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Fungrim entry: d651d1

(z)2k=4k(z2)k(z+12)k\left(z\right)_{2 k} = {4}^{k} \left(\frac{z}{2}\right)_{k} \left(\frac{z + 1}{2}\right)_{k}
Assumptions:zC  and  kZ0z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 0}
\left(z\right)_{2 k} = {4}^{k} \left(\frac{z}{2}\right)_{k} \left(\frac{z + 1}{2}\right)_{k}

z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 0}
Fungrim symbol Notation Short description
RisingFactorial(z)k\left(z\right)_{k} Rising factorial
Powab{a}^{b} Power
CCC\mathbb{C} Complex numbers
ZZGreaterEqualZn\mathbb{Z}_{\ge n} Integers greater than or equal to n
Source code for this entry:
    Formula(Equal(RisingFactorial(z, Mul(2, k)), Mul(Mul(Pow(4, k), RisingFactorial(Div(z, 2), k)), RisingFactorial(Div(Add(z, 1), 2), k)))),
    Variables(z, k),
    Assumptions(And(Element(z, CC), Element(k, ZZGreaterEqual(0)))))

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2021-03-15 19:12:00.328586 UTC