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Fungrim entry: acdce8

p ⁣(n)=k=1n+1(1)k+1(p ⁣(nk(3k1)2)+p ⁣(nk(3k+1)2))p\!\left(n\right) = \sum_{k=1}^{n + 1} {\left(-1\right)}^{k + 1} \left(p\!\left(n - \frac{k \left(3 k - 1\right)}{2}\right) + p\!\left(n - \frac{k \left(3 k + 1\right)}{2}\right)\right)
Assumptions:nZn \in \mathbb{Z}
TeX:
p\!\left(n\right) = \sum_{k=1}^{n + 1} {\left(-1\right)}^{k + 1} \left(p\!\left(n - \frac{k \left(3 k - 1\right)}{2}\right) + p\!\left(n - \frac{k \left(3 k + 1\right)}{2}\right)\right)

n \in \mathbb{Z}
Definitions:
Fungrim symbol Notation Short description
PartitionsPp ⁣(n)p\!\left(n\right) Integer partition function
Sumnf ⁣(n)\sum_{n} f\!\left(n\right) Sum
Powab{a}^{b} Power
ZZZ\mathbb{Z} Integers
Source code for this entry:
Entry(ID("acdce8"),
    Formula(Equal(PartitionsP(n), Sum(Mul(Pow(-1, Add(k, 1)), Add(PartitionsP(Sub(n, Div(Mul(k, Sub(Mul(3, k), 1)), 2))), PartitionsP(Sub(n, Div(Mul(k, Add(Mul(3, k), 1)), 2))))), Tuple(k, 1, Add(n, 1))))),
    Variables(n),
    Assumptions(Element(n, ZZ)))

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2019-08-21 11:44:15.926409 UTC