2023.11.9 闲话

快来解方程：$F(-2z)=\mathrm e^{-2z}+\mathrm e^z\cdot F(z)$（

推歌：クリスマスなんて興味ないけど - かめりあ feat. ななひら .

Candy - S3RL feat. Sara

I'm just like candy,
I can be sweet.
If you're a good boy,
I'll be your treat.

I'm just like candy,
I can be sweet.
If you're a good boy,
I'll be your treat.

It might be nice, but whatever you do,
Don't have too much, I'm no good for you.

Now unwrap me and you will see,
One little taste brings you to your knees.
It might be fun, but whatever you do,
Don't have too much, I'm no good for you.

No good for you, no-no good for you.
No good for you, no-no good for you.

I'm just like candy,
I can be sweet.
If you're a good boy,
I'll be your treat.

It might be nice, but whatever you do,
Don't have too much, I'm no good for you.

I'm no good for you.

Lick me, suck me, and bite me gently.
It doesn't matter if it gets a little messy.
I'll tell you when you can stop
Or when you can have more.

I'm just like candy,
I can be sweet.
If you're a good boy,
I'll be your treat.

It might be nice, but whatever you do,
Don't have too much, I'm no good for you.

Now unwrap me and you will see,
One little taste brings you to your knees.
It might be fun, but whatever you do,
Don't have too much, I'm no good for you.

Lick me, suck me, and bite me gently.
It doesn't matter if it gets a little messy.
I'll tell you when you can stop
Or when you can have more.

Doesn't matter if it gets a little messy.

I'll tell you when you can stop,
Or when you can have more.

I'm just like candy,
I can be sweet.
If you're a good boy,
I'll be your treat.

It might be nice, but whatever you do,
Don't have too much, I'm no good for you.

Now unwrap me and you will see,
One little taste brings you to your knees.
It might be fun, but whatever you do,
Don't have too much, I'm no good for you.

1. 古神相关的知识是人类不应该触及的知识（组合数学也是）
2. 古神相关知识无法被人类理解，看到的人就会失去理智（组合数学也是）
3. 克苏鲁世界观中邪教徒经常写诡异的符号（组合数学中也有诡异的符号）
4. 克苏鲁世界观中有一本书叫《死灵之书》，很可怕，让人读不下去（《具体数学》也很可怕，让人读不下去）
5. 克苏鲁世界观有很多古神（组合数学里也有很多神仙）
6. 有些古神以戏弄人类为乐（有些神仙也以出毒瘤比赛迫害蒟蒻为乐）
7. 古神大多有很多化身（神仙也大多在各个 OJ 和论坛有很多账号）
8. 克苏鲁神话描述的图景很让人绝望（一一五惨案发生后，OI 之路的前景也很让人绝望）

1. 克苏鲁有很多信徒（joke 也有很多信徒）
2. joke 是数学神，又已论证了组合数学是禁忌知识
3. 克苏鲁潜藏海底，世人看不到他的真身（joke 在学校里，我也看不到她的真身）
4. 克苏鲁有克苏鲁的呼唤（joke 也经常在 QQ 上召唤我）
5. 古神应该没有性别（joke 身份证上是男，却又（将要）女装，也分不出性别）
6. 克苏鲁是旧日支配者，很早就来到地球（见下

一目了然 不言而喻

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附：joke 大战 APJ

joke 是邪神克苏鲁，论证完毕 .

$$\tag*{$\square$}$$

Ref. x义x 的博 .

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后文 $\bm p$ 是素数 .

组合数 $\binom nk=[z^0](1+z)^nz^{-k}$ .

那么在 $\bmod p$ 意义下考察之：

$$\begin{aligned}\,[z^0](1+z)^nz^{-k}&\equiv[z^0](1+z)^{n\bmod p}z^{-(k\bmod p)}(1+z^p)^{\lfloor n/p\rfloor}z^{-p\lfloor k/p\rfloor}\\&\equiv[z^0](1+z)^{n\bmod p}z^{-(k\bmod p)}[z^0](1+z)^{p\lfloor n/p\rfloor}z^{-p\lfloor k/p\rfloor}\pmod p\end{aligned}$$

也就是：

$$\dbinom nm\equiv\dbinom{n\bmod p}{m\bmod p}\dbinom{\lfloor n/p\rfloor}{\lfloor m/p\rfloor}\pmod p$$

（Lucas）

其实这种格式的东西都是可以的：

$$a_n=\sum_{k=0}^{\lfloor n/2\rfloor}\dbinom nk\dbinom{n-k}k$$

（A002426）

这个也有 $a_n\equiv a_{n\bmod p}\cdot a_{\lfloor n/p\rfloor}\pmod p$：

$$\begin{aligned}a_n&=\sum_k\dbinom{2k}k\dbinom n{n-2k}\\&=\sum_k\dbinom nk[z^0](1+z^2)^kz^{-k}\\&=[z^0](z^{-1}+1+z)^n\end{aligned}$$

接下来就和组合数的模式一样了，不再展开叙述 .

用另外一个展开 $a_n=[z^n]\dfrac1{\sqrt{1-2z-3z^2}}$ 也能导出结论，可以试试 .

最终版本：

对于 $m$ 阶整数拆分 $\lambda\vdash n$，定义：

$$a_n=\left[\dfrac{z^n}{n!}\right]\prod_{k=1}^m\sum_{i\ge 0}\dfrac{z^{\lambda_ki}}{{i!^{\lambda_k}}}$$

证明：$a_n\equiv a_{n\bmod p}\cdot a_{\lfloor n/p\rfloor}\pmod p$ .

考察：

$$Q(t,x_1,\cdots,x_n)=[z_1^0\cdots z_n^0]\dfrac1{1-t(z_1/z_1+\cdots+z_{\lambda_1-1}/z_{\lambda}+z_{\lambda_1}/x_1)-\cdots}$$

那么可以类似得出：

$$[z_1^0\cdots z_n^0]\dfrac1{Q(t,z_1,\cdots,z_n)}=[z_1^0\cdots z_n^0]\dfrac{Q(t,z_1,\cdots,z_n)^{p-1}}{Q(t^p,z_1^p,\cdots,z_n^p)}$$

提取系数即得 .

Ref. 那些你不要的（1）2021.4.27 ~ 2021.6.12 - Elegia .

赞美 EI！

放个具体数学题吧，也不能光复读 EI：

习题 5.109

对于 Apéry 数：

$$A_n=\sum_k\dbinom nk^2\dbinom{n+k}k^2$$

证明：$A_n\equiv A_{n\bmod p}\cdot A_{\lfloor n/p\rfloor}\pmod p$ .

我们遇到的第一个问题（一般也是最后一个）可能是如何刻画二项式系数平方的 OGF . 事实上，对于我们熟知的 OGF：

$$\begin{aligned}&A_c(z)=\sum_{n\ge 0}\dbinom cnz^n=(1+z)^c\\&B_c(z)=\sum_{n\ge 0}\dbinom{c+n-1}nz^n=\dfrac1{(1-z)^c}\end{aligned}$$

考察 $A_c(z)^2$ 和 $B_c(z)^2$：

$$\begin{aligned}&[z^n]A_c(z)^2=\sum_{i=0}^n\dbinom cn\dbinom c{n-i}=\dbinom{2c}{2n}\\&[z^n]B_c(z)^2=\sum_{i=0}^n\dbinom{c+i-1}{c-1}\dbinom{c+n-i-1}{c-1}=\dbinom{2c+n-1}{2c-1}\end{aligned}$$

那么就可以自然导出了 . 后面的步骤留作习题（？

增强版本：对于

$$a_n=\sum_k\prod_{i=0}^l\dbinom{n+ik}k^{c_i}$$

证明结论 .