By Odlyzko.

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**Additional info for Asymptotic enumeration methods**

**Example text**

8) k=0 is the exponential generating function of the number of objects of size n, while ∂ h(y) = f (x, y) ∂x ∞ x=1 = n=0 yn n! 9) k=0 is the exponential generating function of the sum of the weights of objects of size n. Therefore the average weight of an object of size n is [y n ]h(y) . 10) The wide applicability and power of generating functions come primarily from the structured way in which most enumeration problems arise. Usually the class of objects to be counted is derived from simpler objects through basic composition rules.

2, which gives precise estimates for the tails as well as the mean of the distribution. Information about moments of distribution functions can often be used to obtain the limiting distribution. 92) converges to µ(k) as n → ∞, then there is a limiting measure with distribution function F (x) whose k-th moment is µ(k). If the moments µ(k) do not grow too rapidly, then they determine the distribution function F (x) uniquely, and the F n (x) converge to F (x) (in the weak star sense [50]). A sufficient condition for the µ(k) to determine F (x) uniquely is that the generating function ∞ U (x) = k=0 µ(2k)xk (2k)!

It has the ordinary generating function ∞ ∞ n p(n)z = f (z) = n=0 (1 − z k )−1 . 13) k=1 Let g(s) = log f (e−s ), and consider s > 0, s → 0. There are extremely accurate estimates of g(s). It is known [13, 23], for example, that g(s) = π 2 /(6s) + (log s)/2 − (log 2π)/2 − s/24 + O(exp(−4π 2 /s)) . 15) which yields p(1) + p(2) + · · · + p(n) ≤ 2−3/4 e−1/4 n−1/4 (1 + o(1)) exp(2π6−1/2 n1/2 ) . 16) is too high by a factor of n 1/4 . 16) to bound p(n) alone, we obtain a bound that is too large by a factor of n 3/4 .

### Asymptotic enumeration methods by Odlyzko.

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