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Carlitzによる二重超幾何級数の和公式

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以下は, Carlitzによって1967年に示された公式である.

Carlitz(1967)

$\begin{array}{r} \sum_{0 \leq m, n} \frac{(a)_{m} (a^{'})_{n} (b)_{m + n} (c)_{m} (c^{'})_{n}}{m! n! (b)_{m} (b)_{n} (c + c^{'})_{m + n}} = \frac{Γ (c - a^{'}) Γ (c^{'} - a) Γ (b) Γ (b - a - a^{'}) Γ (c + c^{'})}{Γ (c + c^{'} - a - a^{'}) Γ (b - a) Γ (b - a^{'}) Γ (c) Γ (c^{'})} \end{array}$
が左辺の和が絶対収束するとき, 成立する.

Vandermondeの恒等式より,
$\begin{array}{r} \frac{(b)_{m + n}}{(b)_{m} (b)_{n}} = \sum_{0 \leq k} \frac{(- m, - n)_{k}}{k! (b)_{k}} \end{array}$
であるから,
$\begin{aligned} \sum_{0 \leq m, n} \frac{(a)_{m} (a^{'})_{n} (b)_{m + n} (c)_{m} (c^{'})_{n}}{m! n! (b)_{m} (b)_{n} (c + c^{'})_{m + n}} \\ = \sum_{0 \leq m, n} \frac{(a)_{m} (a^{'})_{n} (c)_{m} (c^{'})_{n}}{m! n! (c + c^{'})_{m + n}} \sum_{0 \leq k} \frac{(- m, - n)_{k}}{k! (b)_{k}} \\ = \sum_{0 \leq m, n, k} \frac{(a)_{m} (a^{'})_{n} (c)_{m} (c^{'})_{n}}{k! (b)_{k} (m - k)! (n - k)! (c + c^{'})_{m + n}} \\ = \sum_{0 \leq k} \frac{(a)_{k} (a^{'})_{k} (c^{'})_{k}}{k! (b)_{k}} \sum_{0 \leq m} \frac{(a + k)_{m} (c)_{m + k}}{m! (c + c^{'})_{m + 2 k}} \sum_{0 \leq n} \frac{(a^{'} + k)_{n} (c^{'} + k)_{n}}{n! (c + c^{'} + 2 k + m)_{n}} \end{aligned}$
ここで, Gaussの超幾何定理より,
$\begin{aligned} \sum_{0 \leq n} \frac{(a^{'} + k)_{n} (c^{'} + k)_{n}}{n! (c + c^{'} + 2 k + m)_{n}} & = \frac{Γ (c + c^{'} + 2 k + m) Γ (c - a^{'} + m)}{Γ (c + c^{'} - a + k + m) Γ (c + k + m)} \\ = \frac{Γ (c + c^{'}) Γ (c - a^{'})}{Γ (c + c^{'} - a^{'}) Γ (c)} \frac{(c + c^{'})_{m + 2 k} (c - a^{'})_{m}}{(c + c^{'} - a^{'})_{m + k} (c)_{m + k}} \end{aligned}$
だから,
$\begin{aligned} \sum_{0 \leq k} \frac{(a)_{k} (a^{'})_{k} (c^{'})_{k}}{k! (b)_{k}} \sum_{0 \leq m} \frac{(a + k)_{m} (c)_{m + k}}{m! (c + c^{'})_{m + 2 k}} \sum_{0 \leq n} \frac{(a^{'} + k)_{n} (c^{'} + k)_{n}}{n! (c + c^{'} + 2 k + m)_{n}} \\ = \frac{Γ (c + c^{'}) Γ (c - a^{'})}{Γ (c + c^{'} - a^{'}) Γ (c)} \sum_{0 \leq k} \frac{(a)_{k} (a^{'})_{k} (c^{'})_{k}}{k! (b)_{k}} \sum_{0 \leq m} \frac{(a + k)_{m}}{m!} \frac{(c - a^{'})_{m}}{(c + c^{'} - a^{'})_{m + k}} \end{aligned}$
である. 再びGaussの超幾何定理より,
$\begin{aligned} \sum_{0 \leq m} \frac{(a + k)_{m}}{m!} \frac{(c - a^{'})_{m}}{(c + c^{'} - a^{'})_{m + k}} & = \frac{1}{(c + c^{'} - a^{'})_{k}} \frac{Γ (c + c^{'} - a + k) Γ (c - a)}{Γ (c + c^{'} - a - a^{'}) Γ (c^{'} + k)} \\ = \frac{Γ (c + c^{'} - a) Γ (c^{'} - a)}{Γ (c + c^{'} - a - a^{'}) Γ (c^{'}) (c^{'})_{k}} \end{aligned}$
だから,
$\begin{aligned} \frac{Γ (c + c^{'}) Γ (c - a^{'})}{Γ (c + c^{'} - a^{'}) Γ (c)} \sum_{0 \leq k} \frac{(a)_{k} (a^{'})_{k} (c^{'})_{k}}{k! (b)_{k}} \sum_{0 \leq m} \frac{(a + k)_{m}}{m!} \frac{(c - a^{'})_{m}}{(c + c^{'} - a^{'})_{m + k}} \\ = \frac{Γ (c + c^{'}) Γ (c - a^{'}) Γ (c^{'} - a)}{Γ (c) Γ (c^{'}) Γ (c + c^{'} - a - a^{'})} \sum_{0 \leq k} \frac{(a)_{k} (a^{'})_{k}}{k! (b)_{k}} \end{aligned}$
最後にGaussの超幾何定理より
$\begin{array}{r} \sum_{0 \leq k} \frac{(a)_{k} (a^{'})_{k}}{k! (b)_{k}} = \frac{Γ (b) Γ (b - a - a^{'})}{Γ (b - a) Γ (b - a^{'})} \end{array}$
であるから定理を得る.