By Odlyzko.

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**Extra resources for Asymptotic enumeration methods**

**Example text**

40) by z n and summing on n ≥ 0 yields 2FA (z) = z −1 (FA (z) − 1) + z −1 GA (z) . 41) We need one more relation, and to obtain it we consider any string B = b 1 b2 · · · bn that does not contain A any place inside. If we let C be the concatenation of A and B, so that C = a1 a2 · · · ak b1 b2 · · · bn , then C starts with A, and may contain other occurrences of A, but only at positions that overlap with the initial A. 42) and this gives the relation FA (z) = z −k CA (z)GA (z) . 38), while GA (z) = zk .

34) Often there is no obvious recurrence for the sequence a n being studied, but there is one involving some other auxiliary function. Usually if one can obtain at least as many recurrences as there are sequences, one can obtain their generating functions by methods similar to those used for a single sequence. The main additional complexity comes from the need to solve a system of linear equations with polynomial coefficients. We illustrate this with the following example. 4. Sequences with forbidden subwords.

88) n=0 where the last equality follows from Eq. 38). Another, more general, way to derive this is to use GA (z). Note that gA (n)2−n is the probability that A appears in the first n coin tosses, but not in the first n − 1. Hence the r-th moment of the time until A appears is ∞ nr gA (n)2−n = z n=0 d dz r GA (z) . 88). 91) , since 1 ≤ CA (1/2) ≤ 2. Higher moments can be used to obtain more detailed information. 2, which gives precise estimates for the tails as well as the mean of the distribution.

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