◆大阪大学理学部研究生の河野 進さんからの解答。
私が趣味で作った公式集のようなものをお送りします。
よく知られている事ばかりかも知れませんし、間違えや曖昧な点も多い事と思いますが、
関連する事などを含めて指摘していただければ嬉しいです。
AMS-LaTeX のファイルです。
(1.0) はヒルゼブルフの教科書に載っていた２項係数の拡張
の方法です。n を不定元としても定義できます。
(1.1) は (1.4) の系ですが先に気付いて、このような証明
を付けましたが、後にもっと一般的な結果 (1.4) が知
られている事を先輩に教えて貰いました。
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{\bf Definition 1.0.} Let $k$ be an integer. Then we set
$$\binom{n}{k} = \left\{
\begin{array}{ll}
n(n-1)\cdots (n-k+1)/k! & (k > 0), \\
\noalign{\vskip0.1cm}
1 & (k = 0), \\
\noalign{\vskip0.1cm}
0 & (k < 0).
\end{array} \right.$$
\vspace*{3ex}
\noindent
(1.1) \begin{it} Let $n$ and $k$ be integers with
$n\geq k > 0$. Then \end{it}
$$\sum_{i=1}^{n} (-1)^{n-i} \binom{n}{i}i^{k} = \left\{
\begin{array}{ll}
0 & (n > k), \\
\noalign{\vskip0.1cm}
n! \hspace{1ex} (= k!) & (n = k).
\end{array} \right.$$
\vspace*{2ex}
Proof. \hspace{.1cm} Set the polynomials $f_{j}(x) \in \BZ [x]$
($0 \leq j \leq n$) inductively by
$$f_{0}(x) = (x-1)^{n},
f_{j+1}(x) = x\cdot (\frac{d}{dx}f_{j})(x) \hspace{.2cm}
(0 \leq j < n).$$
Then
$f_{j}(x) = \sum_{i=1}^{n}(-1)^{n-i}\binom{n}{i}i^{j}x^{i}$
and $f_{j}(1) =$ $\sum_{i=1}^{n}(-1)^{n-i}\binom{n}{i}i^{j}$
($0 < j \leq n$). On the other hand, it is shown by the
induction with respect to $j$ that there exists $g_{j}(x) \in$
$\BZ [x]$ with
\noindent
\hspace*{2.4cm}
$f_{j}(x) = n(n-1)\cdots (n-j+1)x^{j}(x-1)^{n-j}
+(x-1)^{n-j+1}g_{j}(x)$.
\noindent
Thus
\noindent
\hspace*{2.9cm}
$\sum_{i=1}^{n} (-1)^{n-i} \binom{n}{i}i^{k} = f_{k}(1) =
n(n-1)\cdots (n-k+1)0^{n-k}$.
\noindent
This implies (1.1). \hspace{11.6cm} q.e.d.
\vspace*{3ex}
\noindent
(1.2) \begin{it} Let $n \geq 0$ be an integer.
Then, in the polynomial ring $\BZ [x]$, following
equalities hold. \end{it}
\noindent
(1) \hspace{.5cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}
\sum_{j=0}^{i}\binom{i}{j}x^{j}
= \sum_{i=0}^{n}\binom{n}{i}\sum_{j=0}^{i}
(-1)^{i-j}\binom{i}{j}x^{j} = x^{n}$.
\noindent
(2) \hspace{.5cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}
\sum_{j=0}^{i}\binom{i}{j}x^{n-i+j}
= \sum_{i=0}^{n}\binom{n}{i}\sum_{j=0}^{i}
(-1)^{i-j}\binom{i}{j}x^{i} = 1$.
\noindent
(3) \hspace{.5cm}
$\sum_{i=0}^{n}\binom{n}{i}\sum_{j=0}^{i}
\binom{i}{j}x^{j} = \sum_{k=0}^{n}2^{n-k}\binom{n}{k}x^{k}$.
\noindent
(4) \hspace{.5cm}
$\sum_{i=0}^{n}\binom{n}{i}
\sum_{j=0}^{i}\binom{i}{j}x^{n-i+j}
= \sum_{i=0}^{n}\binom{n}{i}\sum_{j=0}^{i}
\binom{i}{j}x^{i} = \sum_{k=0}^{n}2^{k}\binom{n}{k}x^{k}$.
\noindent
(5) \hspace{.5cm}
$\sum_{i=0}^{n}(-2)^{n-i}\binom{n}{i}
\sum_{j=0}^{i}\binom{i}{j}x^{j}
= \sum_{i=0}^{n}\binom{n}{i}\sum_{j=0}^{i}
(-2)^{i-j}\binom{i}{j}x^{j}
= \sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}x^{k}$.
\noindent
(6) \hspace{.5cm}
$\sum_{i=0}^{n}(-2)^{n-i}\binom{n}{i}
\sum_{j=0}^{i}\binom{i}{j}x^{n-i+j}
= \sum_{i=0}^{n}\binom{n}{i}\sum_{j=0}^{i}
(-2)^{i-j}\binom{i}{j}x^{i}
= \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}x^{k}$.
\vspace*{2ex}
Proof. \hspace{.1cm} Set $f(x, y, z) = (x+y+z)^{n}
\in \BZ [x, y, z]$. Then
\noindent
\hspace*{2.2cm}
$f(x, y, z) = \sum_{i=0}^{n}\binom{n}{i}(x+y)^{i}z^{n-i}
= \sum_{i=0}^{n}\binom{n}{i}z^{n-i}\sum_{j=0}^{i}
\binom{i}{j}x^{j}y^{i-j}$.
\noindent
Now, (1) is obtained by $f(x, \pm 1, \mp 1)$,
(2) is obtained by $f(x, 1, -x) =$ $f(x, -x, 1)$,
(3) is obtained by $f(x, 1, 1)$,
(4) is obtained by $f(x, 1, x) =$ $f(x, x, 1)$,
(5) is obtained by $f(x, 1, -2) =$ $f(x, -2, 1)$,
and (6) is obtained by $f(x, 1, -2x) =$ $f(x, -2x, 1)$.
\hspace{3.6cm} q.e.d.
\vspace*{3ex}
{\bf Definition 1.3.} Let $K$ be a finite set with $k$
elements, and $N$ a finite set with $n$ elements.
Then, $M_{k,n}$ denotes the number of elements of the
finite set
\noindent
\hspace*{5.3cm}
$\{ f\colon K \rightarrow N | f$ is a surjection$\}$.
\vspace*{3ex}
\noindent
(1.4) \hspace{1.5cm} $M_{k,n} =
\sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i}i^{k}$.
\vspace*{2ex}
Proof. \hspace{.1cm} Since the number of the elements of the
finite set $\{ f\colon K \rightarrow N\}$ is equal to
$n^{k}$, we obtain
$i^{k} = \sum_{j=0}^{i}\binom{i}{j}M_{k,j} \hspace{.2cm}
(i\geq 0)$.
It follows from (1.2) (1) that
\noindent
\hspace*{2.1cm} $\sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i}i^{k}
= \sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i}
\sum_{j=0}^{i}\binom{i}{j}M_{k,j} = M_{k,n}$. \hspace{2.1cm}
q.e.d.
\vspace*{3ex}
Following three assertions (1.5)-(1.7) are slightly modified
version of that of [3].
\vspace*{3ex}
\noindent
(1.5) \begin{it} Let $n \geq 0$ be an integer.
Then, in the polynomial ring $\BZ [x]$, following
equalities hold. \end{it}
\noindent
(1) \hspace{1ex}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n}{n-i}
\sum_{j=0}^{i}\binom{i}{j}x^{j}
= \sum_{k=0}^{n}(-1)^{n-k}\binom{2n-k-1}{n-k}x^{k}$.
\noindent
(2) \hspace{1ex}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}-\binom{2n-1}{n-i-1})
\sum_{j=0}^{i}\binom{i}{j}x^{j}
= \sum_{k=0}^{n}(-1)^{n-k}(\binom{2n-k-2}{n-k}
-\binom{2n-k-2}{n-k-1})x^{k}$.
\noindent
(3) \hspace{1ex}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n-i-1}{n-i}
\sum_{j=0}^{i}(-1)^{i-j}\binom{i}{j}x^{j}
= \sum_{k=0}^{n}(-1)^{n-k}\binom{2n}{n-k}x^{k}$.
\noindent
(4) \hspace{1ex}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-i-2}{n-i}-\binom{2n-i-2}{n-i-1})
\sum_{j=0}^{i}(-1)^{i-j}\binom{i}{j}x^{j}$
\noindent
\hspace*{6.1cm}
$= \sum_{k=0}^{n}(-1)^{n-k}(\binom{2n-1}{n-k}
-\binom{2n-1}{n-k-1})x^{k}$.
\vspace*{2ex}
Proof. \hspace{.1cm} If $j > l \geq 0$, then
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}\binom{2n-l}{n-i}\binom{i-l}{j-l}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-l-1}{n-i}+\binom{2n-l-1}{n-i-1})
\binom{i-l}{j-l}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-l-1}{n-i}\binom{i-l}{j-l}
+\sum_{i=j}^{n-1}(-1)^{n-i}\binom{2n-l-1}{n-i-1}\binom{i-l}{j-l}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-l-1}{n-i}\binom{i-l-1}{j-l-1}$.
\noindent
This implies that
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}\binom{2n}{n-i}\binom{i}{j}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-j}{n-i}\binom{i-j}{0}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-j-1}{n-i}
+\binom{2n-j-1}{n-i-1})$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-j-1}{n-i}
+\sum_{i=j}^{n-1}(-1)^{n-i}\binom{2n-j-1}{n-i-1}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-j-1}{n-i}
-\sum_{i=j+1}^{n}(-1)^{n-i}\binom{2n-j-1}{n-i}
= (-1)^{n-j}\binom{2n-j-1}{n-j}$.
\noindent
Thus we obtain
\noindent
\hspace*{.1cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n}{n-i}
\sum_{j=0}^{i}\binom{i}{j}x^{j}
= \sum_{j=0}^{n}(\sum_{i=j}^{n}(-1)^{n-i}\binom{2n}{n-i}
\binom{i}{j})x^{j}
= \sum_{j=0}^{n}(-1)^{n-j}\binom{2n-j-1}{n-j}x^{j}$.
\noindent
This proves (1).
If $j > l \geq 0$, then
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-l-1}{n-i}
-\binom{2n-l-1}{n-i-1})\binom{i-l}{j-l}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-l-2}{n-i}
-\binom{2n-l-2}{n-i-1}+\binom{2n-l-2}{n-i-1}
-\binom{2n-l-2}{n-i-2})\binom{i-l}{j-l}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-l-2}{n-i}
-\binom{2n-l-2}{n-i-1})\binom{i-l}{j-l}
+\sum_{i=j}^{n-1}(-1)^{n-i}(\binom{2n-l-2}{n-i-1}
-\binom{2n-l-2}{n-i-2})\binom{i-l}{j-l}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-l-2}{n-i}
-\binom{2n-l-2}{n-i-1})\binom{i-l-1}{j-l-1}$.
\noindent
This implies that
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}
-\binom{2n-1}{n-i-1})\binom{i}{j}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-j-1}{n-i}
-\binom{2n-j-1}{n-i-1})\binom{i-j}{0}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-j-2}{n-i}
-\binom{2n-j-2}{n-i-1})
+\sum_{i=j}^{n-1}(-1)^{n-i}(\binom{2n-j-2}{n-i-1}
-\binom{2n-j-2}{n-i-2})$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-j-2}{n-i}
-\binom{2n-j-2}{n-i-1})
-\sum_{i=j+1}^{n}(-1)^{n-i}(\binom{2n-j-2}{n-i}
-\binom{2n-j-2}{n-i-i})$
$= (-1)^{n-j}(\binom{2n-j-2}{n-j} -\binom{2n-j-2}{n-j-1})$.
\noindent
Thus we obtain
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}
-\binom{2n-1}{n-i-1})\sum_{j=0}^{i}\binom{i}{j}x^{j}$
$= \sum_{j=0}^{n}(\sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}
-\binom{2n-1}{n-i-1})\binom{i}{j})x^{j}$
$= \sum_{j=0}^{n}(-1)^{n-j}(\binom{2n-j-2}{n-j}
-\binom{2n-j-2}{n-j-1})x^{j}$.
\noindent
This proves (2).
It follows from (1.2) (1) and (1.5) (1) that
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n-i-1}{n-i}
\sum_{j=0}^{i}(-1)^{i-j}\binom{i}{j}x^{j}$
$= \sum_{k=0}^{n}(-1)^{n-k}\binom{2n}{n-k}
\sum_{i=0}^{k}\binom{k}{i}\sum_{j=0}^{i}(-1)^{i-j}
\binom{i}{j}x^{j}$
$= \sum_{k=0}^{n}(-1)^{n-k}\binom{2n}{n-k}x^{k}$.
\noindent
This proves (3). Similarly, (1.2) (1) and (1.5) (2)
are used to prove (4). \hspace{2.0cm} q.e.d.
\vspace*{3ex}
\noindent
(1.6) \begin{it} Let $n \geq 0$ be an integer.
Then, in the polynomial ring $\BZ [x]$, following
equalities hold. \end{it}
\noindent
(1) \hspace{1.2cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n}{n-i}
\sum_{j=0}^{i}(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})x^{j}
= x^{n}$.
\noindent
(2) \hspace{1.2cm}
$\sum_{i=0}^{n}(\binom{n+i}{n-i}+\binom{n+i-1}{n-i-1})
\sum_{j=0}^{i}(-1)^{i-j}\binom{2i}{i-j}x^{j}
= x^{n}$.
\noindent
(3) \hspace{1.2cm}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}-\binom{2n-1}{n-i-1})
\sum_{j=0}^{i}\binom{i+j-1}{i-j}x^{j} = x^{n}$.
\noindent
(4) \hspace{1.2cm}
$\sum_{i=0}^{n}\binom{n+i-1}{n-i}
\sum_{j=0}^{i}(-1)^{i-j}(\binom{2i-1}{i-j}-\binom{2i-1}{i-j-1})x^{j}
= x^{n}$.
\noindent
(5) \hspace{1.2cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n-i-1}{n-i}
\sum_{j=0}^{i}(\binom{j}{i-j}+\binom{j-1}{i-j-1})x^{j} = x^{n}$.
\noindent
(6) \hspace{1.2cm}
$\sum_{i=0}^{n}(\binom{i}{n-i}+\binom{i-1}{n-i-1})
\sum_{j=0}^{i}(-1)^{i-j}\binom{2i-j-1}{i-j}x^{j} = x^{n}$.
\noindent
(7) \hspace{1.2cm}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-i-2}{n-i}-\binom{2n-i-2}{n-i-1})
\sum_{j=0}^{i}\binom{j-1}{i-j}x^{j} = x^{n}$.
\noindent
(8) \hspace{1.2cm}
$\sum_{i=0}^{n}\binom{i-1}{n-i}
\sum_{j=0}^{i}(-1)^{i-j}(\binom{2i-j-2}{i-j}-\binom{2i-j-2}{i-j-1})
x^{j} = x^{n}$.
\vspace*{2ex}
Proof. \hspace{0.1cm} Since
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}
\binom{2n}{n-i}(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}+\binom{2n-1}{n-i-1})
(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-1}{n-i}
(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})
+\sum_{i=j}^{n-1}(-1)^{n-i}\binom{2n-1}{n-i-1}
(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-1}{n-i}
(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})
-\sum_{i=j+1}^{n}(-1)^{n-i}\binom{2n-1}{n-i}
(\binom{i+j-1}{i-j-1}+\binom{i+j-2}{i-j-2})$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-1}{n-i}
(\binom{i+j-1}{i-j}+\binom{i+j-2}{i-j-1})$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-2j}{n-i}
(\binom{i-j}{i-j}+\binom{i-j-1}{i-j-1})$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-2j}{n-i}
+\sum_{i=j+1}^{n}(-1)^{n-i}\binom{2n-2j}{n-i}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n-2j}{n-i}
-\sum_{i=j}^{n-1}(-1)^{n-i}\binom{2n-2j}{n-i-1}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-2j}{n-i}
-\binom{2n-2j}{n-i-1})$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-2j-1}{n-i}
-\binom{2n-2j-1}{n-i-1}+\binom{2n-2j-1}{n-i-1}
-\binom{2n-2j-1}{n-i-2})$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-2j-1}{n-i}
-\binom{2n-2j-1}{n-i-1})
+\sum_{i=j}^{n-1}(-1)^{n-i}(\binom{2n-2j-1}{n-i-1}
-\binom{2n-2j-1}{n-i-2})$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-2j-1}{n-i}
-\binom{2n-2j-1}{n-i-1})
-\sum_{i=j+1}^{n}(-1)^{n-i}(\binom{2n-2j-1}{n-i}
-\binom{2n-2j-1}{n-i-1})$
$= (-1)^{n-j}(\binom{2n-2j-1}{n-j}
-\binom{2n-2j-1}{n-j-1}) = 0^{n-j}$,
\noindent
we obtain
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n}{n-i}
\sum_{j=0}^{i}(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})x^{j}$
$= \sum_{j=0}^{n}(\sum_{i=j}^{n}(-1)^{n-i}\binom{2n}{n-i}
(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1}))x^{j}
= x^{n}$.
\noindent
This proves (1).
Since
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}
-\binom{2n-1}{n-i-1})\binom{i+j-1}{i-j}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}
-\binom{2n-1}{n-i-1})\binom{i+j}{i-j}
-\sum_{i=j+1}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}
-\binom{2n-1}{n-i-1})\binom{i+j-1}{i-j-1}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}
-\binom{2n-1}{n-i-1})\binom{i+j}{i-j}
+\sum_{i=j}^{n-1}(-1)^{n-i}(\binom{2n-1}{n-i-1}
-\binom{2n-1}{n-i-2})\binom{i+j}{i-j}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}-\binom{2n-1}{n-i-1}
+\binom{2n-1}{n-i-1}-\binom{2n-1}{n-i-2})\binom{i+j}{i-j}$
$= \sum_{i=j}^{n}(-1)^{n-i}(\binom{2n}{n-i}-\binom{2n}{n-i-1})
\binom{i+j}{i-j}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n}{n-i}\binom{i+j}{i-j}
-\sum_{i=j}^{n-1}(-1)^{n-i}\binom{2n}{n-i-1}\binom{i+j}{i-j}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n}{n-i}\binom{i+j}{i-j}
+\sum_{i=j+1}^{n}(-1)^{n-i}\binom{2n}{n-i}\binom{i+j-1}{i-j-1}$
$= \sum_{i=j}^{n}(-1)^{n-i}\binom{2n}{n-i}(\binom{i+j}{i-j}
+\binom{i+j-1}{i-j-1}) = 0^{n-j}$
\noindent
by the proof above, we obtain
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-1}{n-i}-\binom{2n-1}{n-i-1})
\sum_{j=0}^{i}\binom{i+j-1}{i-j}x^{j}$
$= \sum_{j=0}^{n}(\sum_{i=0}^{n}(-1)^{n-i}
(\binom{2n-1}{n-i}-\binom{2n-1}{n-i-1})
\binom{i+j-1}{i-j})x^{j} = x^{n}$.
\noindent
This proves (3).
If $j > 0$, then
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}\binom{2n-i-1}{n-i}
(\binom{j}{i-j}+\binom{j-1}{i-j-1})$
$= \sum_{i=j}^{2j}(-1)^{n-i}\binom{2n-i-1}{n-i}
\binom{j}{i-j}+\sum_{i=j+1}^{2j}(-1)^{n-i}\binom{2n-i-1}{n-i}
\binom{j-1}{i-j-1}$
$= \sum_{i=0}^{j}(-1)^{n-i-j}\binom{2n-i-j-1}{n-i-j}
\binom{j}{i}+\sum_{i=0}^{j-1}(-1)^{n-i-j-1}\binom{2n-i-j-2}{n-i-j-1}
\binom{j-1}{i}$
$= (-1)^{n-j}\sum_{i=0}^{j}\sum_{k=0}^{j-i}(-1)^{i}
\binom{j-i}{k}\binom{2n-2j-1}{n-i-j-k}\binom{j}{i}$
\noindent
\hspace*{2.6cm} $+(-1)^{n-j+1}\sum_{i=0}^{j-1}\sum_{k=0}^{j-i+1}
(-1)^{i}\binom{j-i-1}{k}\binom{2n-2j-1}{n-i-j-1-k}\binom{j-1}{i}$
$= (-1)^{n-j}\sum_{i=0}^{j}\sum_{k=0}^{i}(-1)^{j-i}
\binom{i}{k}\binom{2n-2j-1}{n-j-(j-i+k)}\binom{j}{i}$
\noindent
\hspace*{2.6cm} $-(-1)^{n-j}\sum_{i=0}^{j-1}\sum_{k=0}^{i}
(-1)^{j-i+1}\binom{i}{k}\binom{2n-2j-1}{n-j-1-(j-1-i+k)}\binom{j-1}{i}$
$= (-1)^{n-j}\binom{2n-2j-1}{n-j} -(-1)^{n-j}\binom{2n-2j-1}{n-j-1}$
\hspace{.3cm} (by (1.2) (2))
$= 0^{n-j}$.
\noindent
In the case $j = 0$,
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n-i-1}{n-i}
(\binom{0}{i}+\binom{-1}{i-1})
= (-1)^{n}(\binom{2n-1}{n}-\sum_{i=1}^{n}\binom{2n-i-1}{n-i})
= 0^{n}$.
\noindent
Thus we obtain
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{2n-i-1}{n-i}
\sum_{j=0}^{i}(\binom{j}{i-j}+\binom{j-1}{i-j-1})x^{j}$
$= \sum_{j=0}^{n}(\sum_{i=j}^{n}(-1)^{n-i}\binom{2n-i-1}{n-i}
(\binom{j}{i-j}+\binom{j-1}{i-j-1}))x^{j} = x^{n}$.
\noindent
This proves (5).
If $j > 0$, then
\noindent
\hspace*{1.4cm}
$\sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-i-2}{n-i}
-\binom{2n-i-2}{n-i-1})\binom{j-1}{i-j}$
$= \sum_{i=j}^{2j-1}(-1)^{n-i}(\binom{2n-i-2}{n-i}
-\binom{2n-i-2}{n-i-1})\binom{j-1}{i-j}$
$= \sum_{i=0}^{j-1}(-1)^{n-i-j}(\binom{2n-i-j-2}{n-i-j}
-\binom{2n-i-j-2}{n-i-j-1})\binom{j-1}{i}$
$= \sum_{i=0}^{j-1}\sum_{k=0}^{j-i-1}(-1)^{n-i-j}
\binom{j-i-1}{k}(\binom{2n-2j-1}{n-i-j-k}
-\binom{2n-2j-1}{n-i-j-1-k})\binom{j-1}{i}$
$= (-1)^{n-j}\sum_{i=0}^{j-1}\sum_{k=0}^{i}(-1)^{j-i-1}
\binom{i}{k}(\binom{2n-2j-1}{n-j-(j-1-j+k)}
-\binom{2n-2j-1}{n-j-1-(j-1-i+k)})\binom{j-1}{i}$
$= (-1)^{n-j}(\binom{2n-2j-1}{n-j} -\binom{2n-2j-1}{n-j-1})$
\hspace{.3cm} (by (1.2) (2))
$= 0^{n-j}$.
\noindent
In the case $j = 0$,
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-i-2}{n-i}
-\binom{2n-i-2}{n-i-1})\binom{-1}{i}$
$= \sum_{i=0}^{n}(-1)^{n}(\binom{2n-i-2}{n-i}
-\binom{2n-i-2}{n-i-1})$
$= (-1)^{n}(\sum_{i=0}^{n}\binom{2n-i-2}{n-i}
-\sum_{i=0}^{n-1}\binom{2n-i-2}{n-i-1})$
$= (-1)^{n}(\binom{2n-1}{n}-\binom{2n-1}{n-1})
= 0^{n}$.
\noindent
Thus we obtain
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}(\binom{2n-i-2}{n-i}
-\binom{2n-i-2}{n-i-1})\sum_{j=0}^{i}\binom{j-1}{i-j})x^{j}$
$= \sum_{j=0}^{n}(\sum_{i=j}^{n}(-1)^{n-i}(\binom{2n-i-2}{n-i}
-\binom{2n-i-2}{n-i-1})\binom{j-1}{i-j-})x^{j} = x^{n}$.
\noindent
This proves (7).
Let $A = (a_{i,j})$ and
$B = (b_{i,j})$ be $((n+1)\times (n+1))$-matrices defined by
\noindent
\hspace*{3.8cm} $a_{i,j} = (-1)^{i-j}\binom{2i-2}{i-j}$,
$b_{i,j} = \binom{i+j-2}{i-j} + \binom{i+j-3}{i-j-1}$.
\noindent
Then (1) implies that $AB = I_{n+1}$, and $B = A^{-1}$.
Thus $BA = I_{n+1}$. This implies (2).
Similarly (4), (6) and (8) follows from (3), (5) and (7)
respectively. \hspace{2.0cm} q.e.d.
\vspace*{3ex}
\noindent
(1.7) \begin{it} Let $n \geq 0$ be an integer.
Then, in the polynomial ring $\BZ [x]$, following
equalities hold. \end{it}
\noindent
(1) \hspace{1.2cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}
\sum_{j=0}^{i}(\binom{i+j}{i-j}+\binom{i+j-1}{i-j-1})x^{j}
= \sum_{k=0}^{n}(\binom{k}{n-k}+\binom{k-1}{n-k-1})x^{k}$.
\noindent
(2) \hspace{1.2cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}
\sum_{j=0}^{i}\binom{i+j-1}{i-j}x^{j}
= \sum_{k=0}^{n}\binom{k-1}{n-k}x^{k}$.
\noindent
(3) \hspace{1.2cm}
$\sum_{i=0}^{n}\binom{n}{i}
\sum_{j=0}^{i}(\binom{j}{i-j}+\binom{j-1}{i-j-1})x^{j}
= \sum_{k=0}^{n}(\binom{n+k}{n-k}+\binom{n+k-1}{n-k-1})x^{k}$.
\noindent
(4) \hspace{1.2cm}
$\sum_{i=0}^{n}\binom{n}{i}\sum_{j=0}^{i}\binom{j-1}{i-j}x^{j}
= \sum_{k=0}^{n}\binom{n+k-1}{n-k}x^{k}$.
\noindent
(5) \hspace{1.2cm}
$\sum_{i=0}^{n}(\binom{n+i}{n-i}+\binom{n+i-1}{n-i-1})
\sum_{j=0}^{i}(-1)^{i-j}\binom{2i-j-1}{i-j}x^{j}
= \sum_{k=0}^{n}\binom{n}{k}x^{k}$.
\noindent
(6) \hspace{1.2cm}
$\sum_{i=0}^{n}\binom{n+i-1}{n-i}\sum_{j=0}^{i}(-1)^{i-j}
(\binom{2i-j-2}{i-j}-\binom{2i-j-2}{i-j-1})x^{j}
= \sum_{k=0}^{n}\binom{n}{k}x^{k}$.
\noindent
(7) \hspace{1.2cm}
$\sum_{i=0}^{n}(\binom{i}{n-i}+\binom{i-1}{n-i-1})
\sum_{j=0}^{i}(-1)^{i-j}\binom{2i}{i-j}x^{j}
= \sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}x^{k}$.
\noindent
(8) \hspace{1.2cm}
$\sum_{i=0}^{n}\binom{i-1}{n-i}\sum_{j=0}^{i}(-1)^{i-j}
(\binom{2i-1}{i-j}-\binom{2i-1}{i-j-1})x^{j}
= \sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}x^{k}$.
\vspace*{2ex}
Proof. \hspace{.1cm} It follows from (1.5) (1) and (1.6) (2)
that we obtain
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(\binom{n+i}{n-i}+\binom{n+i-1}{n-i-1})
\sum_{j=0}^{i}(-1)^{i-j}\binom{2i-j-1}{i-j}x^{j}$
$= \sum_{i=0}^{n}(\binom{n+i}{n-i}+\binom{n+i-1}{n-i-1})
\sum_{k=0}^{i}(-1)^{i-k}\binom{2i}{i-k}
\sum_{j=0}^{k}\binom{k}{j}x^{j} = \sum_{j=0}^{n}\binom{n}{j}x^{j}$.
\noindent
This proves (5).
Similarly, (1.5) (2) and (1.6) (4) are used to obtain (6), (1.5) (3)
and (1.6) (6) are used to obtain (7), (1.5) (4) and (1.6) (8)
are used to obtain (8).
It follows from (5) and (1.6) (5)
that we obtain
\noindent
\hspace*{1.4cm}
$\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}
\sum_{j=0}^{i}(\binom{j}{i-j}+\binom{j-1}{i-j-1})x^{j}$
$= \sum_{k=0}^{n}(\binom{n+k}{n-k}+\binom{n+k-1}{n-k-1})
\sum_{i=0}^{k}(-1)^{k-i}\binom{2k-i-1}{k-i}
\sum_{j=0}^{i}(\binom{j}{i-j}+\binom{j-1}{i-j-1})x^{j}$
$= \sum_{k=0}^{n}(\binom{n+k}{n-k}+\binom{n+k-1}{n-k-1})x^{k}$.
\noindent
This proves (3). Similarly, (6) and (1.6) (7) are used to obtain (4),
(7) and (1.6) (1) are
used to obtain (1), and (8) and (1.6) (3) are used to obtain (2).
\hspace{4.4cm} q.e.d.
\vspace*{3ex}
\noindent
(1.8) \begin{it} Let $n,k$ and $j$ be integers with $0\leq k\leq n$
and $1\leq j\leq n$. Then, following equalities hold. \end{it}
\noindent
(1) \hspace{1.2cm}
$\sum_{i=0}^{2n-1}(-1)^{i}2^{2n-1-i}\binom{i+1}{2k-1}
\binom{j}{i+1-j} = 0$.
\noindent
(2) \hspace{1.2cm}
$\sum_{i=0}^{2n}(-1)^{i}2^{2n-i}\binom{i+1}{2k}
(\binom{j+1}{i+1-j}+\binom{j}{i-j}) = 0$.
\vspace*{2ex}
Proof. \hspace{.1cm} (1) \hspace{.1cm}
$\sum_{i=0}^{2n-1}(-1)^{i}2^{2n-1-i}\binom{i+1}{2k-1}\binom{j}{i+1-j}$
$= \sum_{i=j-1}^{2j-1}(-1)^{i}2^{2n-1-i}\binom{i+1}{2k-1}
\binom{j}{i+1-j}$
$= \sum_{i=0}^{j}(-1)^{j-1-i}2^{2n-j-i}\binom{i+j}{2k-1}
\binom{j}{i}$
$= -2^{2n-2j}\sum_{i=0}^{j}\sum_{m=0}^{i}(-2)^{j-i}\binom{i}{m}
\binom{j}{2k-1-m}\binom{j}{i}$
$= -2^{2n-2j}\sum_{m=0}^{j}(-1)^{j-m}\binom{j}{m}
\binom{j}{2k-1-m}$ \hspace{.3cm} (by (1.2) (5))
$= -2^{2n-2j}\sum_{m=0}^{2k-1}(-1)^{j-m}\binom{j}{m}
\binom{j}{2k-1-m}$
$= -2^{2n-2j}(\sum_{m=0}^{k-1}(-1)^{j-m}\binom{j}{m}
\binom{j}{2k-1-m}+\sum_{m=k}^{2k-1}(-1)^{j-m}\binom{j}{m}
\binom{j}{2k-1-m})$
$= -2^{2n-2j}(\sum_{m=0}^{k-1}(-1)^{j-m}\binom{j}{m}
\binom{j}{2k-1-m}-\sum_{m=0}^{k-1}(-1)^{j-m}\binom{j}{m}
\binom{j}{2k-1-m}) = 0$.
(2) \hspace{.1cm}
$\sum_{i=0}^{2n}(-1)^{i}2^{2n-i}\binom{i+1}{2k}
(\binom{j+1}{i+1-j}+\binom{j}{i-j})$
$= \sum_{i=j-1}^{2j}(-1)^{i}2^{2n-i}\binom{i+1}{2k}
\binom{j+1}{i+1-j}+\sum_{i=j}^{2j}(-1)^{i}2^{2n-i}
\binom{i+1}{2k}\binom{j}{i-j}$
$= \sum_{i=0}^{j+1}(-1)^{j+1-i}2^{2n+1-j-i}\binom{j+i}{2k}
\binom{j+1}{i}+\sum_{i=0}^{j}(-1)^{j-i}2^{2n-j-i}
\binom{j+i+1}{2k}\binom{j}{i}$
$= 2^{2n-2j}(\sum_{i=0}^{j+1}\sum_{m=0}^{i}(-2)^{j+1-i}
\binom{i}{m}\binom{j}{2k-m}\binom{j+1}{i}
+\sum_{i=0}^{j}\sum_{m=0}^{i}(-2)^{j-i}
\binom{i}{m}\binom{j+1}{2k-m}\binom{j}{i})$
$= 2^{2n-2j}(\sum_{m=0}^{j+1}(-1)^{j+1-m}
\binom{j+1}{m}\binom{j}{2k-m}
+\sum_{m=0}^{j}(-1)^{j-m}
\binom{j}{m}\binom{j+1}{2k-m})$ \hspace{.3cm} (by (1.2) (5))
$= 2^{2n-2j}(-\sum_{m=0}^{2k}(-1)^{j-m}
\binom{j+1}{m}\binom{j}{2k-m}
+\sum_{m=0}^{2k}(-1)^{j-m}
\binom{j}{m}\binom{j+1}{2k-m})$
$= 2^{2n-2j}(-\sum_{m=0}^{2k}(-1)^{j-m}
\binom{j+1}{2k-m}\binom{j}{m}
+\sum_{m=0}^{2k}(-1)^{j-m}
\binom{j}{m}\binom{j+1}{2k-m}) = 0$. \hspace{1.8cm} q.e.d.
\vspace*{3ex}
\noindent
(1.9) \begin{it} Let $n$ and $i$ be integers with $n > 0$.
Then, following equalities hold. \end{it}
\noindent
(1) \hspace{1.2cm}
$\binom{n}{i+1} = -(-1)^{n}\binom{0}{i+1-n}
+\sum_{j=0}^{n-1}(\binom{n-j}{j}+\binom{n-1-j}{j-1})
\binom{j}{i+1-j}$.
\noindent
(2) \hspace{1.2cm}
$\binom{n}{i+1} = (-1)^{n}\binom{0}{i+1-n}
+\sum_{j=0}^{n-1}\binom{n-1-j}{j}(\binom{j+1}{i+1-j}
+\binom{j}{i-j})$.
\vspace*{2ex}
Proof. \hspace{.1cm} The proof in the case $n = 1$, $2$ are
not difficult. Suppose $m \geq 2$ and (1.9) is true for
$1\leq n\leq m$. Then
\noindent
\hspace*{1.4cm} $\binom{m+1}{i+1} = \binom{m}{i+1} +\binom{m-1}{i}
+\binom{m-1}{i-1}$
$= -(-1)^{m}\binom{0}{i+1-m} +(-1)^{m}\binom{0}{i+1-m}
+(-1)^{m}\binom{0}{i-m}
+\sum_{j=0}^{m-1}(\binom{m-j}{j}+\binom{m-1-j}{j-1})
\binom{j}{i+1-j}$
\noindent
\hspace*{1.0cm}
$+\sum_{j=0}^{m-2}(\binom{m-1-j}{j}+\binom{m-2-j}{j-1})
\binom{j}{i-j}
+\sum_{j=0}^{m-2}(\binom{m-1-j}{j}+\binom{m-2-j}{j-1})
\binom{j}{i-1-j}$
$= (-1)^{m}\binom{0}{i-m}
+\sum_{j=0}^{m-1}(\binom{m-j}{j}+\binom{m-1-j}{j-1})
\binom{j}{i+1-j}
+\sum_{j=0}^{m-2}(\binom{m-1-j}{j}+\binom{m-2-j}{j-1})
\binom{j+1}{i-j}$
$= (-1)^{m}\binom{0}{i-m}
+\sum_{j=0}^{m-1}(\binom{m-j}{j}+\binom{m-1-j}{j-1})
\binom{j}{i+1-j}
+\sum_{j=0}^{m-1}(\binom{m-j}{j-1}+\binom{m-1-j}{j-2})
\binom{j}{i+1-j}$
$= (-1)^{m}\binom{0}{i-m}
+\sum_{j=0}^{m}(\binom{m+1-j}{j}+\binom{m-j}{j-1})
\binom{j}{i+1-j}$
$= -(-1)^{m+1}\binom{0}{i+1-(m+1)}
+\sum_{j=0}^{m+1-1}(\binom{m+1-j}{j}+\binom{m+1-j-1}{j-1})
\binom{j}{i+1-j}$
\noindent
and
\noindent
\hspace*{1.4cm} $\binom{m+1}{i+1} = \binom{m}{i+1} +\binom{m-1}{i}
+\binom{m-1}{i-1}$
$= (-1)^{m}\binom{0}{i+1-m} -(-1)^{m}\binom{0}{i+1-m}
-(-1)^{m}\binom{0}{i-m}
+\sum_{j=0}^{m-1}\binom{m-1-j}{j}(\binom{j+1}{i+1-j}+
\binom{j}{i-j})$
\noindent
\hspace*{1.0cm}
$+\sum_{j=0}^{m-2}\binom{m-2-j}{j}(\binom{j+1}{i-j}
+\binom{j}{i-1-j})
+\sum_{j=0}^{m-2}\binom{m-2-j}{j}(\binom{j+1}{i-1-j}
+\binom{j}{i-2-j})$
$= (-1)^{m+1}\binom{0}{i-m}
+\sum_{j=0}^{m-1}\binom{m-1-j}{j}(\binom{j+1}{i+1-j}
+\binom{j}{i-j}
+\sum_{j=0}^{m-2}\binom{m-2-j}{j}(\binom{j+2}{i-j}
+\binom{j+1}{i-1-j})$
$= (-1)^{m+1}\binom{0}{i-m}
+\sum_{j=0}^{m-1}(\binom{m-1-j}{j}(\binom{j+1}{i+1-j}
+\binom{j}{i-j})
+\sum_{j=0}^{m-1}\binom{m-1-j}{j-1}(\binom{j+1}{i+1-j}
+\binom{j}{i-j})$
$= (-1)^{m+1}\binom{0}{i-m}
+\sum_{j=0}^{m}\binom{m-j}{j}(\binom{j+1}{i+1-j}
+\binom{j}{i-j})$
$= (-1)^{m+1}\binom{0}{i+1-(m+1)}
+\sum_{j=0}^{m+1-1}\binom{m+1-j-1}{j}(\binom{j+1}{i+1-j}
+\binom{j}{i-j})$.
\noindent
Thus (1.9) is proved by the induction with respect to $n$.
\hspace{4.9cm} q.e.d.
\vspace*{6ex}
\noindent
(2.0) (Steenrod [2])
\begin{it} Let $p$ be a prime, and $m$ and $n$ non-negative
integers. Suppose that $m =$ $\sum_{j=0}^{N}a_{j}p^{j}$ and
$n =$ $\sum_{j=0}^{N}b_{j}p^{j}$, where $a_{j}$ and $b_{j}$
are non-negative integers with $a_{j}\leq p-1$ and $b_{j}\leq p-1$
$(0\leq j\leq N)$. Then
$$\binom{m}{n} \equiv
\prod_{j=0}^{N}\binom{a_{j}}{b_{j}} \pmod{p}.$$ \end{it}
\vspace*{2ex}
Proof. \hspace{.1cm} Let $i$ be an integer with $0 < i < p$. Then
$$\binom{p}{i} = \frac{p(p-1)\cdots (p-i+1)}{1\cdot 2\cdots i}
\equiv 0 \pmod{p}.$$
Therefore, in the polynomial ring $\BZ_{p}[x]$, we have
$(1+x)^{p} =$ $1+x^{p}$. It follows by induction on $j$ that
$(1+x)^{p^{j}} =$ $1+x^{p^{j}}$. Therefore
$$(1+x)^{m} = (1+x)^{\sum_{j=0}^{N}a_{j}p^{j}}
= \prod_{j=0}^{N}(1+x)^{a_{j}p^{j}} =
\prod_{j=0}^{N}\sum_{s=0}^{a_{j}}\binom{a_{j}}{s}x^{sp^{j}}.$$
The coefficient of $x^{n} =$ $x^{\sum_{j=0}^{N}b_{j}p^{j}}$
in the usual expansion of $(1+x)^{m}$ is $\binom{m}{n}$. But,
from the above expansion, we see that it is
$\prod_{j=0}^{N}\binom{a_{j}}{b_{j}}$. (2.0) follows.
\hspace{4.4cm} q.e.d.
\vspace*{3ex}
\noindent
(2.1) (Kobayashi [1])
\begin{it} Let $p$ be a prime, and $m$, $n$, $k$ and
$s$ non-negative integers with $p^{s} \leq$ $k <$ $p^{s+1}$.
Suppose that $\binom{m}{i} \equiv$ $\binom{n}{i} \pmod{p}$ for
$1\leq$ $i\leq k$. Then
\noindent
\hspace*{6.1cm}
$m\equiv n \hspace{.2cm} \pmod{p^{s+1}}$. \end{it}
\vspace*{2ex}
Proof. \hspace{.1cm} Suppose that $m = \sum_{j=0}^{N}a_{j}p^{j}$,
$n = \sum_{j=0}^{N}b_{j}p^{j}$ and
$i = \sum_{j=0}^{N}c_{i,j}p^{j}$ $(1\leq i\leq k)$, where
$a_{j}$, $b_{j}$ and $c_{i,j}$ are non-negative integers with
$a_{j}\leq$ $p-1$, $b_{j}\leq$ $p-1$ and $c_{i,j}\leq$ $p-1$
$(1\leq i\leq k$, $0\leq j\leq N)$. Then we have
$\binom{m}{i}\equiv$ $\prod_{j=0}^{N}\binom{a_{j}}{c_{i,j}} \pmod{p}$
and $\binom{n}{i}\equiv$ $\prod_{j=0}^{N}\binom{b_{j}}{c_{i,j}}$
(mod $p$) for $1\leq$ $i\leq$ $k$. It follows from the hypothesis
that we have $a_{j} =$ $b_{j}$ for $0\leq$ $j\leq$ $s$; that is,
$m\equiv$ $n \pmod{p^{s+1}}$. \hspace{10.1cm} q.e.d.
\vspace*{3ex}
{\bf Definition 2.2.} Let $p$ be a prime, and $s$ a positive
integer. Then $\nu_{p}(s)$ denotes the exponent of $p$ in the
prime power decomposition of $s$.
\vspace*{3ex}
\noindent
(2.3) \begin{it} Let $p$ be an odd prime.
\noindent
$(1)$ \hspace{.1cm} Let $r$, $k$ and $j$ be integers with
$r > 0$ and $k\equiv$ $j\pmod{p}$. Then
\noindent
\hspace*{4.4cm}
$k^{r} -j^{r} \equiv$ $r(k-j)j^{r-1} \hspace{.2cm}
\pmod{p^{\nu_{p}(r)+2}}$.
\noindent
$(2)$ \hspace{.1cm} Let $r$ and $k$ be integers with
$r > 0$, $r \equiv$ $0 \pmod{(p-1)}$ and $k\not\equiv$
$0 \pmod{p}$. Then
\noindent
\hspace*{4.5cm}
$k^{r}-1\equiv$ $r(1-k^{p-1}) \hspace{.2cm} \pmod{p^{\nu_{p}(r)+2}}$.
\end{it}
\vspace*{2ex}
Proof. \hspace{.1cm} (1) \hspace{.1cm} Since $k\equiv j \pmod{p}$,
we have
\noindent
\hspace*{4.4cm} $k^{r}-j^{r} = (k-j)\sum_{i=0}^{r-1}k^{i}j^{r-i-1}$
\noindent
\hspace*{5.8cm} $\equiv (k-j)\sum_{i=0}^{r-1}j^{r-1} \hspace{.2cm}
\pmod{p^{2}}$
\noindent
\hspace*{5.8cm} $= r(k-j)j^{r-1}$.
\noindent
This proves (1) for the case $\nu_{p}(r) = 0$. Moreover
\noindent
\hspace*{.5cm} $\sum_{i=0}^{p-1}(k^{r})^{i}(j^{r})^{p-i-1} \equiv
\sum_{i=0}^{p-1}(r(k-j)j^{r-1}+j^{r})^{i}(j^{r})^{p-i-1}
\hspace{3.5cm} \pmod{p^{2}}$
\noindent
\hspace*{3.9cm} $\equiv (j^{r})^{p-1}+ \sum_{i=1}^{p-1}
(ir(k-j)j^{r-1}(j^{r})^{i-1}+(j^{r})^{i})(j^{r})^{p-i-1}
\hspace{.2cm} \pmod{p^{2}}$
\noindent
\hspace*{3.9cm} $= (p(p-1)/2)r(k-j)j^{r-1}(j^{r})^{p-2}
+p(j^{r})^{p-1}$
\noindent
\hspace*{3.9cm} $\equiv p(j^{r})^{p-1} \hspace{1.2cm} \pmod{p^{2}}$.
\noindent
Assume that
\noindent
\hspace*{4.4cm} $k^{r} -j^{r} \equiv
r(k-j)j^{r-1} \hspace{.2cm} \pmod{p^{\nu_{p}(r)+2}}$.
\noindent
Then we have
\noindent
\hspace*{3.4cm} $k^{pr}-j^{pr} = (k^{r}-j^{r})
\sum_{i=0}^{p-1}(k^{r})^{i}(j^{r})^{p-i-1}$
\noindent
\hspace*{5.1cm} $\equiv r(k-j)j^{r-1}p(j^{r})^{p-1} \hspace{.2cm}
\pmod{p^{\nu_{p}(r)+3}}$
\noindent
\hspace*{5.1cm} $= pr(k-j)j^{pr-1}$.
\noindent
Thus (1) is proved by the induction with respect to
$\nu_{p}(r)$.
(2) \hspace{.1cm} Since $k^{p-1} \equiv 1 \pmod{p}$
for each $k$ prime to $p$, we have
\noindent
\hspace*{3.1cm} $k^{r}-1 = (k^{p-1})^{r/(p-1)} - 1^{r/(p-1)}$
\noindent
\hspace*{4.3cm} $\equiv (r/(p-1))(k^{p-1}-1) \hspace{1.4cm}
\pmod{p^{\nu_{p}(r)+2}}$
\noindent
\hspace*{4.3cm} $\equiv (1-p)(r/(p-1))(k^{p-1}-1) \hspace{.2cm}
\pmod{p^{\nu_{p}(r)+2}}$
\noindent
\hspace*{4.3cm} $= r(1 - k^{p-1})$. \hspace{8.1cm} q.e.d.
\vspace*{3ex}
\noindent
(2.4) \begin{it} Let $r$, $k$, $i$ and $s$ be integers with
$i\geq$ $2$, $k\equiv$ $\pm 1 \pmod{2^{i}}$, $r > 0$ and
$\nu =$ $\nu_{2}(r) \geq$ $s\geq$ $1$. Then we have \end{it}
\noindent
(1) \hspace{2.1cm} $k^{r} -1\equiv (k^{2^{\nu}}-1)(r/2^{\nu})$
\hspace{1.1cm} $\pmod{2^{2\nu +2i}}$.
\noindent
(2) \hspace{2.8cm} $k^{r}\equiv 1$
\hspace{3.6cm} $\pmod{2^{\nu +i}}$.
\noindent
(3) \hspace{2.1cm} $k^{r} -1\equiv (k^{2^{s}}-1)(r/2^{s})$
\hspace{1.1cm} $\pmod{2^{\nu +s+2i-1}}$.
\vspace*{2ex}
Proof. \hspace{.1cm} (1) \hspace{.1cm} Since $k^{2}\equiv 1
\pmod{2^{i+1}}$, we have
\noindent
\hspace*{4.5cm} $k^{r}-1 = (k^{2}-1)\sum_{l=1}^{r/2}(k^{2})^{l-1}$
\noindent
\hspace*{5.7cm} $\equiv (k^{2}-1)(r/2) \hspace{.2cm}
\pmod{2^{2i+2}}$.
\noindent
This proves (3) for the case $s = \nu = 1$. Assume that
\noindent
\hspace*{4.5cm} $k^{r} - 1 \equiv
(k^{2}-1)(r/2) \hspace{.2cm} \pmod{2^{\nu +2i}}$.
\noindent
Then we have
\noindent
\hspace*{2.9cm} $k^{2r}-1 = (k^{r}-1)(k^{r}+1)$
\noindent
\hspace*{4.3cm} $\equiv (k^{2}-1)(r/2)(k^{r}+1) \hspace{2.1cm}
\pmod{2^{\nu +2i+1}}$
\noindent
\hspace*{4.3cm} $\equiv (k^{2}-1)(r/2)(2+(k^{2}-1)(r/2))
\hspace{.2cm} \pmod{2^{2\nu +3i}}$
\noindent
\hspace*{4.3cm} $\equiv (k^{2}-1)(2r/2) \hspace{3.4cm}
\pmod{2^{2\nu +2i}}$.
\noindent
Since $\nu \geq 1$, this implies
\noindent
\hspace*{4.3cm} $k^{2r}-1 \equiv (k^{2}-1)(2r/2) \hspace{.2cm}
\pmod{2^{\nu +1+2i}}$.
\noindent
Thus the case $s = 1$ of (3) is proved by the induction with
respect to $\nu$. This implies (2). In particular, we have
$k^{2^{\nu}} \equiv$ $1 \pmod{2^{\nu + i}}$. This implies
\noindent
\hspace*{4.4cm} $k^{r}-1 = (k^{2^{\nu}}-1)
\sum_{l=1}^{r/2^{\nu}}(k^{2^{\nu}})^{l-1}$
\noindent
\hspace*{5.6cm} $\equiv (k^{2^{\nu}}-1)(r/2^{\nu}) \hspace{.2cm}
\pmod{2^{2\nu +2i}}$.
\noindent
This proves (1), and hence (3) for the case $\nu = s$. Assume
that $k^{r}-1 \equiv$ $(k^{2^{s}}-1)(r/2^{s})$
(mod $2^{\nu +s+2i-1}$). Since $k^{2^{s}} \equiv$
$1 \pmod{2^{s+i}}$ by (2), we have
\noindent
\hspace*{2.9cm} $k^{2r}-1 = (k^{r}-1)(k^{r}+1)$
\noindent
\hspace*{4.3cm} $\equiv (k^{2^{s}}-1)(r/2^{s})(k^{r}+1)
\hspace{2.5cm} \pmod{2^{\nu +s+2i}}$
\noindent
\hspace*{4.3cm} $\equiv (k^{2^{s}}-1)(r/2^{s})
(2+(k^{2^{s}}-1)(r/2^{s})) \hspace{.2cm} \pmod{2^{2\nu +s+3i-1}}$
\noindent
\hspace*{4.3cm} $\equiv (k^{2^{s}}-1)(2r/2^{s}) \hspace{3.7cm}
\pmod{2^{2\nu +2i}}$.
\noindent
Since $\nu \geq s \geq 1$, this implies
\noindent
\hspace*{4.1cm} $k^{2r}-1 \equiv (k^{2^{s}}-1)(2r/2^{s})
\hspace{.2cm} \pmod{2^{\nu +1+s+2i-1}}$.
\noindent
Thus (3) is proved by the induction with respect to $\nu$.
\hspace{5.1cm} q.e.d.
\vspace*{6ex}
\begin{thebibliography}{[1]}
\bibitem[1]{kobayashi1}
T.~Kobayashi:
{\it Stable homotopy types of stunted lens spaces} mod $p^{r}$,
Mem.\ Fac.\ Sci.\ Kochi Univ.\ (Math.)\ {\bf 15} (1994), 9--14.
\bibitem[2]{steenrod1}
N.~E.~Steenrod:
Cohomology operations,
Princeton University Press, 1962.
\bibitem[3]{tamamura1}
A.~Tamamura:
{\it $J$-groups of the suspensions of the stunted lens spaces} mod $2p$,
Osaka J.\ Math.\ {\bf 30} (1993), 581--610.
\end{thebibliography}
\noindent
%==308==
\end{document}
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