2024.3.27 闲话

歌：もうまい - suni feat. 初音ミク .

因为 APJ 有歌词所以发一篇 .

CG5 - HIGH

Okay okay oka-a-ay okay oh

Okay okay okay okay oh

Okay okay oka-a-ay okay oh

Okay okay okay okay oh

You've been on my mind

So meet my melody in the relative major

Flows so well

Like you and me

Goes to the beat

We're so perfect on paper

Think the stars have aligned

I'll tell you why

But they've aligned in our favor

'Cause we just get along

So-o-o can we just get along

La la la la

Take me high what I say

And she takes me high

High

Take me high what I say

And she takes me higher oh oh

Take me high that's what I said

Won't you take me high

High

A little bit of lovin' won't do

Give me all of you now

This is a story I'ma tell right now

Of a life of singularity a life so foul

This man he ran away from love what

Too meticulous particular and never enough

One of the stranger things when you're upside down

In a gauntlet where your thoughts been knockin' you around

It's confliction hey commotion hey confusion hey corruption

Don't wanna suffer the repercussions of a commitment concussion ho

He continued on with caution

A connection crusade

What

But she got through the armor

Cut through with a blade of what

Caring compassion and kindness too

And his only defense was I love you ho

Take me high what I say

And she takes me high hey yeah

Take me higher

Take me high what I say

And she takes me higher

Yeah yeah yeah

Take me high that's what I said

Won't you take me high high

A little bit of lovin' won't do

Give me all of you

Give me ya give me ya

Take me high what I say

And she takes me high hey yeah

Take me higher

Take me high what I say

And she takes me higher

Take me high that's what I said

Won't you take me high

Don't take me down

A little bit of lovin' won't do

Give me all of you

Take me high take me higher

---

估计：

\[f_k(n)=\sum_{d\mid n}\mu(d)\dbinom{n/d}k \]

解答：

令 \(g_k(n)\) 是 \(\{1,2,\cdots,n\}\) 中 GCD 为 1 的 \(k\) 元子集个数，则：

\[\sum_{i=1}^ng_k\left(\left\lfloor\dfrac ni\right\rfloor\right)=\dbinom nk \]

从而：

\[\dbinom nk=g_k(n)+g_k(\lfloor n/2\rfloor)+\sum_{i=3}^ng_k\left(\left\lfloor\dfrac ni\right\rfloor\right)+1\le g_k(n)+\dbinom{\lfloor n/2\rfloor}k+n\dbinom{\lfloor n/3\rfloor}k \]

也即：

\[\dbinom nk-\dbinom{\lfloor n/2\rfloor}k-n\dbinom{\lfloor n/3\rfloor}k\le g_k(n)\le\dbinom nk-\dbinom{\lfloor n/2\rfloor}k \]

然而这又有什么用呢？

注意到 \(f_k(n)\) 即为 \(\{1,2,\cdots,n\}\) 中 GCD 与 \(n\) 互质的 \(k\) 元子集数量，于是：

\[f_k(n)=\sum_{d\perp n}g_k\left(\left\lfloor\dfrac nd\right\rfloor\right) \]

从而和 \(g_k\) 类似分析，展开前几项就可以得到：

\[f_k(n)=\begin{cases}\binom nk-\binom{\lfloor n/2\rfloor}k+O\big(n\binom{\lfloor n/3\rfloor}k\big)&2\mid n\\\binom nk+O\big(n\binom{\lfloor n/3\rfloor}k\big)&2\nmid n\end{cases} \]