Generating function ($A(x)={\sum }_{n\ge 0}{a}_n{x}^n$)

A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

An algebraic generating function is a formal power series $F(x)$ satisfying a nonzero polynomial equation $P(x,F(x))=0$. Over a field of characteristic zero, it is D-finite.

The Euler transform of a sequence $(b_n{)}_{n\ge 1}$ has coefficients $a_n$ defined by ${\sum }_{n\ge 0}{a}_n{x}^n={\prod }_{n\ge 1}(1-x^n{)}^{-b_n}$. In unlabelled counting, $b_n$ may count connected components of size $n$, while $a_n$ counts multisets of components.

A rational generating function is a formal power series equal to a quotient of two polynomials, with denominator nonzero at the origin. Multiplying by the denominator shows that its coefficients eventually satisfy a linear recurrence with constant coefficients.

The binomial transform of a sequence $a_k$ is $b_n={\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)a_k$. When $a_k$ counts objects on a specified set of $k$ positions, $b_n$ counts the choices obtained by selecting those positions from $n$ and filling them with such an object. Its inverse is $a_n={\sum }_{k=0}^n(-1{)}^{n-k}(\genfrac{}{}{0ex}{}{n}{k})b_k$, by binomial inversion.

An exponential generating function encodes a sequence using $A(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$. For labelled counting, $a_n$ is the number of structures on a fixed set of $n$ distinct labels. Labels distinguish otherwise similar objects: changing which person belongs to a group can change the structure. Recover the count by coefficient extraction as $a_n=n![z^n]A(z)$. The raw coefficient is $a_n/n!$, not $a_n$.

The factorial makes multiplication account automatically for distributing labels. To combine an $A$-structure on $k$ labels and a $B$-structure on the remaining $n-k$, choose the first label subset in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways, then choose the two structures. The binomial coefficient counts how the two groups of labels interleave within the full label set. ThusThe normalization works because $1/(k!(n-k)!)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)/n!$. This explains the usefulness of exponential generating functions for labelled counting: an ordinary Cauchy product would omit the choices of labels. See Peter Magyar, Math 482, notes for 2/7, labelled product principle.

For a concrete example, there is one undecorated set on any given collection of labels. Its exponential generating function is the exponential function $E(z)=e^z={\sum }_{n\ge 0}{z}^n/n!$. Combine a red set and a blue set, allowing either to be empty:Every label independently receives one of two colours, giving $2^n$ assignments by the rule of product. With three labels, the counts by number of red labels are $1,3,3,1$, totalling eight. Choosing only the two group sizes would give the four possibilities $(0,3),(1,2),(2,1),(3,0)$; it would forget which labels are red.

Indeed, the ordinary generating function for the constant sequence $a_n=1$ is the geometric series $1/(1-z)$. Its square has coefficient $[z^n](1-z{)}^{-2}=n+1$, counting pairs of nonnegative sizes. Both encodings are correct for their respective questions. The decision is whether identity matters when objects are combined.

For unordered sets of labelled components of positive size, an exponential generating function $B(z)$ becomes $e^{B(z)}$: the term $B(z{)}^m/m!$ forgets the ordering of $m$ components. The existing rooted-tree generating function develops the richer example $T=z{e}^T$. A labelled rooted tree consists of its root of a tree and an unordered set of smaller labelled rooted trees, connecting the same label-distribution mechanism to recursive counting.