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メモ：超幾何級数の公式まとめ

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超幾何級数の公式のまとめです.
いずれもBailey pairで証明できるらしいです.

$\begin{aligned} _{7} F_{6} [\begin{array}{c} a, 1 + \frac{a}{2}, b, d, a - p + 1, p + 1, - N \\ \frac{a}{2}, 1 + a - b, 1 + a - d, p, a - p, 1 + a + N \end{array}; 1] \\ = \frac{(1 + a)_{N} (a - b - d)_{N}}{(1 + a - b)_{N} (1 + a - d)_{N}} \frac{(g + 1)_{N}}{(g)_{N}} \end{aligned}$

ここで
$\begin{array}{r} g = \frac{p (b + d - a) (p - a)}{p (a - p) - d b} \end{array}$

$\begin{aligned} _{9} F_{8} [\begin{array}{c} a, 1 + \frac{a}{2}, b, c, a - p + 1, p + 1, d, e, - N \\ \frac{a}{2}, 1 + a - b, 1 + a - c, p, a - p, 1 + a - d, 1 + a - e, 1 + a + N \end{array}; 1] \\ = \frac{(1 + a)_{N} (1 + a - c - d)_{N}}{(1 + a - c)_{N} (1 + a - d)_{N}}_{5} F_{4} [\begin{array}{c} c, d, a - b - e, h + 1, - N \\ 1 + a - b, 1 + a - e, c + d - a - N, h \end{array}; 1] \end{aligned}$

ここで
$\begin{array}{r} h = \frac{p (b + e - a) (p - a)}{p (a - p) - b e} \end{array}$

$\begin{aligned} _{9} F_{8} [\begin{array}{c} a, 1 + \frac{a}{2}, b, c, d, a - p + 1, p + 1, 2 a - d - c - b + N, - N \\ \frac{a}{2}, 1 + a - b, 1 + a - c, 1 + a - d, p, a - p, 1 + b + d + c - a - N, 1 + a + N \end{array}; 1] \\ = \frac{(1 + a)_{N} (1 + a - c - d)_{N} (a - b - d)_{N} (a - b - c)_{N} (k + 1)_{N}}{(1 + a - c)_{N} (1 + a - d)_{N} (1 + a - b)_{N} (a - b - c - d)_{N} (k)_{N}} \end{aligned}$

ここで
$\begin{aligned} h = \frac{p (a - d - c + N) (p - a)}{p (a - p) - b (2 a - d - c - b + N)} \\ k = \frac{h (b + d - a) (b + c - a)}{c d - h (b + c + d - a)} . \end{aligned}$

$\begin{aligned} _{11} F_{10} [\begin{array}{c} a, 1 + \frac{a}{2}, b, c, d, a - p + 1, p + 1, e, f, g, - N \\ \frac{a}{2}, 1 + a - b, 1 + a - c, 1 + a - d, p, a - p, 1 + a - e, 1 + a - f, 1 + a - g, 1 + a + N \end{array}; 1] \\ = \frac{(1 + s - e)_{N} (1 + s - f)_{N} (1 + s - g)_{N} (1 + s - e - f - g)_{N}}{(1 + s)_{N} (1 + s - e - f)_{N} (1 + s - f - g)_{N} (1 + s - e - g)_{N}} \\ \times_{11} F_{10} [\begin{array}{c} s, 1 + \frac{s}{2}, e, f, g, a - b - c, a - d - c, a - b - d, 1 + \frac{s}{2} + λ, 1 + \frac{s}{2} - λ,, - N \\ \frac{s}{2}, 1 + s - e, 1 + s - f, 1 + s - g, 1 + a - d, 1 + a - c, 1 + a - b, \frac{s}{2} - λ, \frac{s}{2} + λ, 1 + s + N \end{array}; 1] \end{aligned}$

ここで
$\begin{aligned} s = 2 a - b - c - d, 3 a = e + f + g + b + c + d - 1 - N, A = a - c - d, λ^{2} = \frac{s^{2}}{4} - A D \\ D = \frac{- p (p - a) (b + d - a) (b + c - a)}{p (p - a) (b + c + d - a) + b c d} \end{aligned}$

$\begin{aligned} _{7} F_{6} [\begin{array}{c} a, 1 + \frac{a}{2}, b, c, d, a - p + 1, p + 1 \\ \frac{a}{2}, 1 + a - b, 1 + a - c, 1 + a - d, p, a - p \end{array}; 1] \\ = \frac{Γ (1 + a - b) Γ (a + a - c) Γ (1 + a - d) Γ (a - b - c - d) Γ (k)}{Γ (1 + a) Γ (1 + a - c - d) Γ (a - b - d) Γ (a - b - c) Γ (k + 1)} \end{aligned}$
$\begin{aligned} k = \frac{h (b + d - a) (b + c - a)}{c d - h (b + c + d - a)} \\ h = \frac{p (a - p)}{b} \end{aligned}$