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Given the Gauss hypergeometric function ${}_{2}F_{1} (a,b,c,z)$, its $n$th derivative can be written as

$\frac{\mathrm{d}^{n}}{\mathrm{d} z^{n}} {}_{2}F_{1} (a,b,c,z) = \frac{(a)_{n} (b)_{n}}{(c)_{n}} {}_{2}F_{1} (a+n,b+n,c+n,z)$

However, when $c = b+1$, it appears that one can write the $n$th derivative differently. For instance, with Mathematica, I obtain

$\frac{\mathrm{d}}{\mathrm{d} z} {}_{2}F_{1} (a,b,b+1,z) = \frac{b \big( (1-z)^{-a} - {}_{2}F_{1} (a,b,b+1,z) \big)}{z}$

$\frac{\mathrm{d}^{2}}{\mathrm{d} z^{2}} {}_{2}F_{1} (a,b,b+1,z) = \frac{b \left(a (1-z)^{-a-1}-\frac{b \left((1-z)^{-a}-\, _2F_1(a,b;b+1;z)\right)}{z}\right)}{z}-\frac{b \left((1-z)^{-a}-\, _2F_1(a,b;b+1;z)\right)}{z^2}$

$\frac{\mathrm{d}^{3}}{\mathrm{d} z^{3}} {}_{2}F_{1} (a,b,b+1,z) = \ldots$

Although less compact, the interesting thing of this form is that all Gauss hypergeometric functions appearing in the derivatives are the same as the original function. What is the property behind this result? I searched all the properties of the Gauss hypergeometric function in http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/ and in the book "Table of integrals, series and products" but with no luck.

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    ${}_{2}F_{1} (a,b,b+1,z)$ is related to incomplete Beta function. – 2017-01-25
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    If I'm not wrong, ${}_{2}F_{1}(a.b.b+1,z)$ can be written in integral form as $\int_{0}^{1} t^{b-1} (1 - t z)^{a} \mathrm{d} t$, which is quite different from $B(z,a,b)$, even for $z=1$. – 2017-01-26
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    please, see this http://dlmf.nist.gov/8.17.E7 – 2017-01-26
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    Sorry if I don't understand, but how is ${}_{2}F_{1} (a,1-b,a+1,z)$ related to ${}_{2}F_{1} (a,b,b+1,z)$? – 2017-01-26

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