Generating function

The ordinary generating function of a sequence a_n is

$${\displaystyle G(a_{n};x)=\sum _{n=0}^{\infty }a_{n}x^{n}.}$$

When the term generating function is used without qualification, it is usually taken to mean an ordinary generating function.

If a_n is the probability mass function of a discrete random variable, then its ordinary generating function is called a probability-generating function.

The ordinary generating function can be generalized to arrays with multiple indices. For example, the ordinary generating function of a two-dimensional array a_m,n (where n and m are natural numbers) is

$${\displaystyle G(a_{m,n};x,y)=\sum _{m,n=0}^{\infty }a_{m,n}x^{m}y^{n}.}$$

The exponential generating function of a sequence a_n is

$${\displaystyle \operatorname {EG} (a_{n};x)=\sum _{n=0}^{\infty }a_{n}{\frac {x^{n}}{n!}}.}$$

Exponential generating functions are generally more convenient than ordinary generating functions for combinatorial enumeration problems that involve labelled objects.[3]

Another benefit of exponential generating functions is that they are useful in transferring linear recurrence relations to the realm of differential equations. For example, take the Fibonacci sequence {f_n} that satisfies the linear recurrence relation f_n+2 = f_n+1 + f_n. The corresponding exponential generating function has the form

$${\displaystyle \operatorname {EF} (x)=\sum _{n=0}^{\infty }{\frac {f_{n}}{n!}}x^{n}}$$

and its derivatives can readily be shown to satisfy the differential equation EF″(x) = EF′(x) + EF(x) as a direct analogue with the recurrence relation above. In this view, the factorial term n! is merely a counter-term to normalise the derivative operator acting on xⁿ.

The Poisson generating function of a sequence a_n is

$${\displaystyle \operatorname {PG} (a_{n};x)=\sum _{n=0}^{\infty }a_{n}e^{-x}{\frac {x^{n}}{n!}}=e^{-x}\,\operatorname {EG} (a_{n};x).}$$

The Lambert series of a sequence a_n is

$${\displaystyle \operatorname {LG} (a_{n};x)=\sum _{n=1}^{\infty }a_{n}{\frac {x^{n}}{1-x^{n}}}.}$$

The Lambert series coefficients in the power series expansions

$${\displaystyle b_{n}:=[x^{n}]\operatorname {LG} (a_{n};x)}$$

for integers n ≥ 1 are related by the divisor sum

$${\displaystyle b_{n}=\sum _{d|n}a_{d}.}$$

The main article provides several more classical, or at least well-known examples related to special arithmetic functions in number theory.

In a Lambert series the index n starts at 1, not at 0, as the first term would otherwise be undefined.

The Bell series of a sequence a_n is an expression in terms of both an indeterminate x and a prime p and is given by[4]

$${\displaystyle \operatorname {BG} _{p}(a_{n};x)=\sum _{n=0}^{\infty }a_{p^{n}}x^{n}.}$$

Formal Dirichlet series are often classified as generating functions, although they are not strictly formal power series. The Dirichlet series generating function of a sequence a_n is[5]

$${\displaystyle \operatorname {DG} (a_{n};s)=\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}.}$$

The Dirichlet series generating function is especially useful when a_n is a multiplicative function, in which case it has an Euler product expression[6] in terms of the function's Bell series

$${\displaystyle \operatorname {DG} (a_{n};s)=\prod _{p}\operatorname {BG} _{p}(a_{n};p^{-s})\,.}$$

If a_n is a Dirichlet character then its Dirichlet series generating function is called a Dirichlet L-series. We also have a relation between the pair of coefficients in the Lambert series expansions above and their DGFs. Namely, we can prove that

$${\displaystyle [x^{n}]\operatorname {LG} (a_{n};x)=b_{n}}$$

if and only if

$${\displaystyle \operatorname {DG} (a_{n};s)\zeta (s)=\operatorname {DG} (b_{n};s),}$$

where ζ(s) is the Riemann zeta function.[7]

The idea of generating functions can be extended to sequences of other objects. Thus, for example, polynomial sequences of binomial type are generated by

$${\displaystyle e^{xf(t)}=\sum _{n=0}^{\infty }{\frac {p_{n}(x)}{n!}}t^{n}}$$

where p_n(x) is a sequence of polynomials and f(t) is a function of a certain form. Sheffer sequences are generated in a similar way. See the main article generalized Appell polynomials for more information.

Polynomials are a special case of ordinary generating functions, corresponding to finite sequences, or equivalently sequences that vanish after a certain point. These are important in that many finite sequences can usefully be interpreted as generating functions, such as the Poincaré polynomial and others.

A fundamental generating function is that of the constant sequence 1, 1, 1, 1, 1, 1, 1, 1, 1, ..., whose ordinary generating function is the geometric series

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

The left-hand side is the Maclaurin series expansion of the right-hand side. Alternatively, the equality can be justified by multiplying the power series on the left by 1 − x, and checking that the result is the constant power series 1 (in other words, that all coefficients except the one of x⁰ are equal to 0). Moreover, there can be no other power series with this property. The left-hand side therefore designates the multiplicative inverse of 1 − x in the ring of power series.

Expressions for the ordinary generating function of other sequences are easily derived from this one. For instance, the substitution x → ax gives the generating function for the geometric sequence 1, a, a², a³, ... for any constant a:

$${\displaystyle \sum _{n=0}^{\infty }(ax)^{n}={\frac {1}{1-ax}}.}$$

(The equality also follows directly from the fact that the left-hand side is the Maclaurin series expansion of the right-hand side.) In particular,

$${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}x^{n}={\frac {1}{1+x}}.}$$

One can also introduce regular gaps in the sequence by replacing x by some power of x, so for instance for the sequence 1, 0, 1, 0, 1, 0, 1, 0, ... (which skips over x, x³, x⁵, ...) one gets the generating function

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

By squaring the initial generating function, or by finding the derivative of both sides with respect to x and making a change of running variable n → n + 1, one sees that the coefficients form the sequence 1, 2, 3, 4, 5, ..., so one has

$${\displaystyle \sum _{n=0}^{\infty }(n+1)x^{n}={\frac {1}{(1-x)^{2}}},}$$

and the third power has as coefficients the triangular numbers 1, 3, 6, 10, 15, 21, ... whose term n is the binomial coefficient (^{n + 2}
₂), so that

$${\displaystyle \sum _{n=0}^{\infty }{\binom {n+2}{2}}x^{n}={\frac {1}{(1-x)^{3}}}.}$$

More generally, for any non-negative integer k and non-zero real value a, it is true that

$${\displaystyle \sum _{n=0}^{\infty }a^{n}{\binom {n+k}{k}}x^{n}={\frac {1}{(1-ax)^{k+1}}}\,.}$$

Since

$${\displaystyle 2{\binom {n+2}{2}}-3{\binom {n+1}{1}}+{\binom {n}{0}}=2{\frac {(n+1)(n+2)}{2}}-3(n+1)+1=n^{2},}$$

one can find the ordinary generating function for the sequence 0, 1, 4, 9, 16, ... of square numbers by linear combination of binomial-coefficient generating sequences:

$${\displaystyle G(n^{2};x)=\sum _{n=0}^{\infty }n^{2}x^{n}={\frac {2}{(1-x)^{3}}}-{\frac {3}{(1-x)^{2}}}+{\frac {1}{1-x}}={\frac {x(x+1)}{(1-x)^{3}}}.}$$

We may also expand alternately to generate this same sequence of squares as a sum of derivatives of the geometric series in the following form:

$${\displaystyle {\begin{aligned}G(n^{2};x)&=\sum _{n=0}^{\infty }n^{2}x^{n}\\[4px]&=\sum _{n=0}^{\infty }n(n-1)x^{n}+\sum _{n=0}^{\infty }nx^{n}\\[4px]&=x^{2}D^{2}\left[{\frac {1}{1-x}}\right]+xD\left[{\frac {1}{1-x}}\right]\\[4px]&={\frac {2x^{2}}{(1-x)^{3}}}+{\frac {x}{(1-x)^{2}}}={\frac {x(x+1)}{(1-x)^{3}}}.\end{aligned}}}$$

By induction, we can similarly show for positive integers m ≥ 1 that[8][9]

$${\displaystyle n^{m}=\sum _{j=0}^{m}{\begin{Bmatrix}m\\j\end{Bmatrix}}{\frac {n!}{(n-j)!}},}$$

where {ⁿ
_k} denote the Stirling numbers of the second kind and where the generating function

$${\displaystyle \sum _{n=0}^{\infty }{\frac {n!}{(n-j)!}}\,z^{n}={\frac {j!\cdot z^{j}}{(1-z)^{j+1}}},}$$

so that we can form the analogous generating functions over the integral mth powers generalizing the result in the square case above. In particular, since we can write

$${\displaystyle {\frac {z^{k}}{(1-z)^{k+1}}}=\sum _{i=0}^{k}{\binom {k}{i}}{\frac {(-1)^{k-i}}{(1-z)^{i+1}}},}$$

we can apply a well-known finite sum identity involving the Stirling numbers to obtain that[10]

$${\displaystyle \sum _{n=0}^{\infty }n^{m}z^{n}=\sum _{j=0}^{m}{\begin{Bmatrix}m+1\\j+1\end{Bmatrix}}{\frac {(-1)^{m-j}j!}{(1-z)^{j+1}}}.}$$

The ordinary generating function of a sequence can be expressed as a rational function (the ratio of two finite-degree polynomials) if and only if the sequence is a linear recursive sequence with constant coefficients; this generalizes the examples above. Conversely, every sequence generated by a fraction of polynomials satisfies a linear recurrence with constant coefficients; these coefficients are identical to the coefficients of the fraction denominator polynomial (so they can be directly read off). This observation shows it is easy to solve for generating functions of sequences defined by a linear finite difference equation with constant coefficients, and then hence, for explicit closed-form formulas for the coefficients of these generating functions. The prototypical example here is to derive Binet's formula for the Fibonacci numbers via generating function techniques.

We also notice that the class of rational generating functions precisely corresponds to the generating functions that enumerate quasi-polynomial sequences of the form [11]

$${\displaystyle f_{n}=p_{1}(n)\rho _{1}^{n}+\cdots +p_{l}(n)\rho _{l}^{n},}$$

where the reciprocal roots, ρ_i ∈ ℂ, are fixed scalars and where p_i(n) is a polynomial in n for all 1 ≤ i ≤ l.

In general, Hadamard products of rational functions produce rational generating functions. Similarly, if

$${\displaystyle F(s,t):=\sum _{m,n\geq 0}f(m,n)w^{m}z^{n}}$$

is a bivariate rational generating function, then its corresponding diagonal generating function,

$${\displaystyle \operatorname {diag} (F):=\sum _{n=0}^{\infty }f(n,n)z^{n},}$$

is algebraic. For example, if we let[12]

$${\displaystyle F(s,t):=\sum _{i,j\geq 0}{\binom {i+j}{i}}s^{i}t^{j}={\frac {1}{1-s-t}},}$$

then this generating function's diagonal coefficient generating function is given by the well-known OGF formula

$${\displaystyle \operatorname {diag} (F)=\sum _{n=0}^{\infty }{\binom {2n}{n}}z^{n}={\frac {1}{\sqrt {1-4z}}}.}$$

This result is computed in many ways, including Cauchy's integral formula or contour integration, taking complex residues, or by direct manipulations of formal power series in two variables.

Multiplication of ordinary generating functions yields a discrete convolution (the Cauchy product) of the sequences. For example, the sequence of cumulative sums (compare to the slightly more general Euler–Maclaurin formula)

$${\displaystyle (a_{0},a_{0}+a_{1},a_{0}+a_{1}+a_{2},\ldots )}$$

of a sequence with ordinary generating function G(a_n; x) has the generating function

$${\displaystyle G(a_{n};x)\cdot {\frac {1}{1-x}}}$$

because 1/1 − x is the ordinary generating function for the sequence (1, 1, ...). See also the section on convolutions in the applications section of this article below for further examples of problem solving with convolutions of generating functions and interpretations.

For integers m ≥ 1, we have the following two analogous identities for the modified generating functions enumerating the shifted sequence variants of ⟨ g_{n − m} ⟩ and ⟨ g_{n + m} ⟩, respectively:

$${\displaystyle {\begin{aligned}&z^{m}G(z)=\sum _{n=m}^{\infty }g_{n-m}z^{n}\\[4px]&{\frac {G(z)-g_{0}-g_{1}z-\cdots -g_{m-1}z^{m-1}}{z^{m}}}=\sum _{n=0}^{\infty }g_{n+m}z^{n}.\end{aligned}}}$$

We have the following respective power series expansions for the first derivative of a generating function and its integral:

$${\displaystyle {\begin{aligned}G'(z)&=\sum _{n=0}^{\infty }(n+1)g_{n+1}z^{n}\\[4px]z\cdot G'(z)&=\sum _{n=0}^{\infty }ng_{n}z^{n}\\[4px]\int _{0}^{z}G(t)\,dt&=\sum _{n=1}^{\infty }{\frac {g_{n-1}}{n}}z^{n}.\end{aligned}}}$$

The differentiation–multiplication operation of the second identity can be repeated k times to multiply the sequence by n^k, but that requires alternating between differentiation and multiplication. If instead doing k differentiations in sequence, the effect is to multiply by the kth falling factorial:

$${\displaystyle z^{k}G^{(k)}(z)=\sum _{n=0}^{\infty }n^{\underline {k}}g_{n}z^{n}=\sum _{n=0}^{\infty }n(n-1)\dotsb (n-k+1)g_{n}z^{n}\quad {\text{for all }}k\in \mathbb {N} .}$$

Using the Stirling numbers of the second kind, that can be turned into another formula for multiplying by ${\displaystyle n^{k}}$ as follows (see the main article on generating function transformations):

$${\displaystyle \sum _{j=0}^{k}{\begin{Bmatrix}k\\j\end{Bmatrix}}z^{j}F^{(j)}(z)=\sum _{n=0}^{\infty }n^{k}f_{n}z^{n}\quad {\text{for all }}k\in \mathbb {N} .}$$

A negative-order reversal of this sequence powers formula corresponding to the operation of repeated integration is defined by the zeta series transformation and its generalizations defined as a derivative-based transformation of generating functions, or alternately termwise by and performing an integral transformation on the sequence generating function. Related operations of performing fractional integration on a sequence generating function are discussed here.

In this section we give formulas for generating functions enumerating the sequence {f_{an + b}} given an ordinary generating function F(z) where a, b ∈ ℕ, a ≥ 2, and 0 ≤ b < a (see the main article on transformations). For a = 2, this is simply the familiar decomposition of a function into even and odd parts (i.e., even and odd powers):

$${\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }f_{2n}z^{2n}&={\frac {F(z)+F(-z)}{2}}\\[4px]\sum _{n=0}^{\infty }f_{2n+1}z^{2n+1}&={\frac {F(z)-F(-z)}{2}}.\end{aligned}}}$$

More generally, suppose that a ≥ 3 and that ω_a = exp 2πi/a denotes the ath primitive root of unity. Then, as an application of the discrete Fourier transform, we have the formula[13]

$${\displaystyle \sum _{n=0}^{\infty }f_{an+b}z^{an+b}={\frac {1}{a}}\sum _{m=0}^{a-1}\omega _{a}^{-mb}F\left(\omega _{a}^{m}z\right).}$$

For integers m ≥ 1, another useful formula providing somewhat reversed floored arithmetic progressions — effectively repeating each coefficient m times — are generated by the identity[14]

$${\displaystyle \sum _{n=0}^{\infty }f_{\left\lfloor {\frac {n}{m}}\right\rfloor }z^{n}={\frac {1-z^{m}}{1-z}}F(z^{m})=\left(1+z+\cdots +z^{m-2}+z^{m-1}\right)F(z^{m}).}$$

A formal power series (or function) F(z) is said to be holonomic if it satisfies a linear differential equation of the form[15]

$${\displaystyle c_{0}(z)F^{(r)}(z)+c_{1}(z)F^{(r-1)}(z)+\cdots +c_{r}(z)F(z)=0,}$$

where the coefficients c_i(z) are in the field of rational functions, ℂ(z). Equivalently, F(z) is holonomic if the vector space over ℂ(z) spanned by the set of all of its derivatives is finite dimensional.

Since we can clear denominators if need be in the previous equation, we may assume that the functions, c_i(z) are polynomials in z. Thus we can see an equivalent condition that a generating function is holonomic if its coefficients satisfy a P-recurrence of the form

$${\displaystyle {\widehat {c}}_{s}(n)f_{n+s}+{\widehat {c}}_{s-1}(n)f_{n+s-1}+\cdots +{\widehat {c}}_{0}(n)f_{n}=0,}$$

for all large enough n ≥ n₀ and where the ĉ_i(n) are fixed finite-degree polynomials in n. In other words, the properties that a sequence be P-recursive and have a holonomic generating function are equivalent. Holonomic functions are closed under the Hadamard product operation ⊙ on generating functions.

The functions e^z, log z, cos z, arcsin z, √1 + z, the dilogarithm function Li₂(z), the generalized hypergeometric functions _pF_q(...; ...; z) and the functions defined by the power series

$${\displaystyle \sum _{n=0}^{\infty }{\frac {z^{n}}{(n!)^{2}}}}$$

and the non-convergent

$${\displaystyle \sum _{n=0}^{\infty }n!\cdot z^{n}}$$

are all holonomic.

Examples of P-recursive sequences with holonomic generating functions include f_n ≔ 1/n + 1 (²ⁿ
_n) and f_n ≔ 2ⁿ/n² + 1, where sequences such as √n and log n are not P-recursive due to the nature of singularities in their corresponding generating functions. Similarly, functions with infinitely many singularities such as tan z, sec z, and Γ(z) are not holonomic functions.

Tools for processing and working with P-recursive sequences in Mathematica include the software packages provided for non-commercial use on the RISC Combinatorics Group algorithmic combinatorics software site. Despite being mostly closed-source, particularly powerful tools in this software suite are provided by the Guess package for guessing P-recurrences for arbitrary input sequences (useful for experimental mathematics and exploration) and the Sigma package which is able to find P-recurrences for many sums and solve for closed-form solutions to P-recurrences involving generalized harmonic numbers.[16] Other packages listed on this particular RISC site are targeted at working with holonomic generating functions specifically.

When the series converges absolutely,

$${\displaystyle G\left(a_{n};e^{-i\omega }\right)=\sum _{n=0}^{\infty }a_{n}e^{-i\omega n}}$$

is the discrete-time Fourier transform of the sequence a₀, a₁, ....

In calculus, often the growth rate of the coefficients of a power series can be used to deduce a radius of convergence for the power series. The reverse can also hold; often the radius of convergence for a generating function can be used to deduce the asymptotic growth of the underlying sequence.

For instance, if an ordinary generating function G(a_n; x) that has a finite radius of convergence of r can be written as

$${\displaystyle G(a_{n};x)={\frac {A(x)+B(x)\left(1-{\frac {x}{r}}\right)^{-\beta }}{x^{\alpha }}}}$$

where each of A(x) and B(x) is a function that is analytic to a radius of convergence greater than r (or is entire), and where B(r) ≠ 0 then

$${\displaystyle a_{n}\sim {\frac {B(r)}{r^{\alpha }\Gamma (\beta )}}\,n^{\beta -1}\left({\frac {1}{r}}\right)^{n}\sim {\frac {B(r)}{r^{\alpha }}}{\binom {n+\beta -1}{n}}\left({\frac {1}{r}}\right)^{n}={\frac {B(r)}{r^{\alpha }}}\left(\!\!{\binom {\beta }{n}}\!\!\right)\left({\frac {1}{r}}\right)^{n}\,,}$$

using the Gamma function, a binomial coefficient, or a multiset coefficient.

Often this approach can be iterated to generate several terms in an asymptotic series for a_n. In particular,

$${\displaystyle G\left(a_{n}-{\frac {B(r)}{r^{\alpha }}}{\binom {n+\beta -1}{n}}\left({\frac {1}{r}}\right)^{n};x\right)=G(a_{n};x)-{\frac {B(r)}{r^{\alpha }}}\left(1-{\frac {x}{r}}\right)^{-\beta }\,.}$$

The asymptotic growth of the coefficients of this generating function can then be sought via the finding of A, B, α, β, and r to describe the generating function, as above.

Similar asymptotic analysis is possible for exponential generating functions; with an exponential generating function, it is a_n/n! that grows according to these asymptotic formulae. Generally, if the generating function of one sequence minus the generating function of a second sequence has a radius of convergence that is larger than the radius of convergence of the individual generating functions then the two sequences have the same asymptotic growth.

As derived above, the ordinary generating function for the sequence of squares is

$${\displaystyle G(n^{2};x)={\frac {x(x+1)}{(1-x)^{3}}}.}$$

With r = 1, α = −1, β = 3, A(x) = 0, and B(x) = x + 1, we can verify that the squares grow as expected, like the squares:

$${\displaystyle a_{n}\sim {\frac {B(r)}{r^{\alpha }\Gamma (\beta )}}\,n^{\beta -1}\left({\frac {1}{r}}\right)^{n}={\frac {1+1}{1^{-1}\,\Gamma (3)}}\,n^{3-1}\left({\frac {1}{1}}\right)^{n}=n^{2}.}$$

The ordinary generating function for the Catalan numbers is

$${\displaystyle G(C_{n};x)={\frac {1-{\sqrt {1-4x}}}{2x}}.}$$

With r = 1/4, α = 1, β = −1/2, A(x) = 1/2, and B(x) = −1/2, we can conclude that, for the Catalan numbers,

$${\displaystyle C_{n}\sim {\frac {B(r)}{r^{\alpha }\Gamma (\beta )}}\,n^{\beta -1}\left({\frac {1}{r}}\right)^{n}={\frac {-{\frac {1}{2}}}{\left({\frac {1}{4}}\right)^{1}\Gamma \left(-{\frac {1}{2}}\right)}}\,n^{-{\frac {1}{2}}-1}\left({\frac {1}{\,{\frac {1}{4}}\,}}\right)^{n}={\frac {4^{n}}{n^{\frac {3}{2}}{\sqrt {\pi }}}}.}$$

One can define generating functions in several variables for arrays with several indices. These are called multivariate generating functions or, sometimes, super generating functions. For two variables, these are often called bivariate generating functions.

For instance, since (1 + x)ⁿ is the ordinary generating function for binomial coefficients for a fixed n, one may ask for a bivariate generating function that generates the binomial coefficients (ⁿ
_k) for all k and n. To do this, consider (1 + x)ⁿ itself as a sequence in n, and find the generating function in y that has these sequence values as coefficients. Since the generating function for aⁿ is

$${\displaystyle {\frac {1}{1-ay}},}$$

the generating function for the binomial coefficients is:

$${\displaystyle \sum _{n,k}{\binom {n}{k}}x^{k}y^{n}={\frac {1}{1-(1+x)y}}={\frac {1}{1-y-xy}}.}$$

Expansions of (formal) Jacobi-type and Stieltjes-type continued fractions (J-fractions and S-fractions, respectively) whose hth rational convergents represent 2h-order accurate power series are another way to express the typically divergent ordinary generating functions for many special one and two-variate sequences. The particular form of the Jacobi-type continued fractions (J-fractions) are expanded as in the following equation and have the next corresponding power series expansions with respect to z for some specific, application-dependent component sequences, {ab_i} and {c_i}, where z ≠ 0 denotes the formal variable in the second power series expansion given below:[17]

$${\displaystyle {\begin{aligned}J^{[\infty ]}(z)&={\cfrac {1}{1-c_{1}z-{\cfrac {{\text{ab}}_{2}z^{2}}{1-c_{2}z-{\cfrac {{\text{ab}}_{3}z^{2}}{\ddots }}}}}}\\[4px]&=1+c_{1}z+\left({\text{ab}}_{2}+c_{1}^{2}\right)z^{2}+\left(2{\text{ab}}_{2}c_{1}+c_{1}^{3}+{\text{ab}}_{2}c_{2}\right)z^{3}+\cdots \end{aligned}}}$$

The coefficients of ${\displaystyle z^{n}}$, denoted in shorthand by j_n ≔ [zⁿ] J^[∞](z), in the previous equations correspond to matrix solutions of the equations

$${\displaystyle {\begin{bmatrix}k_{0,1}&k_{1,1}&0&0&\cdots \\k_{0,2}&k_{1,2}&k_{2,2}&0&\cdots \\k_{0,3}&k_{1,3}&k_{2,3}&k_{3,3}&\cdots \\\vdots &\vdots &\vdots &\vdots \end{bmatrix}}={\begin{bmatrix}k_{0,0}&0&0&0&\cdots \\k_{0,1}&k_{1,1}&0&0&\cdots \\k_{0,2}&k_{1,2}&k_{2,2}&0&\cdots \\\vdots &\vdots &\vdots &\vdots \end{bmatrix}}\cdot {\begin{bmatrix}c_{1}&1&0&0&\cdots \\{\text{ab}}_{2}&c_{2}&1&0&\cdots \\0&{\text{ab}}_{3}&c_{3}&1&\cdots \\\vdots &\vdots &\vdots &\vdots \end{bmatrix}},}$$

where j₀ ≡ k_0,0 = 1, j_n = k_0,n for n ≥ 1, k_r,s = 0 if r > s, and where for all integers p, q ≥ 0, we have an addition formula relation given by

$${\displaystyle j_{p+q}=k_{0,p}\cdot k_{0,q}+\sum _{i=1}^{\min(p,q)}{\text{ab}}_{2}\cdots {\text{ab}}_{i+1}\times k_{i,p}\cdot k_{i,q}.}$$

For h ≥ 0 (though in practice when h ≥ 2), we can define the rational hth convergents to the infinite J-fraction, J^[∞](z), expanded by

$${\displaystyle \operatorname {Conv} _{h}(z):={\frac {P_{h}(z)}{Q_{h}(z)}}=j_{0}+j_{1}z+\cdots +j_{2h-1}z^{2h-1}+\sum _{n=2h}^{\infty }{\widetilde {j}}_{h,n}z^{n}}$$

component-wise through the sequences, P_h(z) and Q_h(z), defined recursively by

$${\displaystyle {\begin{aligned}P_{h}(z)&=(1-c_{h}z)P_{h-1}(z)-{\text{ab}}_{h}z^{2}P_{h-2}(z)+\delta _{h,1}\\Q_{h}(z)&=(1-c_{h}z)Q_{h-1}(z)-{\text{ab}}_{h}z^{2}Q_{h-2}(z)+(1-c_{1}z)\delta _{h,1}+\delta _{0,1}.\end{aligned}}}$$

Moreover, the rationality of the convergent function Conv_h(z) for all h ≥ 2 implies additional finite difference equations and congruence properties satisfied by the sequence of j_n, and for M_h ≔ ab₂ ⋯ ab_{h + 1} if h ‖ M_h then we have the congruence

$${\displaystyle j_{n}\equiv [z^{n}]\operatorname {Conv} _{h}(z){\pmod {h}},}$$

for non-symbolic, determinate choices of the parameter sequences {ab_i} and {c_i} when h ≥ 2, that is, when these sequences do not implicitly depend on an auxiliary parameter such as q, x, or R as in the examples contained in the table below.

The next table provides examples of closed-form formulas for the component sequences found computationally (and subsequently proved correct in the cited references[18]) in several special cases of the prescribed sequences, j_n, generated by the general expansions of the J-fractions defined in the first subsection. Here we define 0 < |a|, |b|, |q| < 1 and the parameters R, α ∈ ℤ⁺ and x to be indeterminates with respect to these expansions, where the prescribed sequences enumerated by the expansions of these J-fractions are defined in terms of the q-Pochhammer symbol, Pochhammer symbol, and the binomial coefficients.

${\displaystyle j_{n}}$	${\displaystyle c_{1}}$	${\displaystyle c_{i}(i\geq 2)}$	${\displaystyle \mathrm {ab} _{i}(i\geq 2)}$
${\displaystyle q^{n^{2}}}$	${\displaystyle q}$	${\displaystyle q^{2h-3}\left(q^{2h}+q^{2h-2}-1\right)}$	${\displaystyle q^{6h-10}\left(q^{2h-2}-1\right)}$
${\displaystyle (a;q)_{n}}$	${\displaystyle 1-a}$	${\displaystyle q^{h-1}-aq^{h-2}\left(q^{h}+q^{h-1}-1\right)}$	${\displaystyle aq^{2h-4}\left(aq^{h-2}-1\right)\left(q^{h-1}-1\right)}$
${\displaystyle \left(zq^{-n};q\right)_{n}}$	${\displaystyle {\frac {q-z}{q}}}$	${\displaystyle {\frac {q^{h}-z-qz+q^{h}z}{q^{2h-1}}}}$	${\displaystyle {\frac {\left(q^{h-1}-1\right)\left(q^{h-1}-z\right)\cdot z}{q^{4h-5}}}}$
${\displaystyle {\frac {(a;q)_{n}}{(b;q)_{n}}}}$	${\displaystyle {\frac {1-a}{1-b}}}$	${\displaystyle {\frac {q^{i-2}\left(q+abq^{2i-3}+a(1-q^{i-1}-q^{i})+b(q^{i}-q-1)\right)}{\left(1-bq^{2i-4}\right)\left(1-bq^{2i-2}\right)}}}$	${\displaystyle {\frac {q^{2i-4}\left(1-bq^{i-3}\right)\left(1-aq^{i-2}\right)\left(a-bq^{i-2}\right)\left(1-q^{i-1}\right)}{\left(1-bq^{2i-5}\right)\left(1-bq^{2i-4}\right)^{2}\left(1-bq^{2i-3}\right)}}}$
${\displaystyle \alpha ^{n}\cdot \left({\frac {R}{\alpha }}\right)_{n}}$	${\displaystyle R}$	${\displaystyle R+2\alpha (i-1)}$	${\displaystyle (i-1)\alpha {\bigl (}R+(i-2)\alpha {\bigr )}}$
${\displaystyle (-1)^{n}{\binom {x}{n}}}$	${\displaystyle -x}$	${\displaystyle -{\frac {(x+2(i-1)^{2})}{(2i-1)(2i-3)}}}$	${\displaystyle {\begin{cases}-{\dfrac {(x-i+2)(x+i-1)}{4\cdot (2i-3)^{2}}}&{\text{for }}i\geq 3;\\[4px]-{\frac {1}{2}}x(x+1)&{\text{for }}i=2.\end{cases}}}$
${\displaystyle (-1)^{n}{\binom {x+n}{n}}}$	${\displaystyle -(x+1)}$	${\displaystyle {\frac {{\bigl (}x-2i(i-2)-1{\bigr )}}{(2i-1)(2i-3)}}}$	${\displaystyle {\begin{cases}-{\dfrac {(x-i+2)(x+i-1)}{4\cdot (2i-3)^{2}}}&{\text{for }}i\geq 3;\\[4px]-{\frac {1}{2}}x(x+1)&{\text{for }}i=2.\end{cases}}}$

The radii of convergence of these series corresponding to the definition of the Jacobi-type J-fractions given above are in general different from that of the corresponding power series expansions defining the ordinary generating functions of these sequences.

As an example of a Lambert series identity not given in the main article, we can show that for |x|, |xq| < 1 we have that [19]

$${\displaystyle \sum _{n=1}^{\infty }{\frac {q^{n}x^{n}}{1-x^{n}}}=\sum _{n=1}^{\infty }{\frac {q^{n}x^{n^{2}}}{1-qx^{n}}}+\sum _{n=1}^{\infty }{\frac {q^{n}x^{n(n+1)}}{1-x^{n}}},}$$

where we have the special case identity for the generating function of the divisor function, d(n) ≡ σ₀(n), given by

$${\displaystyle \sum _{n=1}^{\infty }{\frac {x^{n}}{1-x^{n}}}=\sum _{n=1}^{\infty }{\frac {x^{n^{2}}\left(1+x^{n}\right)}{1-x^{n}}}.}$$

$${\displaystyle \operatorname {DG} \left(n^{2};s\right)=\sum _{n=1}^{\infty }{\frac {n^{2}}{n^{s}}}=\zeta (s-2),}$$

using the Riemann zeta function.

The sequence a_k generated by a Dirichlet series generating function (DGF) corresponding to:

$${\displaystyle \operatorname {DG} (a_{k};s)=\zeta (s)^{m}}$$

where ζ(s) is the Riemann zeta function, has the ordinary generating function:

$${\displaystyle \sum _{k=1}^{k=n}a_{k}x^{k}=x+{\binom {m}{1}}\sum _{2\leq a\leq n}x^{a}+{\binom {m}{2}}{\underset {ab\leq n}{\sum _{a=2}^{\infty }\sum _{b=2}^{\infty }}}x^{ab}+{\binom {m}{3}}{\underset {abc\leq n}{\sum _{a=2}^{\infty }\sum _{c=2}^{\infty }\sum _{b=2}^{\infty }}}x^{abc}+{\binom {m}{4}}{\underset {abcd\leq n}{\sum _{a=2}^{\infty }\sum _{b=2}^{\infty }\sum _{c=2}^{\infty }\sum _{d=2}^{\infty }}}x^{abcd}+\cdots }$$

Multivariate generating functions arise in practice when calculating the number of contingency tables of non-negative integers with specified row and column totals. Suppose the table has r rows and c columns; the row sums are t₁, t₂ ... t_r and the column sums are s₁, s₂ ... s_c. Then, according to I. J. Good,[20] the number of such tables is the coefficient of

$${\displaystyle x_{1}^{t_{1}}\cdots x_{r}^{t_{r}}y_{1}^{s_{1}}\cdots y_{c}^{s_{c}}}$$

in

$${\displaystyle \prod _{i=1}^{r}\prod _{j=1}^{c}{\frac {1}{1-x_{i}y_{j}}}.}$$

In the bivariate case, non-polynomial double sum examples of so-termed "double" or "super" generating functions of the form

$${\displaystyle G(w,z):=\sum _{m,n\geq 0}g_{m,n}w^{m}z^{n}}$$

include the following two-variable generating functions for the binomial coefficients, the Stirling numbers, and the Eulerian numbers:[21]

$${\displaystyle {\begin{aligned}e^{z+wz}&=\sum _{m,n\geq 0}{\binom {n}{m}}w^{m}{\frac {z^{n}}{n!}}\\[4px]e^{w(e^{z}-1)}&=\sum _{m,n\geq 0}{\begin{Bmatrix}n\\m\end{Bmatrix}}w^{m}{\frac {z^{n}}{n!}}\\[4px]{\frac {1}{(1-z)^{w}}}&=\sum _{m,n\geq 0}{\begin{bmatrix}n\\m\end{bmatrix}}w^{m}{\frac {z^{n}}{n!}}\\[4px]{\frac {1-w}{e^{(w-1)z}-w}}&=\sum _{m,n\geq 0}\left\langle {\begin{matrix}n\\m\end{matrix}}\right\rangle w^{m}{\frac {z^{n}}{n!}}\\[4px]{\frac {e^{w}-e^{z}}{we^{z}-ze^{w}}}&=\sum _{m,n\geq 0}\left\langle {\begin{matrix}m+n+1\\m\end{matrix}}\right\rangle {\frac {w^{m}z^{n}}{(m+n+1)!}}.\end{aligned}}}$$

Generating functions give us several methods to manipulate sums and to establish identities between sums.

The simplest case occurs when s_n = ∑ⁿ
_{k = 0} a_k. We then know that S(z) = A(z)/1 − z for the corresponding ordinary generating functions.

For example, we can manipulate

$${\displaystyle s_{n}=\sum _{k=1}^{n}H_{k}\,,}$$

where H_k = 1 + 1/2 + ⋯ + 1/k are the harmonic numbers. Let

$${\displaystyle H(z)=\sum _{n=1}^{\infty }{H_{n}z^{n}}}$$

be the ordinary generating function of the harmonic numbers. Then

$${\displaystyle H(z)={\frac {1}{1-z}}\sum _{n=1}^{\infty }{\frac {z^{n}}{n}}\,,}$$

and thus

$${\displaystyle S(z)=\sum _{n=1}^{\infty }{s_{n}z^{n}}={\frac {1}{(1-z)^{2}}}\sum _{n=1}^{\infty }{\frac {z^{n}}{n}}\,.}$$

Using

$${\displaystyle {\frac {1}{(1-z)^{2}}}=\sum _{n=0}^{\infty }(n+1)z^{n}\,,}$$

convolution with the numerator yields

$${\displaystyle s_{n}=\sum _{k=1}^{n}{\frac {n+1-k}{k}}=(n+1)H_{n}-n\,,}$$

which can also be written as

$${\displaystyle \sum _{k=1}^{n}{H_{k}}=(n+1)(H_{n+1}-1)\,.}$$

As another example of using generating functions to relate sequences and manipulate sums, for an arbitrary sequence ⟨ f_n ⟩ we define the two sequences of sums

$${\displaystyle {\begin{aligned}s_{n}&:=\sum _{m=0}^{n}{\binom {n}{m}}f_{m}3^{n-m}\\[4px]{\tilde {s}}_{n}&:=\sum _{m=0}^{n}{\binom {n}{m}}(m+1)(m+2)(m+3)f_{m}3^{n-m}\,,\end{aligned}}}$$

for all n ≥ 0, and seek to express the second sums in terms of the first. We suggest an approach by generating functions.

First, we use the binomial transform to write the generating function for the first sum as

$${\displaystyle S(z)={\frac {1}{1-3z}}F\left({\frac {z}{1-3z}}\right).}$$

Since the generating function for the sequence ⟨ (n + 1)(n + 2)(n + 3) f_n ⟩ is given by

$${\displaystyle 6F(z)+18zF'(z)+9z^{2}F''(z)+z^{3}F'''(z)}$$

we may write the generating function for the second sum defined above in the form

$${\displaystyle {\tilde {S}}(z)={\frac {6}{(1-3z)}}F\left({\frac {z}{1-3z}}\right)+{\frac {18z}{(1-3z)^{2}}}F'\left({\frac {z}{1-3z}}\right)+{\frac {9z^{2}}{(1-3z)^{3}}}F''\left({\frac {z}{1-3z}}\right)+{\frac {z^{3}}{(1-3z)^{4}}}F'''\left({\frac {z}{1-3z}}\right).}$$

In particular, we may write this modified sum generating function in the form of

$${\displaystyle a(z)\cdot S(z)+b(z)\cdot zS'(z)+c(z)\cdot z^{2}S''(z)+d(z)\cdot z^{3}S'''(z),}$$

for a(z) = 6(1 − 3z)³, b(z) = 18(1 − 3z)³, c(z) = 9(1 − 3z)³, and d(z) = (1 − 3z)³, where (1 − 3z)³ = 1 − 9z + 27z² − 27z³.

Finally, it follows that we may express the second sums through the first sums in the following form:

$${\displaystyle {\begin{aligned}{\tilde {s}}_{n}&=[z^{n}]\left(6(1-3z)^{3}\sum _{n=0}^{\infty }s_{n}z^{n}+18(1-3z)^{3}\sum _{n=0}^{\infty }ns_{n}z^{n}+9(1-3z)^{3}\sum _{n=0}^{\infty }n(n-1)s_{n}z^{n}+(1-3z)^{3}\sum _{n=0}^{\infty }n(n-1)(n-2)s_{n}z^{n}\right)\\[4px]&=(n+1)(n+2)(n+3)s_{n}-9n(n+1)(n+2)s_{n-1}+27(n-1)n(n+1)s_{n-2}-(n-2)(n-1)ns_{n-3}.\end{aligned}}}$$

In this example, we reformulate a generating function example given in Section 7.3 of Concrete Mathematics (see also Section 7.1 of the same reference for pretty pictures of generating function series). In particular, suppose that we seek the total number of ways (denoted U_n) to tile a 3-by-n rectangle with unmarked 2-by-1 domino pieces. Let the auxiliary sequence, U_n, be defined as the number of ways to cover a 3-by-n rectangle-minus-corner section of the full rectangle. We seek to use these definitions to give a closed form formula for U_n without breaking down this definition further to handle the cases of vertical versus horizontal dominoes. Notice that the ordinary generating functions for our two sequences correspond to the series

$${\displaystyle {\begin{aligned}U(z)=1+3z^{2}+11z^{4}+41z^{6}+\cdots ,\\V(z)=z+4z^{3}+15z^{5}+56z^{7}+\cdots .\end{aligned}}}$$

If we consider the possible configurations that can be given starting from the left edge of the 3-by-n rectangle, we are able to express the following mutually dependent, or mutually recursive, recurrence relations for our two sequences when n ≥ 2 defined as above where U₀ = 1, U₁ = 0, V₀ = 0, and V₁ = 1: