Analytic combinatorics uses generating functions and analysis to explain how large families of discrete objects grow and how their features behave. It starts with a precise counting model, translates that model into a function, then studies the function to recover information about its coefficients. The gold standard is to connect every approximation back to the objects, justify the analytic conditions and check whether the approximation is useful at the sizes that matter.
This guide develops that connection through words, tiled strips, balanced brackets and marked features. You will learn why the nearest singularity often controls exponential growth, why its type affects the accompanying power of size, and why a plausible growth rate can still give a poor numerical answer. All examples are mathematical models, not measurements of real systems.
Use the Gold Standard master to explore the collection, the Education guide for transferable learning, and the How X Works hub for wider mechanisms. The Enumerative Combinatorics root develops exact counting. Here the central question is what those counts and related statistics do as size increases.
Choose your analytic combinatorics route
Chapters 7–9: Check singularities
Chapters 10–12: Mark and diagnose
Chapter index
Build the model · Chapters 1–3
1. Keep the object, function and approximation separate
Read poles · Chapters 4–6
4. Read a simple pole through an exact example
Check singularities · Chapters 7–9
7. Find every dominant singularity before ignoring oscillation
Mark and diagnose · Chapters 10–12
10. Mark a feature to obtain more than a count
Transfer and continue · Chapters 13–14
CHAPTER 1 OF 14 · Build the model
1. Keep the object, function and approximation separate
Suppose a strip of length n is tiled with pieces of length one and two. Tiles cannot overlap, the strip must be completely covered, and their left-to-right order matters. These rules define a finite collection for each n. They do not describe unordered coin selections or arrangements in a circle.
Let tₙ count the tilings. There is one empty tiling, so t₀=1. There is one tiling of length one. For n≥2, a tiling begins with either a length-one tile followed by a tiling of length n−1, or a length-two tile followed by a tiling of length n−2. Thus tₙ=tₙ₋₁+tₙ₋₂.
The generating function T(z)=Σtₙzⁿ satisfies T(z)=1+zT(z)+z²T(z), giving T(z)=1/(1−z−z²). This is an exact encoding. A later estimate of tₙ for large n is a third layer, built from the function. Confusing these layers is a common source of overclaiming.
Identify what each layer guarantees
The object specification determines what is counted. The generating-function identity holds coefficient by coefficient because of a proved decomposition. An asymptotic expression describes limiting behaviour and may require additional analytic hypotheses. It usually does not give an exact answer at every finite n.
For example, a model may correctly predict exponential growth in the number of possible configurations while saying nothing about how frequently people choose them. Likewise, the number of possible inputs to a program does not by itself determine the program’s running time. A cost model or distribution must be added.
A useful working document therefore records the family, size parameter, generating-function convention and desired conclusion separately. If the question asks for an exact count at n=20, do not silently replace it with a growth classification. If it asks how the family scales, an enormous exact integer may be less informative than a justified approximation.
CHAPTER 2 OF 14 · Build the model
2. Translate constructions before manipulating formulas
For ordinary generating functions, a disjoint choice between two classes produces a sum. An ordered pair of objects, with total size equal to the sum of their sizes, produces a product. A sequence of components from a class B gives 1/(1−B(z)), provided B has no size-zero objects and the combinatorial decomposition is unambiguous.
For the tiled strip, one component has generating function B(z)=z+z². A sequence of such components has generating function 1+B(z)+B(z)²+…, which equals 1/(1−z−z²). This agrees with the first-tile recurrence through a different description of the same objects.
The prohibition on size-zero components matters. If a freely repeatable component had size zero, infinitely many component sequences could describe objects of the same total size. The formal geometric-series construction would then fail to represent a finite count at each size.
A change of order changes the function
For unordered selections of unlimited one-unit and two-unit coins, the function is 1/((1−z)(1−z²)). At total four, the selections are four ones, two ones plus one two, and two twos: three possibilities. Ordered tilings of length four have five possibilities because the position of a two-unit tile matters.
Both functions contain the same tile or coin sizes. The difference lies in how components combine. Manipulating a rational expression accurately cannot repair a mistaken constructor. Explain why addition, multiplication or sequence formation matches the original objects before using any analytic machinery.
The existing Generating Functions, Coefficients and Recursive Counting guide gives further practice with these translations. The Algebraic Combinatorics guide explores other structural encodings.
For a new model, calculate at least the first three nontrivial coefficients directly from the objects. The purpose is to test the specification, not to prove the eventual asymptotic theorem. A discrepancy at size two is usually easier to understand than a wrong exponential constant discovered after pages of analysis.
CHAPTER 3 OF 14 · Build the model
3. Distinguish a formal series from an analytic function
A formal power series is a sequence of coefficients with algebraic rules. In that setting, z is a bookkeeping symbol. You can multiply series, compare coefficients and derive recurrences without assigning a numerical value to z or proving convergence at that value.
An analytic power series is also interpreted as a function of a complex variable inside its disk of convergence. This interpretation introduces geometric information: where the function is analytic, how far its series extends from the origin, and what prevents further extension at boundary points.
For 1/(1−2z), the formal expansion is Σ2ⁿzⁿ. Numerically it converges when |z|<1/2. The expression 1/(1−2z) is defined at z=1, where its value is −1, but the series 1+2+4+… does not converge there. A rational expression’s value outside the disk is not automatically the sum of the original series.
Why the complex plane enters a counting problem
A power series’ radius of convergence depends on the nearest non-removable singularities in the complex plane, not just the positive real axis. A singularity is a point where the relevant analytic continuation fails to be analytic. Poles and branch points are two important types, but they are not interchangeable.
The general coefficient relation is 1/R=limsup |aₙ|^(1/n), where R is the radius of convergence. It captures exponential scale, with appropriate conventions for R=0 or infinity. A limsup is used because zeros or oscillations may prevent an ordinary limit.
For a nonnegative counting sequence, a finite positive radius is therefore deeply informative. Still, knowing R alone does not determine the leading constant, polynomial factor, periodic gaps or finite-size error. Those require more information about the function.
When learning the subject, keep a two-column distinction in your notes: coefficient identities that are formal, and analytic deductions that require convergence or continuation. This prevents a valid formal derivation from being treated as permission to use every analytic theorem without checking its hypotheses.
Consider words over an alphabet containing two symbols. There are 2ⁿ words of length n, including one empty word. Their ordinary generating function is W(z)=1/(1−2z). It has a simple pole at ρ=1/2, and its coefficients are exactly ρ⁻ⁿ=2ⁿ.
Now prepend a compulsory symbol of length one to every word. The new function is z/(1−2z). For n≥1, the count is 2ⁿ⁻¹. The pole remains at 1/2, but the multiplicative constant changes. This tiny example demonstrates why locating the singularity is only part of the calculation.
More generally, if a rational function P(z)/Q(z) has a unique smallest-modulus pole ρ, that pole is simple, and P(ρ) is nonzero, its leading contribution is Aρ⁻ⁿ with A=−P(ρ)/(ρQ′(ρ)). Other poles contribute smaller exponential scales, possibly accompanied by powers of n.
Derive the constant rather than memorising its sign
Near a simple zero ρ, Q(z) is approximately Q′(ρ)(z−ρ). Since z−ρ=−ρ(1−z/ρ), the singular part becomes −P(ρ)/(ρQ′(ρ)) times 1/(1−z/ρ). The coefficient of the latter function is ρ⁻ⁿ. The minus sign comes from the change of form.
For W(z), P=1 and Q=1−2z. With ρ=1/2 and Q′=−2, A=1. For zW(z), P(ρ)=1/2, giving A=1/2. Both match the exact counts above.
Check cancellation before applying the rule. The expression (1−2z)/(1−2z) is the constant function one after removing the apparent singularity. Treating its denominator zero as a genuine pole would predict nonzero coefficients at every size, while all positive-size coefficients are actually zero.
This is a dependable first exercise because every analytic claim can be checked against exact coefficients. It teaches the location, order and amplitude of a pole without requiring an unfamiliar combinatorial family at the same time.
Return to T(z)=1/(1−z−z²). Write φ=(1+√5)/2 and ψ=(1−√5)/2. The denominator factors as (1−φz)(1−ψz). Its zeros are 1/φ≈0.6180 and 1/ψ≈−1.6180; the positive zero is closer to the origin.
Partial fractions give the exact formula tₙ=(φⁿ⁺¹−ψⁿ⁺¹)/√5. Consequently tₙ is asymptotic to φⁿ⁺¹/√5. Here “asymptotic to” means that the ratio of the exact count to the proposed approximation tends to one as n tends to infinity.
The omitted term is −ψⁿ⁺¹/√5. Since |ψ|<1, its absolute magnitude tends to zero. This exceptionally strong behaviour comes from the exact two-pole structure. It should not be assumed for every asymptotic formula in combinatorics.
Check one finite size
At n=10, the recurrence gives t₁₀=89. The leading expression is approximately 88.99775. Its absolute error is about 0.00225, and it underestimates the count by about 0.00253 percent. The sign of the correction alternates with n because ψ is negative.
If you retain only the base φ and write φ¹⁰, you obtain approximately 122.99. That is the right exponential base but a poor estimate of this count. The shift by one and division by √5 are part of the leading constant, not expendable decoration.
At n=20, the exact count is 10,946. A practical implementation should generate this integer from the recurrence rather than rely on floating-point exponentiation and rounding. The exact expression explains the growth; the recurrence is often the simpler tool for computing moderate-size values reliably.
This example also illustrates the difference between an asymptotic statement and a certified finite-size estimate. Here the second term gives an explicit error bound. Elsewhere, a statement that a ratio tends to one may not specify how large n must be for a chosen tolerance. That extra requirement deserves its own calculation.
The generating function 1/(1−2z)² has coefficients (n+1)2ⁿ. One interpretation is an ordered pair of binary words whose combined length is n. Choose the length i of the first word in n+1 ways, then choose all n symbols in 2ⁿ ways.
The closest singularity remains 1/2, but it is now a double pole. Compared with the simple-pole example, an additional factor of approximately n appears. The exponential scale alone cannot distinguish these two families.
For a positive integer k, the coefficient of (1−z/ρ)⁻ᵏ is C(n+k−1,k−1)ρ⁻ⁿ. For fixed k and large n, the binomial factor behaves like nᵏ⁻¹/(k−1)!. This exact identity explains the polynomial factor associated with poles of different orders.
Compare models at the same size
At n=10, a binary word family contains 1,024 objects. An ordered pair of binary words with total length ten contains 11,264 objects. Both have exponential base two; their counts differ by a factor of eleven. Describing both simply as “exponential growth” loses information that matters for enumeration.
The pair model permits an empty first or second word. If both must be nonempty, there are n−1 possible splits for n≥2, giving (n−1)2ⁿ. Its leading asymptotic term remains n2ⁿ, even though its exact small-size behaviour differs.
This is another reason to retain the original specification. Two distinct models can share the same leading asymptotic expression. The approximation is not an identifier of the family, nor does matching it prove that two constructions are equivalent.
When reporting growth, distinguish the exponential factor, the power of n and the leading constant. Then state whether lower-order terms have been computed. This vocabulary allows meaningful comparisons without presenting a rough scale as an exact count or obscuring finite-size differences that the reader needs to know.
CHAPTER 7 OF 14 · Check singularities
7. Find every dominant singularity before ignoring oscillation
Take A(z)=1/(1−z²). Its coefficient is one at every even index and zero at every odd index. The radius of convergence is one, but a claim that aₙ tends to a positive constant would be false along the odd indices.
Partial fractions expose the mechanism: A(z)=1/(2(1−z))+1/(2(1+z)). Both poles, at 1 and −1, lie on the circle of convergence. Their coefficient contributions are 1/2 and (−1)ⁿ/2. Together they produce the exact parity pattern.
Keeping only the positive pole would predict one half for every coefficient. This is not a small relative error on odd indices; the exact coefficient there is zero. The missing singularity contributes at the same exponential scale and cannot be discarded.
Periodicity has a combinatorial explanation
A sequence of size-two components can only have even total size. The function’s symmetry records that restriction. If the only permitted component sizes share a common divisor, inspect whether the resulting family occupies only particular residue classes.
For B(z)=1/(1−4z²), the even-index coefficient b₂ₘ is 4ᵐ and the odd-index coefficient is zero. The poles are ±1/2. Their combined contribution is (2ⁿ+(-2)ⁿ)/2. Along even indices the exponential base in n is two, while the odd indices remain impossible.
Adding a permitted size-one component can remove this particular gap, but it does not justify assuming every generating function has a single dominant singularity. Inspect the actual equation and its roots. Positivity of coefficients helps locate a positive-real singularity when the radius is finite, but does not by itself eliminate others of equal modulus.
A useful numerical diagnostic is to inspect counts separately by parity or residue class before taking consecutive ratios. Ratios involving zeros are undefined, and oscillating ratios may reflect genuine structure rather than numerical noise. Changing the sampling interval can reveal a clean pattern that a single unqualified ratio conceals.
CHAPTER 8 OF 14 · Check singularities
8. See a square-root singularity through balanced brackets
Let Cₙ count correctly nested strings of n bracket pairs, and let C(z)=ΣCₙzⁿ. Every nonempty valid string has a unique first matching pair, containing one valid string and followed by another. Counting the outer pair as one unit gives C(z)=1+zC(z)².
Solving the quadratic and selecting the branch with C(0)=1 gives C(z)=(1−√(1−4z))/(2z). The alternative sign does not have the required finite expansion at zero. The apparent division by z in the chosen expression is removable at zero; the branch point at z=1/4 is not.
The exact coefficients are Catalan numbers Cₙ=C(2n,n)/(n+1). Their leading growth is Cₙ∼4ⁿ/(√π n³ᐟ²). NIST’s Catalan reference records the exact formula, generating function and limiting form.
Compare with a pole at the same location
The function 1/(1−4z) also has radius 1/4, but its coefficients are exactly 4ⁿ. The Catalan coefficients are smaller by a factor of order n³ᐟ². The location determines the exponential scale; the local type distinguishes the power correction.
At n=10, the Catalan count is 16,796. The leading approximation is about 18,707.9, roughly 11.38 percent too high. This is much less accurate than the tiled-strip approximation at the same index. Both are mathematically valid asymptotic equivalents; their rates of approach differ.
The contrast is a useful antidote to the idea that “large n” has a universal numerical meaning. Ten may be enough for one model and inadequate for another, depending on the error tolerance. A correct limiting formula is a starting point for numerical judgement, not its replacement.
For a parser, tree generator or bracket exercise, confirm what n measures. Counting pairs, symbols, leaves or internal nodes can introduce shifts and factors. Even a correct Catalan formula answers the wrong question if its size parameter is silently changed halfway through the calculation.
CHAPTER 9 OF 14 · Check singularities
9. Understand what a transfer theorem must justify
Singularity analysis turns local behaviour of a generating function near a dominant singularity into asymptotic behaviour of its coefficients. A basic model is (1−z/ρ)⁻ᵅ, whose coefficients behave like ρ⁻ⁿnᵅ⁻¹/Γ(α), for fixed α outside the nonpositive integers. The gamma function extends factorials: Γ(k)=(k−1)! for positive integers k.
To apply a corresponding rule to a more general function, one must establish suitable analytic continuation around the dominant point, an appropriate local expansion and control of the remainder. A common setting uses a dented neighbourhood that excludes a wedge beyond the singularity, often called a Delta-domain.
The Flajolet–Sedgewick treatment of singularity analysis explicitly makes continuation and local-expansion conditions part of the process. A real-axis approximation alone is not an automatic substitute for these hypotheses.
Use the theorem’s conditions as questions
Where are the smallest-modulus singularities? Is there one or several? Which branch of an algebraic function corresponds to the counting series? Is the local expansion valid in the complex region required by the theorem? What does the remainder become after coefficient extraction?
For a square-root term, α=−1/2 and Γ(−1/2)=−2√π. The coefficient contribution therefore contains a factor n⁻³ᐟ², with a sign that combines with the multiplier of the square root. Ignoring that multiplier or sign can yield an impossible negative leading count.
An analytic part of a local expansion also needs careful treatment. One cannot indiscriminately apply the displayed power rule to every integer power and divide by a gamma value at a pole. The standard transfer statements distinguish these cases and the accompanying remainder estimates.
A beginner need not reproduce the full complex-analysis proof to use an elementary rational example responsibly. Partial fractions already provide a complete route there. For more advanced algebraic or logarithmic cases, state the theorem being used and verify its assumptions rather than describing a successful pattern match as a proof.
CHAPTER 10 OF 14 · Mark and diagnose
10. Mark a feature to obtain more than a count
Introduce a second variable u to record a statistic. For tiled strips, let a length-two tile contribute an extra factor u. The bivariate generating function is T(z,u)=1/(1−z−uz²). The coefficient of zⁿuᵏ counts length-n tilings containing exactly k length-two tiles.
There are n−2k length-one tiles and k length-two tiles, for a total of n−k tiles. Choosing which k positions in that tile sequence are long gives C(n−k,k) tilings. This is valid for 0≤k≤floor(n/2); outside that range the count is zero.
At length six, the counts for k=0,1,2,3 are 1,5,6,1. Their sum is thirteen. If all thirteen tilings are equally likely, the expected number of long tiles is (0×1+1×5+2×6+3×1)/13=20/13.
Differentiation counts marked occurrences
Differentiating with respect to u and setting u=1 gives z²/(1−z−z²)². Its coefficient of zⁿ is the total number of long-tile occurrences across all length-n tilings. Dividing by tₙ produces the mean under the uniform distribution on that size class.
A second derivative records k(k−1), not k². At n=6, the total is 2×6+6×1=18. Therefore E[K(K−1)]=18/13 and E[K²]=38/13. Subtracting (20/13)² gives variance 94/169.
The size-n probability generating function is [zⁿ]T(z,u)/[zⁿ]T(z,1). This normalisation is essential. Evaluating the unnormalised bivariate function at arbitrary z does not directly give the distribution at a fixed size.
The example shows why analytic combinatorics includes statistics of structures, not only their total number. It also shows the boundary of the claim: these are probabilities under a specified uniform model. If an application generates tilings with unequal probabilities, the count-based mean may not describe its actual outputs.
CHAPTER 11 OF 14 · Mark and diagnose
11. Relate a moving singularity to an average density
In the marked tiling model, the positive singularity ρ(u) solves 1−ρ(u)−uρ(u)²=0. Near u=1, it varies smoothly and remains the dominant simple pole. The leading coefficient form has the structure A(u)ρ(u)⁻ⁿ, with a prefactor A(u).
Taking a logarithmic derivative with respect to u suggests a mean of −nρ′(1)/ρ(1) plus a bounded contribution from the prefactor. In this rational example, the explicit roots and partial fractions justify the needed uniform local analysis; it is not merely a formal differentiation of an arbitrary asymptotic statement.
Implicit differentiation gives ρ′(u)=−ρ(u)²/(1+2uρ(u)). At u=1, ρ=(√5−1)/2, so the limiting mean number of long tiles per unit strip length is ρ/(1+2ρ)=(1−1/√5)/2, approximately 0.276393.
Interpret the denominator of a density
This density is long tiles divided by strip length. It is not long tiles divided by total tile count. Each long tile occupies two units. Thus approximately 55.28 percent of the strip’s units belong to long tiles in the limiting mean sense, while the number of long tiles per unit length is about 27.64 percent.
At length six, the exact mean is 20/13≈1.53846, whereas the leading linear term is about 1.65836. The difference is not a contradiction: the asymptotic mean includes a bounded correction. Small sizes need the exact distribution or a more detailed approximation.
For a practical feature, name the statistic and denominator together. A tree’s leaves per node, a word’s pattern occurrences per symbol and a program’s comparisons per input element are distinct quantities. An unexplained percentage can be numerically correct yet conceptually misleading.
More advanced limit laws can describe fluctuations around an average, but a mean alone does not establish concentration or a normal distribution. Those conclusions require their own hypotheses and analysis. Marking is a powerful organising device; it does not remove the obligation to state exactly what has been proved.
CHAPTER 12 OF 14 · Mark and diagnose
12. Diagnose an attractive but unreliable approximation
Begin a diagnostic by comparing the approximation with exact values over several sizes. Record both absolute and relative error. An error of a thousand may be negligible for a trillion-sized count and disastrous for a count near ten. Neither error measure is universally sufficient on its own.
Next separate the exponential base from the prefactor. If an estimate has the right base but differs by a growing factor of n, suspect a missed pole order or singular exponent. If it differs by an approximately constant factor, inspect residues, indexing shifts and the definition of size.
If errors alternate or whole residue classes vanish, inspect multiple dominant singularities and periodic support. If the predicted count is negative, check branch selection and the signs of local terms. If a denominator zero seems to imply growth absent from the coefficients, check for cancellation.
Preserve the distinction between evidence and proof
Fitting an exponential to twenty values does not prove the family has that exponential growth rate. Several different asymptotic forms can look similar over a short range. A polynomial correction may be difficult to distinguish from a slightly changed exponential base in small data.
Likewise, observed ratios approaching a number are evidence for a conjecture, not a proof of the analytic conditions behind it. Use the exact generating function, a justified recurrence or another structural argument to establish the conclusion. Computation should challenge and quantify the analysis, not quietly replace it.
Numerical stability also matters. Factorials and large powers can overflow ordinary machine-number formats. Exact integer recurrences are often the safest source of benchmark counts. Logarithms help compare very large magnitudes, but taking logarithms of a sequence containing zeros requires separate treatment.
Finally, do not invent a finite-size guarantee from an unspecified error term. A statement aₙ=bₙ(1+O(1/n)) implies the existence of constants in an eventual bound, but does not give their numerical values. If a decision requires one-percent accuracy at n=50, obtain a quantified bound or compare with a reliable exact computation at that size.
CHAPTER 13 OF 14 · Transfer and continue
13. Transfer checks with changed assumptions and answers
First, change the tiling pieces to lengths two and four while continuing to measure size in unit cells. There are no tilings of odd length. At length 2m, the count equals the original length-m tiling count because halving every tile and the strip gives a bijection. The function is 1/(1−z²−z⁴).
The dominant singularities now include both positive and negative roots of modulus √ρ, where ρ=(√5−1)/2. A single positive-root estimate cannot represent all coefficients. Along even sizes, use the original estimate with index m; along odd sizes, the exact answer is zero.
Second, give every length-one and length-two tile one of two colours. A sequence of coloured tiles has generating function 1/(1−2z−2z²). It is not obtained by multiplying the original tₙ by 2ⁿ, because different tilings of the same strip length contain different numbers of tiles.
Work the new model instead of adjusting by intuition
At length two, two single tiles have four colourings and one long tile has two, giving six. The recurrence becomes cₙ=2cₙ₋₁+2cₙ₋₂, with c₀=1 and c₁=2. The dominant exponential base is 1+√3, from the reciprocal of the positive denominator root.
Third, take an ordered pair of arbitrary binary words with total length n and require the first word to be nonempty. There are n allowable split positions for n≥1, so the count is n2ⁿ. The generating function is 2z/(1−2z)². It has the same leading n2ⁿ behaviour as (n+1)2ⁿ but different exact coefficients.
Fourth, ask whether the approximation 4ⁿ alone is adequate for Catalan counts. It captures the exponential scale but its ratio to the exact count grows like √π n³ᐟ². Its relative error therefore does not tend to zero. “Correct exponential base” and “asymptotic equivalent” are different standards.
For each change, identify which layer changed: the objects, the generating function, the singularities or the requested statistic. Then say which earlier result remains valid. This is the transfer test that matters: understanding the mechanism well enough to repair the argument when a familiar formula no longer applies.
CHAPTER 14 OF 14 · Transfer and continue
14. Build a practical study route and state the limits
Start with exact counting and coefficient extraction. Learn to derive a recurrence from a disjoint decomposition, turn it into a rational function and verify the first coefficients. Partial fractions then provide a complete bridge from functions to asymptotic growth for a useful class of examples.
Next study complex power series, analytic continuation and Cauchy’s coefficient formula. These explain why the geometry of singularities affects coefficients. After that, move to square-root expansions, transfer theorems and marked generating functions. The sequence matters: notation is easier to manage when each step solves a question you already understand.
The Flajolet–Sedgewick book resources provide a primary route into the subject. NIST’s gamma-function asymptotics support factorial and gamma estimates, while its Catalan entry supplies a compact exact-versus-asymptotic example. Read theorem conditions alongside formulas, especially when several singularities compete.
Is analytic combinatorics just approximate counting?
No. Its starting identities can be exact, and its conclusions include growth estimates, distributions and structural statistics. Approximation is one outcome of the method. The model and the symbolic translation remain exact mathematical statements when properly justified.
Must every generating function have a finite singularity?
No. Entire functions have no finite singularities, and their coefficient analysis may call for other methods, such as saddle-point techniques. Even with finite singularities, the elementary pole and square-root examples here do not cover every possible behaviour. Select a method that matches the function rather than forcing the function into a familiar rule.
Does faster growth mean an algorithm is slower?
Not by itself. A program may avoid enumerating the entire family, exploit shared structure or stop after finding one object. To infer running time, specify what operations the algorithm performs and how often. Counting potential objects and analysing computational cost are connected but distinct tasks.
What should a finished analysis contain?
It should identify the counted objects and size parameter, derive the generating function, locate the relevant singularities, justify the coefficient estimate and report its practical limitations. Include exact small cases and finite-size error checks where useful. A reader should be able to follow the chain from a concrete object to a trustworthy conclusion without relying on an unexplained formula.