約数に関する研究2　負の約数を根にもつ多項式について

定義$2$から$d$を$n$の正の約数とすると$f_n(-d)=0$が成立する．よって，
$$\sum _{d|n,0<d}{f}_n(-d)=0$$
となり，

$$\begin{array}{rl}\sum _{d|n,0<d}{f}_n(-d)& =\sum _{d|n,0<d}\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)(-d{)}^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)(-1{)}^k\sum _{d|n,0<d}{d}^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}(-1{)}^k{c}_n(k){\sigma }_k(n)\end{array}$$

より，
$$\sum _{k=0}^{{\sigma }_0(n)}(-1{)}^k{c}_n(k){\sigma }_k(n)=0$$

定義$2$において右辺の展開を考えると，$c_n(k)$は${\sigma }_0(n)-k$個の互いに異なる$n$の正の約数の積の総和だとわかる．

まず次の二つの値を求める．
(1)
$$\sum _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}1=?$$
前で説明したように，$c_n(k)$は${\sigma }_0(n)-k$個の互いに異なる$n$の正の約数の積の総和を計算しているため，${\sigma }_0(n)$個の中から${\sigma }_0(n)-k$個を取る取り方に等しくなる．ゆえに
$$\sum _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}1={}_{{\sigma }_0(n)}{\mathrm{C}}_{{\sigma }_0(n)-k}={}_{{\sigma }_0(n)}{\mathrm{C}}_k$$
となる．

2. $$\prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}{d}_1{d}_2\cdots d_{{\sigma }_0(n)-k}=?$$
   普通に式変形していく．
   $$\begin{array}{rl}\prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}{d}_1{d}_2\cdots d_{{\sigma }_0(n)-k}& =\prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}\frac{n}{d_1}\cdot \frac{n}{d_2}\cdot \cdots \cdot \frac{n}{d_{{\sigma }_0(n)-k}}\\ & =\prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}\frac{n^{{\sigma }_0(n)-k}}{d_1{d}_2\cdots d_{{\sigma }_0(n)-k}}\\ & =n^{({\sigma }_0(n)-k){}_{{\sigma }_0(n)}{\mathrm{C}}_k}\cdot (\prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}{d}_1{d}_2\cdots d_{{\sigma }_0(n)-k}{)}^{-1}\\ (\prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}{d}_1{d}_2\cdots d_{{\sigma }_0(n)-k}{)}^2& =n^{({\sigma }_0(n)-k){}_{{\sigma }_0(n)}{\mathrm{C}}_k}\\ \therefore \prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}{d}_1{d}_2\cdots d_{{\sigma }_0(n)-k}& =n^{\frac{{\sigma }_0(n)-k}{2}{}_{{\sigma }_0(n)}{\mathrm{C}}_k}\end{array}$$

これで準備が整った．
$$\sum _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}1=a,\prod _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}{d}_1{d}_2\cdots d_{{\sigma }_0(n)-k}=b$$
とおけば相加相乗平均の関係から，

$$\begin{array}{rl}{c}_n(k)& \ge a\sqrt[a]{b}\\ & ={}_{{\sigma }_0(n)}{\mathrm{C}}_k\sqrt[{}_{{\sigma }_0(n)}{\mathrm{C}}_k]{n^{\frac{{\sigma }_0(n)-k}{2}{}_{{\sigma }_0(n)}{\mathrm{C}}_k}}\\ & ={}_{{\sigma }_0(n)}{\mathrm{C}}_k\cdot n^{\frac{{\sigma }_0(n)-k}{2}}\\ \therefore c_n(k)& \ge {}_{{\sigma }_0(n)}{\mathrm{C}}_k\cdot n^{\frac{{\sigma }_0(n)-k}{2}}\end{array}$$

をえる．

定義$3$と定理$3$から，

$$\begin{array}{rl}{f}_n(x)& =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)x^k\\ & \ge \sum _{k=0}^{{\sigma }_0(n)}{}_{{\sigma }_0(n)}{\mathrm{C}}_k{n}^{\frac{{\sigma }_0(n)-k}{2}}{x}^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{}_{{\sigma }_0(n)}{\mathrm{C}}_k(\sqrt{n}{)}^{{\sigma }_0(n)-k}{x}^k\\ & =(\sqrt{n}+x{)}^{{\sigma }_0(n)}\end{array}$$

定理$2$から，$d$が$n$の正の約数なら${\displaystyle \frac{n}{d}}$も$n$の正の約数であるから，

$$\begin{array}{rl}{c}_n(k)& =\sum _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}{d}_1{d}_2\cdots d_{{\sigma }_0(n)-k}\\ \frac{1}{n^{{\sigma }_0(n)-k}}{c}_n(k)& =\sum _{\mathrm{lcm}(d_1,d_2,\cdots ,d_{{\sigma }_0(n)-k})|n,0<d_1<d_2<\cdots <d_{{\sigma }_0(n)-k}}\frac{1}{d_1{d}_2\cdots d_{{\sigma }_0(n)-k}}\\ \frac{n^{\frac{{\sigma }_0(n)}{2}}}{n^{{\sigma }_0(n)-k}}{c}_n(k)& =\sum _{\mathrm{lcm}(d_1,d_2,\cdots ,d_k)|n,0<d_1<d_2<\cdots <d_k}{d}_1{d}_2\cdots d_k\\ n^{k-\frac{{\sigma }_0(n)}{2}}{c}_n(k)& =c_n({\sigma }_0(n)-k)\end{array}$$

定義$3$から

$$\begin{array}{rl}\frac{1}{x^{{\sigma }_0(n)}}{f}_n(x)& =\frac{1}{x^{{\sigma }_0(n)}}\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)x^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)x^{k-{\sigma }_0(n)}\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n({\sigma }_0(n)-k)x^{-k}\\ & =\sum _{k=0}^{{\sigma }_0(n)}{n}^{k-\frac{{\sigma }_0(n)}{2}}{c}_n(k)x^{-k}\\ & =n^{-\frac{{\sigma }_0(n)}{2}}\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)(\frac{n}{x}{)}^k\\ & =n^{-\frac{{\sigma }_0(n)}{2}}{f}_n(\frac{n}{x})\\ \therefore f_n(\frac{n}{x})& =(\frac{\sqrt{n}}{x}{)}^{{\sigma }_0(n)}{f}_n(x)\end{array}$$

定義$3$から
$$\begin{gathered} c_n(0)=f_n(0) \\ {c}_n(1)=f_n^{\prime }(0) \\ {c}_n(2)={\displaystyle \frac{1}{2}}{f}_n^{\prime \prime }(0) \\ ⋮ \\ {c}_n(k)={\displaystyle \frac{1}{k!}}{f}_n^{(k)}(0) \end{gathered}$$
と書ける．

$$\begin{array}{rl}f\ast g(n)& =\sum _{d|n,0<d}f(d)g({\displaystyle \frac{n}{d}})\\ & =\sum _{d|n,0<d}f({\displaystyle \frac{n}{d}})g(d)\\ & =g\ast f(n)\end{array}$$

メビウス反転公式から明らか．

定理$8$で$f(n)=\phi (n)$とおけば，例$3$から導出できる．

$$\begin{array}{rl}{S}_n(N)& =\sum _{k=1}^N{f}_n(k)\\ & =\sum _{k=1}^N\sum _{i=0}^{{\sigma }_0(n)}{c}_n(i)k^i\\ & =\sum _{i=0}^{{\sigma }_0(n)}{c}_n(i)\sum _{k=1}^N{k}^i\\ & \ge \sum _{i=0}^{{\sigma }_0(n)}{c}_n(i)\cdot N\sqrt[N]{\prod _{k=1}^N{k}^i}\\ & =N\sum _{i=0}^{{\sigma }_0(n)}{c}_n(i)(\sqrt[N]{N!}{)}^i\\ & =N{f}_n(\sqrt[N]{N!})\end{array}$$

よって
$$S_n(N)\ge N{f}_n(\sqrt[N]{N!})$$
をえる．

$$\begin{array}{rl}D{f}_n(n)& =\sum _{d|n,0<d}{f}_n(d)\\ & =\sum _{d|n,0<d}\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)d^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)\sum _{d|n,0<d}{d}^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k){\sigma }_k(n)\\ & \ge \sum _{k=0}^{{\sigma }_0(n)}{}_{{\sigma }_0(n)}{C}_k\cdot n^{\frac{{\sigma }_0(n)-k}{2}}(n^k+1+({\sigma }_0(n)-2)n^{\frac{k}{2}})\\ & =n^{\frac{{\sigma }_0(n)}{2}}\sum _{k=0}^{{\sigma }_0(n)}{}_{{\sigma }_0(n)}{C}_k\cdot n^{\frac{k}{2}}+n^{\frac{{\sigma }_0(n)}{2}}\sum _{k=0}^{{\sigma }_0(n)}{}_{{\sigma }_0(n)}{C}_k\cdot n^{-\frac{k}{2}}+({\sigma }_0(n)-2)n^{\frac{{\sigma }_0(n)}{2}}\sum _{k=0}^{{\sigma }_0(n)}{}_{{\sigma }_0(n)}{C}_k\\ & =(n+\sqrt{n}{)}^{{\sigma }_0(n)}+(\sqrt{n}+1{)}^{{\sigma }_0(n)}+2^{{\sigma }_0(n)}({\sigma }_0(n)-2)n^{\frac{{\sigma }_0(n)}{2}}\end{array}$$

途中までは命題$11$の証明と同じ．
$$\begin{array}{rl}D{f}_n(n)& =\sum _{d|n,0<d}{f}_n(d)\\ & =\sum _{d|n,0<d}\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)d^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)\sum _{d|n,0<d}{d}^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k){\sigma }_k(n)\\ & =c_n(0){\sigma }_0(n)+c_n(1){\sigma }_1(n)+\sum _{k=2}^{{\sigma }_0(n)}{c}_n(k){\sigma }_k(n)\\ & \le c_n(0){\sigma }_0(n)+c_n(1){\sigma }_1(n)+\sum _{k=2}^{{\sigma }_0(n)}{c}_n(k)n^{\frac{3}{2}k}\\ & =c_n(0)({\sigma }_0(n)-1)+c_n(1)({\sigma }_1(n)-n\sqrt{n})+\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)n^{\frac{3}{2}k}\\ & =c_n(0)({\sigma }_0(n)-1)+c_n(1)({\sigma }_1(n)-n\sqrt{n})+f_n(n\sqrt{n})\end{array}$$
ここで，定理$2$と定理$5$から
$$c_n(0)=n^{\frac{{\sigma }_0(n)}{2}},c_n(1)=n^{\frac{{\sigma }_0(n)}{2}-1}{c}_n({\sigma }_0(n)-1)=n^{\frac{{\sigma }_0(n)}{2}}{\sigma }_{-1}(n)$$
であるから
$$D{f}_n(n)\le f_n(n\sqrt{n})+n^{\frac{{\sigma }_0(n)}{2}}({\sigma }_0(n)-1)-n^{\frac{{\sigma }_0(n)}{2}}{\sigma }_1(n)(\sqrt{n}-{\sigma }_{-1}(n))$$
となる．

命題$11$の証明から，
$$D{f}_n(n)=\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k){\sigma }_k(n)$$
また，命題$10$の証明から
$$\begin{array}{rl}{S}_n(n)& =\sum _{i=0}^{{\sigma }_0(n)}{c}_n(i)\sum _{k=1}^n{k}^i\\ & \ge \sum _{k=0}^{{\sigma }_0(n)}{c}_n(k)\sum _{d|n,0<d}{d}^k\\ & =\sum _{k=0}^{{\sigma }_0(n)}{c}_n(k){\sigma }_k(n)\end{array}$$
であるから，
$$S_n(n)\ge D{f}_n(n)$$
をえる．

定理$6$から
$$\begin{array}{rl}{f}_n\ast {\displaystyle \frac{1}{f_n}}(n)& =\sum _{d|n,0<d}{\displaystyle \frac{f_n(d)}{f_n({\displaystyle \frac{n}{d}})}}\\ & =\sum _{d|n,0<d}({\displaystyle \frac{d}{\sqrt{n}}}{)}^{{\sigma }_0(n)}\\ & =n^{-\frac{{\sigma }_0(n)}{2}}{\sigma }_{{\sigma }_0(n)}(n)\\ & =n^{\frac{{\sigma }_0(n)}{2}}{\sigma }_{-{\sigma }_0(n)}(n)\end{array}$$