred (1971). Linear Recursion and Fibonacci Sequences. Fibonacci Association.
Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren (1994). Concrete Mathematics: A Foundation for Computer Science (2 ed.). Addison-Welsey. ISBN 0-201-55802-5.

External links

"OEIS Index Rec". OEIS index to a few thousand examples of linear recurrences, sorted by order (number of terms) and signature (vector of values of the constant coefficients)

[1] Greene, Daniel H.; Knuth, Donald E. (1982), "2.1.1 Constant coefficients – A) Homogeneous equations", Mathematics for the Analysis of Algorithms (2nd ed.), Birkhäuser, p. 17.

[2] Martino, Ivan; Martino, Luca (2013-11-14). "On the variety of linear recurrences and numerical semigroups". Semigroup Forum. 88 (3): 569–574. doi:10.1007/s00233-013-9551-2. ISSN 0037-1912.

[3] Chiang, Alpha C., Fundamental Methods of Mathematical Economics, third edition, McGraw-Hill, 1984.

[4] Papanicolaou, Vassilis, "On the asymptotic stability of a class of linear difference equations," Mathematics Magazine 69(1), February 1996, 34–43.

[5] Maurer, Stephen B.; Ralston, Anthony (1998), Discrete Algorithmic Mathematics (2nd ed.), A K Peters, p. 609, ISBN 9781568810911.

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