dna049's blog

By dna049, history, 14 months ago,

zscoder introduced Generating functions in Competitive Programming in his blog, which give a method to get the formula of general term of a array by it's generating function.

But sometime the generating function is somewhat complicated, for instance, the array http://oeis.org/A054726 which has generating function:

$1 + \frac{3}{2} z - z^2 - \frac{z}{2} \sqrt{1 - 12 z + 4 z^2}$

and we can calculate the first n term by poly sqrt with fft in $O(n \log n)$

But this array have a recurrence formula:

$a_{n} = \frac{ (12 n - 30) a_{n-1} - (4 n - 16) a_{n-2} }{n-1}$

which will be much easy to calculate.

My question is how can I get recurrence formula by it's generating function in general(or some special form) ?

I know how to do this now, thanks qwaszx who give me a hint.

If $f(z) = p(z)^{ \frac{1}{m} }$ is generating function of $a_n$, where $p(z)$ is a polynomial. then

$p(z) f'(z) = \frac{1}{m} p(z) p'(z)$

compare the coefficient of $z^n$， and we will get the recurrence formula of $a_n$

Let us calculate the above example: $f(z) = 1 + \frac{3}{2} z - z^2 - \frac{z}{2} \sqrt{1 - 12 z + 4 z^2}$

let $b_n = a_{n+1}$ and $g(z)$ is generating function of $b_n$, then

$g(z) = \sum_{n = 0} ^{\infty} a_{n+1} z ^ {n} = \frac{ f(z) - f(0)}{z} = \frac{3}{2} - z - \frac{1}{2} \sqrt{1 - 12 z + 4 z^2}$

so we get differential of $g(z)$

$g'(z) = -1 - \frac{2z - 3}{\sqrt{1 - 12 z + 4 z^2}} = \frac{3-2z - \sqrt{1 - 12 z + 4 z^2}}{\sqrt{1 - 12 z + 4 z^2}}$

and

\begin{align} (1 - 12z + 4 z^2) g'(z) &= (3-2z - \sqrt{1 - 12 z + 4 z^2}) (\sqrt{1 - 12 z + 4 z^2}) \\ &= (3 - 2z) (3-2z - 2g(z)) - (1 - 12z + 4 z^2) \\ &= (4z - 6) g(z) + 8 \end{align}

compare the coefficient of $z^n$, we have $(n + 1)b_{n+1} - 12 n b_n + 4(n-1)b_{n-1} = 4 b_{n-1} - 6 b_n$, so

$(n+1)a_{n+2} = (12 n - 6) a_{n+1} - (4n - 8) a_n$

so we archive our goal: $a_{n} = \frac{ (12 n - 30) a_{n-1} - (4 n - 16) a_{n-2} }{n-1}$