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Fungrim entry: 599417

n=0p ⁣(n)qn=1ϕ ⁣(q)\sum_{n=0}^{\infty} p\!\left(n\right) {q}^{n} = \frac{1}{\phi\!\left(q\right)}
Assumptions:qCandq<1q \in \mathbb{C} \,\mathbin{\operatorname{and}}\, \left|q\right| < 1
\sum_{n=0}^{\infty} p\!\left(n\right) {q}^{n} = \frac{1}{\phi\!\left(q\right)}

q \in \mathbb{C} \,\mathbin{\operatorname{and}}\, \left|q\right| < 1
Fungrim symbol Notation Short description
Sumnf ⁣(n)\sum_{n} f\!\left(n\right) Sum
PartitionsPp ⁣(n)p\!\left(n\right) Integer partition function
Powab{a}^{b} Power
Infinity\infty Positive infinity
EulerQSeriesϕ ⁣(q)\phi\!\left(q\right) Euler's q-series
CCC\mathbb{C} Complex numbers
Absz\left|z\right| Absolute value
Source code for this entry:
    Formula(Equal(Sum(Mul(PartitionsP(n), Pow(q, n)), Tuple(n, 0, Infinity)), Div(1, EulerQSeries(q)))),
    Assumptions(And(Element(q, CC), Less(Abs(q), 1))))

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2019-08-21 11:44:15.926409 UTC