(1)

漸化式より、\(S_{1}\left(0,k\right)=S_{1}\left(1,k+1\right)\)となる。
\(k\ne0\)のとき\(S_{1}\left(0,k\right)=0\)となる。
\(k=0\)のとき、\(S_{1}\left(0,0\right)=S_{1}\left(1,1\right)=1\)となる。
これより、\(S_{1}\left(0,k\right)=\delta_{0k}\)となる。

(1)-2

\[ P\left(x,n\right)=\sum_{k=0}^{\infty}S_{1}\left(n,k\right)x^{k} \] に\(n=0\)を代入して、
\begin{align*} \sum_{k=0}^{\infty}S_{1}\left(0,k\right)x^{k} & =P\left(x,0\right)\\ & =1\\ & =\sum_{k=0}^{\infty}\delta_{0,k}x^{k} \end{align*} これより、
\[ \sum_{k=0}^{\infty}\left(S_{1}\left(0,k\right)-\delta_{0,k}\right)x^{k} \] 任意の\(x\)について成り立つためには、
\[ S_{1}\left(0,k\right)=\delta_{0,k} \]

(2)

\[ P\left(x,n\right)=\sum_{k=0}^{\infty}S_{1}\left(n,k\right)x^{k} \] に\(n=1\)を代入すると、
\begin{align*} \sum_{k=0}^{\infty}S_{1}\left(1,k\right)x^{k} & =P\left(x,1\right)\\ & =x\\ & =\sum_{k=0}^{\infty}\delta_{1,k}x^{k} \end{align*} これより、
\[ \sum_{k=0}^{\infty}\left(S_{1}\left(1,k\right)-\delta_{1,k}\right)x^{k}=0 \] となるので、
\[ S_{1}\left(1,k\right)=\delta_{1,k} \] となる。

(3)

漸化式より、\(S_{1}\left(n,0\right)=S_{1}\left(n+1,1\right)+nS_{1}\left(n,1\right)\)となる。
\(n\ne0\)のとき、\(S_{1}\left(n,0\right)=S_{1}\left(n+1,1\right)+nS_{1}\left(n,1\right)=\left(-1\right)^{n+2}n!+\left(-1\right)^{n+1}n(n-1)!=\left(-1\right)^{n}\left(n!-n!\right)=0\)となる。
\(n=0\)のとき、\(S_{1}\left(0,0\right)=S_{1}\left(1,1\right)+0S_{1}\left(0,1\right)=1\)となる。
これより、\(S_{1}\left(n,0\right)=\delta_{n0}\)となる。

(3)-2

\begin{align*} Q\left(x,n\right) & =\left(-1\right)^{n}P\left(-x,n\right)\\ & =\left(-1\right)^{n}\sum_{k=0}^{n}S_{1}\left(n,k\right)\left(-x\right)^{k} \end{align*} で\(x\rightarrow0\)とすると左辺は、
\begin{align*} \lim_{x\rightarrow0}Q\left(x,n\right) & =\lim_{x\rightarrow0}\frac{\Gamma\left(x+n\right)}{\Gamma\left(x\right)}\\ & =\delta_{0,n} \end{align*} 右辺は
\begin{align*} \left(-1\right)^{n}\lim_{x\rightarrow0}\sum_{k=0}^{n}S_{1}\left(n,k\right)\left(-x\right)^{k} & =\left(-1\right)^{n}\sum_{k=0}^{n}S_{1}\left(n,k\right)\delta_{0,k}\\ & =\left(-1\right)^{n}S_{1}\left(n,0\right) \end{align*} となるので、
\begin{align*} S_{1}\left(n,0\right) & =\left(-1\right)^{n}\delta_{0,n}\\ & =\delta_{0,n} \end{align*} となる。

(4)

\(n\in\mathbb{N}\)のとき、

\begin{align*} S_{1}\left(n,1\right) & =\left(n-1\right)!\sum_{j=0}^{n-1}\frac{\left(-1\right)^{n-j-1}}{j!}S_{1}\left(j,0\right)\\ & =\left(n-1\right)!\sum_{j=0}^{n-1}\frac{\left(-1\right)^{n-j-1}}{j!}\delta_{0,j}\\ & =\left(-1\right)^{n-1}\left(n-1\right)! \end{align*}

\(n=0\)のとき、

\(S_{1}\left(0,1\right)=0\)となる。

-

これらより題意は成り立つ。

(4)-2

\begin{align*} Q\left(x,n\right) & =\left(-1\right)^{n}P\left(-x,n\right)\\ & =\left(-1\right)^{n}\sum_{k=0}^{n}S_{1}\left(n,k\right)\left(-x\right)^{k} \end{align*} より、両辺を微分して\(x\rightarrow0\)とすると左辺は、
\[ \left[\frac{dQ\left(x,n\right)}{dx}\right]_{x\rightarrow0}=\begin{cases} \Gamma\left(n\right) & n\in\mathbb{N}\\ 0 & n=0 \end{cases} \] 右辺は
\begin{align*} \left[\left(-1\right)^{n}\frac{d}{dx}\sum_{k=0}^{\infty}S_{1}\left(n,k\right)\left(-x\right)^{k}\right]_{x\rightarrow0} & =\left(-1\right)^{n+1}\sum_{k=0}^{\infty}kS_{1}\left(n,k\right)\left[\left(-x\right)^{k-1}\right]_{x\rightarrow0}\\ & =\left(-1\right)^{n+1}\sum_{k=0}^{\infty}kS_{1}\left(n,k\right)\delta_{0,k-1}\\ & =\left(-1\right)^{n+1}S_{1}\left(n,1\right) \end{align*} となるので、
\begin{align*} S_{1}\left(n,1\right) & =\begin{cases} \left(-1\right)^{n+1}\left(n-1\right)! & n\in\mathbb{N}\\ 0 & n=0 \end{cases} \end{align*}

(5)

\(n\in\mathbb{N}\)のとき、

\begin{align*} S_{1}\left(n,2\right) & =\left(n-1\right)!\sum_{k=1}^{n-1}\frac{\left(-1\right)^{n-k-1}}{k!}S_{1}\left(k,1\right)\\ & =\left(n-1\right)!\sum_{k=1}^{n-1}\frac{\left(-1\right)^{n-k-1}}{k!}\left(-1\right)^{k-1}\left(k-1\right)!\\ & =\left(-1\right)^{n}\left(n-1\right)!\sum_{k=1}^{n-1}\frac{1}{k}\\ & =\left(-1\right)^{n}\left(n-1\right)!H_{n-1} \end{align*}

\(n=0\)のとき、

\(S_{1}\left(0,2\right)=0\)となる。

-

これらより題意は成り立つ。

(5)-2

\begin{align*} Q\left(x,n\right) & =\left(-1\right)^{n}P\left(-x,n\right)\\ & =\left(-1\right)^{n}\sum_{k=0}^{\infty}S_{1}\left(n,k\right)\left(-x\right)^{k} \end{align*} より、両辺を2回微分して\(x\rightarrow0\)とすると左辺は、
\[ \left[\frac{d^{2}Q\left(x,n\right)}{dx^{2}}\right]_{x\rightarrow0}=\begin{cases} 2\left(n-1\right)!H_{n-1} & n\in\mathbb{N}\\ 0 & n=0 \end{cases} \] 右辺は、
\begin{align*} \left[\left(-1\right)^{n}\frac{d^{2}}{dx^{2}}\sum_{k=0}^{\infty}S_{1}\left(n,k\right)\left(-x\right)^{k}\right]_{x\rightarrow0} & =\left(-1\right)^{n+2}\sum_{k=0}^{\infty}k\left(k-1\right)S_{1}\left(n,k\right)\left[\left(-x\right)^{k-2}\right]_{x\rightarrow0}\\ & =\left(-1\right)^{n}\sum_{k=0}^{\infty}k\left(k-1\right)S_{1}\left(n,k\right)\delta_{0,k-2}\\ & =\left(-1\right)^{n}2S_{1}\left(n,2\right) \end{align*} となる。
これより、
\[ S_{1}\left(n,2\right)=\begin{cases} \left(-1\right)^{n}\left(n-1\right)!H_{n-1} & n\in\mathbb{N}\\ 0 & n=0 \end{cases} \]

(6)

\(n\in\mathbb{N}\)のとき、

\begin{align*} S_{1}\left(n,3\right) & =\left(n-1\right)!\sum_{k=2}^{n-1}\frac{\left(-1\right)^{n-k-1}}{k!}S_{1}\left(k,2\right)\\ & =\left(n-1\right)!\sum_{k=2}^{n-1}\frac{\left(-1\right)^{n-k-1}}{k!}\left(-1\right)^{k}\left(k-1\right)!H_{k-1}\\ & =\left(-1\right)^{n-1}\left(n-1\right)!\sum_{k=2}^{n-1}\frac{H_{k-1}}{k}\\ & =\left(-1\right)^{n-1}\left(n-1\right)!\sum_{k=1}^{n-1}\frac{H_{k-1}}{k}\\ & =\left(-1\right)^{n-1}\left(n-1\right)!\sum_{k=1}^{n-1}\frac{H_{k}-\frac{1}{k}}{k}\\ & =\left(-1\right)^{n-1}\left(n-1\right)!\left(\sum_{k=1}^{n-1}\frac{H_{k}}{k}-\sum_{k=1}^{n-1}\frac{1}{k^{2}}\right)\\ & =\left(-1\right)^{n-1}\left(n-1\right)!\left(\frac{1}{2}\left(H_{n-1}^{2}+H_{n-1,2}\right)-H_{n-1,2}\right)\cmt{\sum_{k=1}^{n-1}\frac{H_{k}}{k}=\frac{1}{2}\left(H_{n-1}^{2}+H_{n-1,2}\right)}\\ & =\left(-1\right)^{n-1}\frac{1}{2}\left(n-1\right)!\left(H_{n-1}^{2}-H_{n-1,2}\right) \end{align*}

\(n=0\)のとき、

\(S_{1}\left(0,3\right)=0\)となる。

-

これらより題意は成り立つ。

(6)-2

\begin{align*} Q\left(x,n\right) & =\left(-1\right)^{n}P\left(-x,n\right)\\ & =\left(-1\right)^{n}\sum_{k=0}^{\infty}S_{1}\left(n,k\right)\left(-x\right)^{k} \end{align*} より、両辺を3回微分して\(x\rightarrow0\)とすると左辺は、
\[ \left[\frac{d^{3}Q\left(x,n\right)}{dx^{3}}\right]_{x\rightarrow0}=\begin{cases} 3\left(n-1\right)!\left(H_{n-1}^{2}-H_{n-1,2}\right) & n\in\mathbb{N}\\ 0 & n=0 \end{cases} \] 右辺は
\begin{align*} \left[\left(-1\right)^{n}\frac{d^{3}}{dx^{3}}\sum_{k=0}^{\infty}S_{1}\left(n,k\right)\left(-x\right)^{k}\right]_{x\rightarrow0} & =\left(-1\right)^{n+3}\sum_{k=0}^{\infty}k\left(k-1\right)\left(k-2\right)S_{1}\left(n,k\right)\left[\left(-x\right)^{k-3}\right]_{x\rightarrow0}\\ & =\left(-1\right)^{n+1}\sum_{k=0}^{\infty}k\left(k-1\right)\left(k-2\right)S_{1}\left(n,k\right)\delta_{0,k-3}\\ & =\left(-1\right)^{n+1}3!S_{1}\left(n,3\right) \end{align*} となる。
これより、
\[ S_{1}\left(n,3\right)=\begin{cases} \left(-1\right)^{n+1}\frac{1}{2}\left(n-1\right)!\left(H_{n-1}^{2}-H_{n-1,2}\right) & n\in\mathbb{N}\\ 0 & n=0 \end{cases} \]

(7)

\(n\in\mathbb{N}\)のとき、

\begin{align*} S_{1}\left(n,4\right) & =\left(n-1\right)!\sum_{k=3}^{n-1}\frac{\left(-1\right)^{n-k-1}}{k!}S_{1}\left(k,3\right)\\ & =\left(n-1\right)!\sum_{k=3}^{n-1}\frac{\left(-1\right)^{n-k-1}}{k!}\left(-1\right)^{k+1}\frac{1}{2}\left(k-1\right)!\left(H_{k-1}^{2}-H_{k-1,2}\right)\\ & =\left(-1\right)^{n}\left(n-1\right)!\sum_{k=3}^{n-1}\frac{1}{2k}\left(H_{k-1}^{2}-H_{k-1,2}\right)\\ & =\left(-1\right)^{n}\left(n-1\right)!\sum_{k=1}^{n-1}\frac{1}{2k}\left(H_{k-1}^{2}-H_{k-1,2}\right)\\ & =\left(-1\right)^{n}\left(n-1\right)!\sum_{k=1}^{n-1}\frac{1}{2k}\left(\left(H_{k}-\frac{1}{k}\right)^{2}-\left(H_{k,2}-\frac{1}{k^{2}}\right)\right)\\ & =\left(-1\right)^{n}\left(n-1\right)!\sum_{k=1}^{n-1}\frac{1}{2k}\left(H_{k}^{2}-H_{k,2}-2\frac{H_{k}}{k}+\frac{2}{k^{2}}\right)\\ & =\left(-1\right)^{n}\frac{1}{2}\left(n-1\right)!\sum_{k=1}^{n-1}\left(\frac{H_{k}^{2}}{k}-\frac{H_{k,2}}{k}-2\frac{H_{k}}{k^{2}}+\frac{2}{k^{3}}\right)\\ & =\left(-1\right)^{n}\frac{1}{2}\left(n-1\right)!\left(\sum_{k=1}^{n-1}\frac{H_{k}^{2}}{k}-\left(H_{n-1}H_{n-1,2}+H_{n-1,3}-\sum_{m=1}^{n-1}\frac{H_{m}}{m^{2}}\right)-2\sum_{k=1}^{n-1}\frac{H_{k}}{k^{2}}+2H_{n-1,3}\right)\cmt{\sum_{k=1}^{n-1}\frac{H_{k,2}}{k}=H_{n-1}H_{n-1,2}+H_{n-1,3}-\sum_{m=1}^{n-1}\frac{H_{m}}{m^{2}}}\\ & =\left(-1\right)^{n}\frac{1}{2}\left(n-1\right)!\left(\sum_{k=1}^{n-1}\frac{H_{k}^{2}}{k}-H_{n-1}H_{n-1,2}+H_{n-1,3}-\sum_{k=1}^{n-1}\frac{H_{k}}{k^{2}}\right)\\ & =\left(-1\right)^{n}\frac{1}{2}\left(n-1\right)!\left(\frac{1}{3}\left(H_{n-1}^{3}-H_{n-1,3}+3\sum_{k=1}^{n}\frac{H_{k}}{k^{2}}\right)-H_{n-1}H_{n-1,2}+H_{n-1,3}-\sum_{k=1}^{n-1}\frac{H_{k}}{k^{2}}\right)\cmt{\sum_{k=1}^{n-1}\frac{H_{k}^{2}}{k}=\frac{1}{3}\left(H_{n-1}^{3}-H_{n-1,3}+3\sum_{k=1}^{n}\frac{H_{k}}{k^{2}}\right)}\\ & =\left(-1\right)^{n}\frac{1}{3!}\left(n-1\right)!\left(H_{n-1}^{3}-H_{n-1,3}-3H_{n-1}H_{n-1,2}+3H_{n-1,3}\right)\\ & =\left(-1\right)^{n}\frac{1}{3!}\left(n-1\right)!\left(H_{n-1}^{3}+2H_{n-1,3}-3H_{n-1}H_{n-1,2}\right) \end{align*}

\(n=0\)のとき、

\(S_{1}\left(0,4\right)=0\)となる。

-

これらより題意は成り立つ。

(8)

区別の出来るn個のものをn個の区別の出来ない巡回列に分割する方法は1通りであるので\(S_{1}\left(n,n\right)=\left(-1\right)^{n+n}1=1\)となる。

(8)-2

\[ P\left(x,n\right)=\sum_{k=0}^{\infty}S_{1}\left(n,k\right)x^{k} \] の\(x\)の\(n\)次の項の係数より、\(S_{1}\left(n,n\right)=1\)となる。

(9)

\begin{align*} S_{1}\left(n,n-1\right) & =\left(-1\right)^{n+n-1}\sum_{i=0}^{n-1}i\\ & =-\frac{n\left(n-1\right)}{2}\\ & =-\frac{1}{2}P\left(n,2\right) \end{align*}

(9)-2

\[ P\left(x,n\right)=\sum_{k=0}^{\infty}S_{1}\left(n,k\right)x^{k} \] より、\(x\)の\(n-1\)次の項の係数を比較すると、
\begin{align*} S_{1}\left(n,n-1\right) & =-\sum_{k=1}^{n-1}k\\ & =-\frac{\left(1+n-1\right)\left(n-1\right)}{2}\\ & =-\frac{n\left(n-1\right)}{2}\\ & =-\frac{1}{2}P\left(n,2\right) \end{align*} となる。

(10)

\begin{align*} S_{1}\left(n,n-2\right) & =\left(-1\right)^{n+n-2}\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}ij\\ & =\sum_{i=0}^{n-1}iC\left(i,2\right)\\ & =\frac{1}{2!}\sum_{i=0}^{n-1}iP\left(i,2\right)\\ & =\frac{1}{2!}\sum_{i=0}^{n-1}\left\{ \left(i-2\right)P\left(i,2\right)+2P\left(i,2\right)\right\} \\ & =\frac{1}{2!}\sum_{i=0}^{n-1}\left\{ P\left(i,3\right)+2P\left(i,2\right)\right\} \\ & =\frac{1}{2!}\left\{ \frac{1}{4}P\left(n,4\right)+\frac{2}{3}P\left(n,3\right)\right\} \\ & =\frac{1}{8}P\left(n,4\right)+\frac{1}{3}P\left(n,3\right) \end{align*}

(11)

\begin{align*} S_{1}\left(n,n-3\right) & =\left(-1\right)^{n+n-3}\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}\sum_{k=0}^{j-1}ijk\\ & =-\sum_{i=0}^{n-1}i\left\{ \frac{1}{8}P\left(i,4\right)+\frac{1}{3}P\left(i,3\right)\right\} \\ & =-\sum_{i=0}^{n-1}\left\{ \frac{P\left(i,5\right)+4P\left(i,4\right)}{8}+\frac{P\left(i,4\right)+3P\left(i,3\right)}{3}\right\} \\ & =-\sum_{i=0}^{n-1}\left\{ \frac{1}{8}P\left(i,5\right)+\frac{5}{6}P\left(i,4\right)+P\left(i,3\right)\right\} \\ & =-\left\{ \frac{1}{8}\frac{P\left(n,6\right)}{6}+\frac{5}{6}\frac{P\left(n,5\right)}{5}+\frac{P\left(n,4\right)}{4}\right\} \\ & =-\left\{ \frac{1}{48}P\left(n,6\right)+\frac{1}{6}P\left(n,5\right)+\frac{1}{4}P\left(n,4\right)\right\} \end{align*}

(12)

\begin{align*} S_{2}\left(0,k\right) & =\frac{1}{k!}\sum_{j=0}^{\infty}\left(-1\right)^{k-j}C\left(k,j\right)j^{0}\\ & =\frac{\left(-1\right)^{k}}{k!}\sum_{j=0}^{\infty}\left(-1\right)^{j}C\left(k,j\right)\\ & =\frac{\left(-1\right)^{k}}{k!}\delta_{0,k}\\ & =\delta_{0,k} \end{align*}

(13)

\begin{align*} S_{2}\left(1,k\right) & =\frac{1}{k!}\sum_{j=0}^{\infty}\left(-1\right)^{k-j}C\left(k,j\right)j\\ & =\frac{\left(-1\right)^{k}}{k!}k\sum_{j=0}^{\infty}\left(-1\right)^{j}C\left(k-1,j-1\right)\\ & =\frac{\left(-1\right)^{k-1}}{\left(k-1\right)!}\sum_{j=-1}^{\infty}\left(-1\right)^{j}C\left(k-1,j\right)\\ & =\frac{\left(-1\right)^{k-1}}{\left(k-1\right)!}\delta_{0,k-1}\\ & =\delta_{1,k} \end{align*}

(14)

\begin{align*} S_{2}\left(n,0\right) & =\sum_{j=0}^{\infty}\left(-1\right)^{-j}C\left(0,j\right)j^{n}\\ & =C\left(0,0\right)0^{n}\\ & =\delta_{0,n} \end{align*}

(15)

\begin{align*} S_{2}\left(n,1\right) & =\sum_{j=0}^{\infty}\left(-1\right)^{1-j}C\left(1,j\right)j^{n}\\ & =-\sum_{j=0}^{\infty}\left(-1\right)^{j}C\left(1,j\right)j^{n}\\ & =-\left(C\left(1,0\right)\delta_{0,n}-C\left(1,1\right)\right)\\ & =1-\delta_{0,n} \end{align*}

(16)

\begin{align*} S_{2}\left(n,2\right) & =\frac{1}{2!}\sum_{j=0}^{2}\left(-1\right)^{2-j}C\left(2,j\right)j^{n}\\ & =\frac{1}{2}\left(C\left(2,0\right)0^{n}-C\left(2,1\right)1^{n}+C\left(2,2\right)2^{n}\right)\\ & =\frac{1}{2}\left(\delta_{0,n}-2+2^{n}\right)\\ & =2^{n-1}-1+\frac{1}{2}\delta_{0,n} \end{align*}

(17)

\begin{align*} S_{2}\left(n,3\right) & =\frac{1}{3!}\sum_{j=0}^{3}\left(-1\right)^{3-j}C\left(3,j\right)j^{n}\\ & =\frac{1}{3!}\left(-C\left(3,0\right)0^{n}+C\left(3,1\right)1^{n}-C\left(3,2\right)2^{n}+C\left(3,3\right)3^{n}\right)\\ & =\frac{1}{3!}\left(-\delta_{0,n}+3-3\cdot2^{n}+3^{n}\right)\\ & =\frac{1}{2}\left(3^{n-1}-2^{n}+1-\frac{1}{3}\delta_{0,n}\right) \end{align*}

(18)

\begin{align*} S_{2}\left(n,4\right) & =\frac{1}{4!}\sum_{j=0}^{4}\left(-1\right)^{4-j}C\left(4,j\right)j^{n}\\ & =\frac{1}{4!}\left(C\left(4,0\right)0^{n}-C\left(4,1\right)1^{n}+C\left(4,2\right)2^{n}-C\left(4,3\right)3^{n}+C\left(4,4\right)4^{n}\right)\\ & =\frac{1}{4!}\left(\delta_{0,n}-4+6\cdot2^{n}-4\cdot3^{n}+4^{n}\right)\\ & =\frac{1}{3!}\left(4^{n-1}-3^{n}+3\cdot2^{n-1}-1+\frac{1}{4}\delta_{0,n}\right) \end{align*}

(19)

\begin{align*} S_{2}\left(n,n\right) & =\sum_{a_{1}+a_{2}+\cdots+a_{n}=0}1^{a_{1}}2^{a_{2}}\cdots n^{a_{n}}\\ & =\sum_{a_{1}+a_{2}+\cdots+a_{n}=0}1^{0}2^{0}\cdots k^{0}\\ & =1 \end{align*}

(19)-2

\begin{align*} x^{n} & =\sum_{j=0}^{n}S_{2}\left(n,j\right)P\left(x,j\right)\\ & =\sum_{j=0}^{n}S_{2}\left(n,j\right)\prod_{m=0}^{j-1}\left(x-m\right) \end{align*} より、\(x^{n}\)の係数を比較すると \(S_{2}\left(n,n\right)=1\)となる。

(20)

\begin{align*} S_{2}\left(n,n-1\right) & =\sum_{a_{1}+a_{2}+\cdots+a_{n-1}=1}1^{a_{1}}2^{a_{2}}\cdots\left(n-1\right)^{a_{n-1}}\\ & =\sum_{k=0}^{n-1}k\\ & =\frac{n\left(n-1\right)}{2}\\ & =C\left(n,2\right) \end{align*}

(20)-2

\begin{align*} x^{n} & =\sum_{j=0}^{n}S_{2}\left(n,j\right)P\left(x,j\right)\\ & =\sum_{j=0}^{n}S_{2}\left(n,j\right)\prod_{m=0}^{j-1}\left(x-m\right) \end{align*} より、\(x^{n-1}\)の係数を比較すると 、
\begin{align*} S_{2}\left(n,n-1\right) & =\sum_{k=0}^{n-1}k\\ & =\frac{n\left(n-1\right)}{2}\\ & =C\left(n,2\right) \end{align*} となる。

(21)

\begin{align*} S_{2}\left(n,n-2\right) & =\sum_{a_{1}+a_{2}+\cdots+a_{n-2}=2}1^{a_{1}}2^{a_{2}}\cdots\left(n-2\right)^{a_{n-2}}\\ & =\sum_{j=1}^{n-2}\sum_{k=j}^{n-2}jk\\ & =\sum_{j=1}^{n-2}j\frac{\left(n-2-j+1\right)\left(j+n-2\right)}{2}\\ & =\sum_{j=1}^{n-2}j\frac{-\left(j-\left(n-2\right)-1\right)\left(j+n-2\right)}{2}\\ & =\sum_{j=1}^{n-2}j\frac{-j^{2}+\left(n-2\right)^{2}+j+n-2}{2}\\ & =\sum_{j=1}^{n-2}j\frac{-j^{2}+j+\left(n-2\right)\left(n-1\right)}{2}\\ & =\frac{1}{2}\sum_{j=1}^{n-2}\left(-j^{3}+j^{2}+\left(n-2\right)\left(n-1\right)j\right)\\ & =\frac{1}{2}\left(-\left(\frac{\left(n-2\right)\left(n-1\right)}{2}\right)^{2}+\frac{\left(n-2\right)\left(n-1\right)\left(2n-3\right)}{6}+\left(n-2\right)\left(n-1\right)\frac{\left(n-2\right)\left(n-1\right)}{2}\right)\\ & =\frac{1}{2\cdot4!}\left(n-2\right)\left(n-1\right)\left\{ -6\left(n-2\right)\left(n-1\right)+4\left(2n-3\right)+12\left(n-2\right)\left(n-1\right)\right\} \\ & =\frac{1}{2\cdot4!}\left(n-2\right)\left(n-1\right)\left\{ 6\left(n-2\right)\left(n-1\right)+4\left(2n-3\right)\right\} \\ & =\frac{1}{2\cdot4!}\left(n-2\right)\left(n-1\right)\left(6n^{2}-10n\right)\\ & =\frac{n}{4!}\left(n-2\right)\left(n-1\right)\left(3n-5\right)\\ & =\frac{n}{4!}\left(3n^{3}-14n^{2}+21n-10\right) \end{align*}

(22)

\begin{align*} S_{2}\left(n,n-3\right) & =\sum_{a_{1}+a_{2}+\cdots+a_{n-3}=3}1^{a_{1}}2^{a_{2}}\cdots\left(n-3\right)^{a_{n-3}}\\ & =\sum_{i=1}^{n-3}\sum_{j=i}^{n-3}\sum_{k=j}^{n-3}ijk\\ & =\sum_{i=1}^{n-3}i\sum_{j=i}^{n-3}j\frac{\left(n-3-j+1\right)\left(n-3+j\right)}{2}\\ & =\sum_{i=1}^{n-3}i\sum_{j=i}^{n-3}j\frac{-\left(j-\left(n-3\right)-1\right)\left(j+n-3\right)}{2}\\ & =\frac{1}{2}\sum_{i=1}^{n-3}i\sum_{j=i}^{n-3}j\left(-j^{2}+\left(n-3\right)^{2}+j+n-3\right)\\ & =\frac{1}{2}\sum_{i=1}^{n-3}i\sum_{j=i}^{n-3}\left(-j^{3}+j^{2}+\left(n-3\right)\left(n-2\right)j\right)\\ & =\frac{1}{4!}\sum_{i=1}^{n-3}i\left\{ 3i^{4}-10i^{3}-3\left(2n^{2}+10n-9\right)i^{2}+3\left(3n^{2}-15n+17\right)i+3n^{4}-26n^{3}+81n^{2}-106n+48\right\} \\ & =\frac{n}{2\cdot4!}\left(n-3\right)^{2}\left(n-2\right)^{2}\left(n-1\right)\\ & =\frac{n}{2\cdot4!}\left(n^{5}-11n^{4}+47n^{3}-97n^{2}+96n-36\right) \end{align*}