Another solution of CF#641 F2

Revision en2, by Elegia, 2020-05-12 19:02:40

We will get the bivariate generating function $$$F(z, t) = \sum_{n\ge 1}\sum_{k=1}^n A_{n,k}\frac{z^nt^k}{n!}$$$, where $$$A_{n,k}$$$ is the sum of occurrences of $$$k$$$ in all good sequence of length $$$n$$$. Consider the inclusion-exclution principle. For all contribution with $$$\max k = m$$$, we have

$$$ \sum_{j=1}^m \sum_{S \subseteq \{1, 2, \cdots m-1\}} (-1)^{|S|} Q(S, z, t, j) $$$

This means that for all $$$i \in S$$$, we force all recurrences of $$$i$$$ are before $$$i + 1$$$, and the occurrences of $$$j$$$ are counted.

Now our goal is to calculate

$$$ [z^n]\frac{t(\mathrm e^{z(1-t)}-1)}{(1-z) (1-t \mathrm e^{z(1-t)})} $$$

We consider $$$\left([z^n]\frac{t(\mathrm e^{z(1-t)}-1)}{(1-z) (1-t \mathrm e^{z(1-t)})}\right) + 1 = (1-t)[z^n] \frac 1{(1-z)(1-t\mathrm e^{z(1-t)})}$$$, which looks simpler.

$$$ \begin{align}[z^n]\frac 1{(1-z)(1-t\mathrm e^{z(1-t)})} & = \sum_{k=0}^\infty t^k [z^n] \frac{\mathrm e^{kz(1-t)}}{1-z}\\&= \sum_{k=0}^\infty t^k \sum_{j=0}^n \frac{(k(1-t))^j}{j!}\\&= \sum_{j=0}^n (1-t)^j\sum_{k=0}^\infty t^k \frac{k^j}{j!} \\&= \sum_{j=0}^n (1-t)^j\sum_{k=0}^\infty t^k [u^j] \mathrm e^{ku}\\&= \sum_{j=0}^n (1-t)^j [u^j] \frac1{1-\mathrm e^u t}\\&= \sum_{j=0}^n (1-t)^{n-j} [u^{n-j}] \frac1{1-\mathrm e^u t}\\&= [u^n] \sum_{j=0}^\infty (1-t)^{n-j} u^j \frac1{1-\mathrm e^u t}\\&= [u^n] (1-t)^n \frac 1{1-\frac u{1-t}}\frac1{1-\mathrm e^u t}\end{align} $$$

Hence we have

$$$ \begin{aligned}&= [u^n]\frac{(1-t)^{n+2}}{(1-\frac{t}{1-u})(1-t\mathrm e^u)(1-u)}\\&= (1-t)^{n+2} [u^n] \left(\frac{1-\mathrm e^u}{\left(\mathrm e^u u-\mathrm e^u+1\right) \left(1-t \mathrm e^u\right)}+\frac{\frac{u}{1-u}}{\left(\mathrm e^u u-\mathrm e^u+1\right) (1-\frac{t}{1-u})}\right)\end{aligned} $$$

Noticed that $$$[u^m]\mathrm{e}^{ku} = \frac{k^m}{m!}, [u^m] \frac1{(1-u)^k} = \binom{n+k-1}{n}$$$, this could be calculated straightly through multipoint evaluation with time complexity of $$$\Theta(n\log^2 n)$$$. It will pass if your implementation doesn't have a big constant.

It could also be improved to $$$\Theta(n\log n)$$$ through the Lagrange Inversion formula similar to the original solution, I leave this as an exercise.

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en10 English Elegia 2020-05-19 19:30:35 69
en9 English Elegia 2020-05-19 19:29:25 1057 (published)
en8 English Elegia 2020-05-19 19:12:46 1930 (saved to drafts)
en7 English Elegia 2020-05-17 04:44:25 176
en6 English Elegia 2020-05-13 16:57:10 585
en5 English Elegia 2020-05-12 19:18:18 0 (published)
en4 English Elegia 2020-05-12 19:12:23 2
en3 English Elegia 2020-05-12 19:05:08 169
en2 English Elegia 2020-05-12 19:02:40 45
en1 English Elegia 2020-05-12 18:56:10 2116 Initial revision (saved to drafts)